Category: Class 12th

TS Grewal Accountancy Class 12th Solutions Part 1

January 16, 2019
TS Grewal Accountancy Class 12th Solutions
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January 16, 2019
The Invisible Man Chapter Wise Summary & Notes Class 12th English

The Invisible Man

Protagonist and Antagonist

The story contains both external and internal conflict. In either case, both the protagonist and the antagonist is Griffin himself as he has made himself his own worst enemy. The external conflicts that Griffin causes are between Griffin and various members of the town as his invisibility is gradually discovered. People react with fear and then with terror as Griffin aggravates the situation by lashing out against people as soon as they figure him out. The people accept his existence with surprising lack of suspicion about the possibility of such an occurrence, which may be a lack on the author’s part. Once they believe that he exists, the primary goal is to apprehend and imprison him. Although motives are not elaborated upon, it would seem that different people in the town have different notions of what they might do when and if they could capture the man. Griffin also ultimately sees Kemp as an enemy although he had at first believed that Kemp would be both sympathetic and cooperative.

The most important conflict is internal as Griffin himself struggles to live with his situation. He rationalizes his crimes rather than making any sane attempt to get people to understand his predicament. He uses force to get people to help him and goes from bad to worse in his attempts to replenish his research materials for experiments in reversing the process that rendered him invisible. There is no real depth of character. Griffin simply runs from place to place trying to survive by increasingly decadent methods.

Climax

The climax occurs when Griffin returns to Kemp’s house intending to make an example of Kemp for having betrayed him. Kemp escapes out the window but is soon followed by Griffin who can see him although he can’t see Griffin. The entire town is soon involved in the chase.

Outcome

The resolution is the death of Griffin. Once Kemp realizes what is happening he slows down and allows Griffin to catch him. Although Kemp is buffeted about a good bit for his efforts, Griffin is weaker than usual due to his injuries. Some of the men of the town are able to grasp invisible wrists and ankles and hold him down until the effort is no longer necessary.

Synopsis

The plot is simple and straightforward. Griffin, having rendered himself invisible with an earlier experiment, enters a town and sets up a lab in an inn where he works night and day to come up with a formula that will reverse his invisibility. When he slips up and accidentally reveals himself, he engages in immature and violent actions until he is forced

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to run and find a new hiding place. As more people become aware of his existence, his situation becomes more perilous. Finally, he stumbles into the home of a former college professor whom he assumes will be interested in his experiments and willing to help him. The doctor, Mr. Kemp, however, reads newspaper accounts of Griffin’s insane actions against people in the town and betrays his trust. Griffin is hunted down, caught and killed, whereupon he becomes visible again. The little, inconspicuous victim of some of Griffin’s behavior is left with the stolen money and the documents that explain Griffin’s experiments. The story closes with the suggestion that Marvel himself might try the experiments if only he could figure them out.

CHAPTER 1 The Strange Man’s Arrival

A stranger arrives in Bramblehurst railway station. He is bundled from head to foot with only the tip of his nose showing. He enters the Coach & Horses Inn and demands a room and a fire. Mrs. Hall, the owner prepares a supper for him and offers to take his coat and hat, but he refuses to take them off. When he finally removes the hat, his entire head is swathed in a bandage. Mrs. Hall thinks he has endured some accident. She tries to get him to talk about himself, but he is taciturn with her, although not particularly rude.

Notes – This introduction to the Invisible Man through the eyes of the town people is actually about midway through his own story. He has already gone from place to place trying to keep his cover and has committed two acts of violence, one against his own father and the other against the proprietor of a costume shop whom he tied and gagged in order to be able to steal clothing and money. Nevertheless, his intention at this point is simply to find a quiet place and work as quickly as possible to find an antidote to the invisibility. The primary thread of the story-that of the growing rumors and suspicions, which eventually contribute to his exposure-is begun.

CHAPTER 2 Mr. Teddy Henfrey’s First Impressions

Teddy Henfrey, a clock repairman, comes to the inn for tea. Mrs. Hall asks him to “repair the clock” in the stranger’s room. Teddy deliberately takes as long as he can with the clock, taking it apart and reassembling it for no reason. The stranger finally gets him to hurry up and leave. Offended, Teddy talks himself into believing that the stranger is someone of a suspicious nature, perhaps even wanted by the police and is wrapped up to conceal his identity. Teddy runs into Mr. Hall and warns him about the stranger, informing him that a “lot of luggage” will be coming. It would seem that the stranger intends to stay awhile.

Mr. Hall goes home intending to investigate the stranger, but is put off by the short- tempered demeanor of his wife.

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Notes – Mrs. Hall, although not a major character, is revealed as rather devious in a harmless sort of way. She really wants to know what the man’s disfigurement is; she assumes he has been in a horrible accident, and the motherly side of her wants to know how to express sympathy. She is a very good innkeeper under the circumstances. While she is not above using Teddy to pry for information, she does not contribute to the spread of rumors. In fact, we are told later that she defends him as long as he is faithful about paying his bill. Teddy is a character typical of the other people of the town. He wants to know the man’s story, and when he is rebuffed for his persistence, he begins to imagine all sorts of things. His imagination soon becomes fact to him, and he spreads his new knowledge to anyone who will listen.

CHAPTER 3 The Thousand and One Bottles

A stranger arrives in Bramblehurst railway station. He is bundled from head to foot with only the tip of his nose showing. He enters the Coach & Horses Inn and demands a room and a fire. Mrs. Hall, the owner prepares a supper for him and offers to take his coat and hat, but he refuses to take them off. When he finally removes the hat, his entire head is swathed in a bandage. Mrs. Hall thinks he has endured some accident. She tries to get him to talk about himself, but he is taciturn with her, although not particularly rude.

The stranger’s luggage arrives at the inn. Numerous crates fill the deliveryman’s cart, some of them containing bottles packaged in straw. Fearenside, the cartman, owns a dog that starts to growl when the stranger comes down the steps to help with the boxes. The dog jumps for the stranger’s hand, but misses and sinks his teeth in a pant leg. The dog tears open the trouser leg, whereupon the stranger goes quickly back into the inn and to his room.

Concerned about the possibility of injury, Mr. Hall goes to the stranger’s room. He gets a glimpse of what seems like a white mottled face before he is shoved by an unseen force back through the door. The stranger soon reappears at the door, his trousers changed, and gives orders for the rest of his luggage. The stranger unpacks 6 crates of bottles, which he arranges across the windowsill and all the available table and shelf space in the inn’s parlor-a space he seems to have commandeered for himself.

Mrs. Hall enters later to tend to his needs and catches a quick glimpse of him without his glasses. His eyes seem hollow; he quickly puts his glasses on. She starts to complain about the straw on the floor, but he tells her to put it on the bill and to knock before entering his rooms. She points out that he could lock his door if he doesn’t want to be bothered, advice that he takes. He then works behind the locked door all afternoon. At one point, Mrs. Hall hears him raving about not being able to “go on.” She hears a sound like a bottle being broken. Later she takes him tea and notes the broken glass and a stain on the floor. He again tells her to “put it on the bill.”

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Meanwhile Fearenside talks in the beer shop of Iping Hangar. Fearenside says that the stranger is a “black man,” an assumption derived from the absence of “pink flesh” when the trouser leg was ripped open. When reminded of the pink nose, Fearenside claims that the man must therefore be a “piebald,” or a part white, part black creature.

Notes – Fearenside is more observant than even he realizes. Of course, Griffin knows that a close look at his torn pant leg will reveal a “missing” leg, but he also needs to get away from the dog until they can get the animal under control. Subtle differences among characters of the town are beginning to be revealed. Mrs. Hall notices a “hollow” look to the guest’s eyes, an appearance masked by the dark glasses he usually wears. His frustration is over the failure of his experiments; she notes the mess he makes but cleans up after him with minimal complaint when he gives her extra money. Fearenside, on the other hand, liberally discusses the “discoveries” he has made as a result of the brief encounter. Fearenside refers to horses as an example of the “patchy” color that can happen when black and white are mixed.

CHAPTER 4 Mr. Cuss Interviews the Stranger

The stranger works diligently in his room until the end of April with only occasional skirmishes with Mrs. Hall. Whenever she disapproves of anything he does, he quiets her with additional payment. He rarely goes out during the day, but goes out nearly every night, muffled up regardless of the weather.

His identity becomes a topic of speculation in the town. Mrs. Hall defends him, repeating his own words that he is an “experimental investigator.” The view of the town is that he is a criminal trying to escape justice. Mr. Gould, the probationary assistant imagines that the man must be an “anarchist” who is preparing explosives.

Another group of people believe he is a piebald and could make a lot of money if he chose to show himself at the fairs. All agree, however, that due to his habits of secrecy, they dislike him. The young men begin to mock his bearing; a song called “Bogey Man” becomes popular and children follow at a distance calling out “Bogey Man.”

The curiosity of a general practitioner named Cuss is aroused, and he contrives for an interview. During the interview the stranger accidentally removes his hand from his pocket. Cuss is able to see down the empty sleeve to the elbow. Cuss questions him about “moving an empty sleeve.” The stranger laughs, then extends the empty sleeve toward Cuss’s face and pinches his nose. Cuss leaves in terror and tells his story to Bunting, the vicar.

Notes – In spite of Hall’s defense, Griffin will be the cause of his own destruction. Perhaps it is the frustration of always having to guard his secret that causes him to act offensively when challenged, but in any case, he could have handled the situation differently. The deliberate pinching of Cuss’s nose is not only an unnecessary

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affront, but is also a mark of Griffin’s immaturity. Bringing pain upon others for the sake of his own amusement, however, will soon deteriorate to performing criminal acts. In fact, although Bunting is about to become Griffin’s new victim, Griffin has already been foraging at night for places that he could rob in order to maintain his materials and keep up with his rent.

This chapter nudges the plot forward a bit by bringing in Bunting the vicar. The actions which will follow begin to bring the town together in an awareness of a stranger in their midst.

CHAPTER 5 The Burglary and the Vicarage

Mrs. Bunting, the vicar’s wife, wakes up at the sound of bare feet walking through her house. She wakes her husband and the two watch and listen as a candle is lit and papers are rustled in the study. When they hear the telltale clink of money, Rev. Bunting rushes into the study with a raised poker, but the room appears to be empty. Their money disappears and at one point they hear a sneeze in the hallway but are unable to locate or see the intruder.

Notes – Due to the necessity of running about naked, Griffin has caught a cold, which he is unable to completely hide. His sneezes begin to give him away even though people don’t yet understand what they are hearing. In robbing the Buntings, Griffin also sets himself up for accusations and criminal charges. Thus when his presence is discovered, it is inevitable that people will begin to expect the worst and will be concentrating on apprehending him rather than helping him.

CHAPTER 6 The Furniture that Went Mad

The Halls arise very early in the morning on Whit-Monday in order to take care of some private business having something to do with their wine cellar. In passing by the guest’s room, Mr. Hall notices that the door is ajar. A few minutes later, he sees that the bolts on the front door of the house are unlocked although he remembers shutting them on the previous night. The guest is not in his room, but his clothes, shoes, and even his hat are scattered about. As the Halls are investigating, the bed-clothes suddenly gather themselves into a bundle and toss themselves over the bottom rail. Then a chair flies toward Mrs. Hall. The legs of the chair are brought to rest against her back, propelling her out of the room. The door slams and is locked behind them. The Halls decide that the stranger is a spirit.

They send for Sandy Wadgers, the blacksmith who is also supposed to be an exorcist. Wadgers is joined by Huxter, and together they ponder the likelihood of witchcraft and contemplate the propriety of breaking through the door in order to examine the situation more closely. However, before they can carry out any such action, the door opens and the stranger emerges, wrapped and bundled as usual. He distracts them long enough to enter

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the parlor and slam the door against them. When Mr. Hall raps on the door and demands an explanation, the stranger tells him to “go to the devil” and “shut the door after you.”

Notes – The panic is building for Griffin, while characterization is enhanced for the people in the town. Wadgers delays “breaking” into the room, using the excuse of propriety when the real and very human reason is apprehension. While they may talk of spirits and witchcraft in their leisure, it is clear that they have no real experience with such. The growing impression is that the Invisible Man is something evil. Griffin helps the idea along with his continued offenses.

CHAPTER 7 The Unveiling of the Stranger

The stranger remains locked in the parlor all morning. He rings his bell for Mrs. Hall several times, but she does not answer it. About noon, he emerges and demands to know why his meals have not been brought to him. Mrs. Hall tells him that his bill has not been paid in five days. She refuses to accept the excuse that he is waiting for a remittance. When he produces some money, she refuses it, saying she first wants to know why he doesn’t enter by doorways and move about like normal people.

For his answer, the stranger removes all his head wrappings, including his nose and moustache. He thus looks like a person with a missing head. At the sound of screams a crowd of people run toward the inn. “Eye-witnesses” suddenly babble hysterical stories of the man attacking the servant girl, and brandishing a knife. Bobby Jaffers, the village constable, appears with a warrant.

The stranger slaps Jaffers with his glove, but then says he will surrender. He will not accept handcuffs, however. As the constable, Halls and others watch, the man removes the rest of his clothes, becoming invisible before them. He tells them that he is invisible. Jaffers wants to take him in for questioning on suspicion of robbing the Bunting home. A scuffle ensues, and the stranger, now known as the “Invisible Man,” escapes.

Notes – This is the last chapter in which Mrs. Hall has a significant presence, but the reader is left with the image of a very courageous, and spunky lady. She has, just a day before, been shoved out of one of her own rooms with a floating chair; she knows the man has entered and left by some mysterious means and yet she rejects his money and demands an explanation. Griffin’s own actions are quickly becoming offensive, violent and deliberately geared toward creating reactions of fear and terror in his victims. There seems to be no sense of humanity left in him; everything he does is first for survival, then for the sheer thrill of striking terror-simply because he can. He is like an evil schoolboy who enjoys pulling the legs off of flies just to see them squirm. It never occurs to him to try to solve his problem by any means other than violence and terror.

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CHAPTER 8 In Transit

An amateur naturalist named Gibbins is relaxing out on the downs and hears someone coughing, sneezing and swearing. Frightened, Gibbins gets up and runs home.

Notes – This chapter simply indicates the passage of the Invisible Man through the countryside.

CHAPTER 9 Mr. Thomas Marvel

Marvel is an eccentric bachelor and local tramp who likes to be comfortable and take his time about things. He has come across a pair of boots in a ditch. He has tried them on and found them too big, and is occupied in contemplating the boots when he hears a voice nearby. Marvel talks about boots with the voice for several minutes before turning to see his visitor and finding no one there.

First Marvel tells himself that he has had too much to drink, then that his imagination has played some sort of trick on him. The Invisible Man begins throwing things at Marvel to convince him that he is not just imagining the presence. Eventually the Man convinces Marvel that he is real and is in need of an accomplice who will first give him food, water and shelter. He delivers an unfinished threat of what he will do if Marvel betrays him.

Notes – Marvel appears eccentric, unassuming and something of a loner, which would be bait to Griffin. He has no family, and apparently little money as he is first found contemplating whether or not he wants to keep a set of cast-off boots. He is fat, red faced, slow moving and doesn’t seem terribly bright, but that is merely the effect of Griffin having the advantage over him. As soon as he realizes his predicament, he begins to look for any possible means of escape. As for Griffin, he is “making use” of Marvel in the same way that he did the Halls, the stray cat, and even his own father. Whatever means he deems necessary to his purpose is enacted without thought or conscience.

CHAPTER 10 Mr. Marvel’s Visit to Iping

Iping has nearly recovered its earlier holiday atmosphere. As only a few people had actually made contact with the Invisible Man, the general population is soon able to reason him away as some trick of an overactive, holiday imagination.

Around 4:00, Mr. Marvel enters town and is observed by Huxter to behave rather strangely. He makes his way down the street almost reluctantly. He stops at the foot of the steps to the Coach & Horses and seems to undergo a great struggle before finally entering. A few minutes later, he re-emerges, apparently having had a drink, and walks as if he is trying

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to act nonchalant. Soon he disappears into the yard and re-emerges with a bundle wrapped in a tablecloth. Huxter thinks some robbery has taken place and tries to follow Marvel when he is tripped in a mysterious fashion and sent sprawling.

Notes – Griffin has used Marvel to attempt to get his belongings out of the Coach & Horses. Marvel’s resistance manages to get attention, but not the attention he wants. Huxter thinks that Marvel has committed the robbery.

CHAPTER 11 In the Coach Horses

The narrator backtracks to explain what happened inside the Coach & Horses. Mr. Cuss and Mr. Bunting were in the parlor going through the belongings of the Invisible Man. Three large books labeled “Diary” are written in a cipher or code they do not understand.

Suddenly the inn door opens and Mr. Marvel enters. They disregard him and begin studying the books again when an unseen force grabs each of them by the neck and begins pounding their heads on the table between questions about what they are doing with his things. The man demands his belongings, saying he wants his books and some clothes.

Notes – Griffin is on the verge of insanity. He is probably terrified on two counts. One would be lest someone tamper with his notes or other belongings related to his experiments. The other would be lest someone should actually be able to decipher his records.

CHAPTER 12 The Invisible Man Loses His Temper

Mr. Hall and Teddy Henfrey are involved in a discussion behind the hotel bar when they hear a thump on the parlor door. They hear strange sounds as of things being thrown against the door and some bizarre conversation. Doors open and shut and they see Marvel taking off with Huxter trying to follow him. Suddenly Huxter executes a complicated leap in the air. Seconds later, Hall lands on the ground as if he had been attacked by a football player.

Several other individuals are shoved aside or sent sprawling in the streets. Mr. Cuss calls for help, telling people that the “Man” has all of the vicar’s clothes. After breaking all the windows in the Coach & Horses and thrusting a chair through the parlor window of another citizen’s house, the Invisible Man disappears from Iping.

Notes – Marvel has taken advantage of the situation, and rather than carrying Griffin’s material for him, has run off with it. The intervention of Huxter and the other individuals almost enables Marvel to get away with the precious books. Cuss quickly catches on to the fact that Griffin will be visible so long as he is carrying the bundle, but he is unaware of the existence of Marvel. The narrator tells us that “perhaps” the Invisible Man only intended to use the vicar’s clothes to cover his

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retreat, but that at some chance blow he has “gone completely over the edge.” He throws or upends benches, chairs and boards, along with breaking windows. Eventually he catches up with Marvel and they head for the next town.

CHAPTER 13 Mr. Marvel discusses His Resignation

Mr. Marvel, propelled by the unrelenting shoulder grip and vocal threats of the Invisible Man, arrives in Bramblehurst. Marvel tries to reason his way out of the situation to no avail. The Invisible man needs a normal person to carry his books and is determined to make use of the fat, red-faced little man.

Notes – This brief chapter serves to track Griffin’s movement to the next location and to show his crude behavior toward Marvel. Marvel tries reasoning, whining, and even suggesting that he may in the long run be a failure and thus “mess up” Griffin’s plans. Nothing works. For the moment, Griffin needs Marvel. If Marvel should drop in accordance with his professed heart condition, it would mean nothing to Griffin.

CHAPTER 14 At Port Stowe

Marvel arrives in Port Stowe and is seen resting on a bench outside of town. He has the books with him, but the bundle of clothing has been abandoned in the woods. As he sits there, an elderly mariner, carrying a newspaper, sits down beside him. Citing the paper, the mariner brings up the topic of an Invisible man.

According to the newspaper, the man afflicted injuries on the constable at Iping. Certain evidence indicates that he took the road to Port Stowe. The mariner ponders the strange things such a man might be able to do-trespass, rob or even slip through a cordon of policeman.

Marvel begins to confide in the mariner, saying he knows some things about this Invisible Man. Suddenly Marvel is interrupted by an attack of some kind of pain. He says it is a toothache, then goes on to say that the Invisible Man is a hoax. Marvel begins to move off, walking sideways with violent forward jerks.

Later the mariner hears another fantastic story-that of money floating along a wall in butterfly fashion. The story is true, however. All about the neighborhood, money has been making off by the handful and depositing itself in the pockets of Mr. Marvel.

Notes – Marvel tries to take advantage of a short respite to let someone else know about the Invisible Man, but he is caught by Griffin before he can complete his story. This chapter gives us a little insight as to how Griffin has been surviving to this point. He has been stealing money wherever he could find it. Now that he is obliged to remain invisible, however, he has to use Marvel as a repository for his ill-gotten gain. The irony is that although Griffin can steal unlimited amounts, he has no way to use the

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money in his invisible condition. And Marvel, who is for a time nothing more than a helpless victim, will be the one to benefit in the end.

CHAPTER 15 The Man Who Was Running

Dr. Kemp happens to be day-dreaming out his window when he spots a short, fat man running down the hill as fast as he can go. The doctor notices that the man is running “heavy” as if his pockets are “full of lead.” Kemp’s reaction is one of contempt, but the people on the street who see him approaching react a bit differently. The running man is Marvel; his expression is one of terror. A short distance behind him, people hear the sound of panting and a pad like hurrying bare feet. Soon cries of “The Invisible Man is coming” are heard in the streets along with the slamming of doors as people bolt into their houses.

Notes – This chapter simply introduces Kemp into the story. Kemp’s attitude is representative of the average established, self-confident, and self-sufficient individual. He sees a man in trouble, but his reaction in contemptuous instead of concern. He has heard warning cries about an Invisible Man, but clearly doesn’t believe any of it. He is a man who keeps himself apart form the concerns of the general public, is buried in his work, interested only in what award it will ultimately bring him.

CHAPTER 16 In the lolly Cricketers

The Jolly Cricketers is a tavern. The barkeep, a cabman, an American and an off-duty policeman are engaged in idle chat when marvel bursts through the door. Marvel begs for help, claiming the Invisible Man is after him.

A pounding begins at the door and then a window is broken in. The Invisible Man doesn’t come in immediately, however. The barman checks the other doors, but by the time he realizes the yard door is open, the Invisible Man is already inside. Marvel, who is hiding behind the bar, is caught and dragged into the kitchen. The policeman rushes in and grips the invisible wrist of the hand that holds onto Marvel, but is abruptly hit in the face.

People stumble over and into each other as all try to catch the Invisible Man. He yelps when the policeman steps on his foot, then flails wildly about with his Invisible fists and finally gives them the slip. The American fires five cartridges from his gun, sweeping his gun in a circular pattern as he fires. The chapter ends with the men feeling around for an invisible body.

Notes – Griffin is injured in this chapter. He is thus forced to find shelter and help in the nearest possible place. But now, enough people have been involved in Griffin’s mayhem that it will be relatively easy to round up a posse of believers when the time comes to do so.

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CHAPTER 17 Doctor Kemp’s Visitor

Doctor Kemp is still working in his study when he hears the shots fired in the Cricketers. He opens his window and watches the crowd at the bottom of the hill for a few minutes, then returns to his writing desk. A few minutes later, he hears his doorbell ring, but his housekeeper says it was only a “runaway” ring.

The doctor is at his work until 2 AM when he decides to go downstairs for a drink. On the way, he notices a spot of drying blood on his linoleum floor. Then he finds more blood on the doorknob of his own bedroom. In his room, his bedspread is smeared with blood, his sheet is torn, and bedclothes are depressed as if someone has been sitting there.

The Invisible Man introduces himself to Kemp. He is Griffin, of University College. He explains that he made himself Invisible, but is wounded and desperately in need of shelter, clothes and food.

Kemp loans him a dressing gown along with some drawers, socks and slippers. Griffin eats everything Kemp can rustle up and finally asks for a cigar. He promises to tell Kemp the story of his bizarre situation but insists that he must sleep first as he has had no sleep in nearly three days.

Notes – Kemp’s reaction is in stark contrast to Marvel’s original reaction to Griffin. Although he finds the story hard to believe, he is too well educated and too intelligent to deny the evidence of his own eyes. Nor is he prey to hysterics or to working class superstitions. The idea of a spirit or witchcraft doesn’t even occur to him. His cool demeanor as he helps Griffin to the things he needs could be an indication of hope for the Invisible Man.

CHAPTER 18 The Invisible Man Sleeps

Griffin examines the windows of the room, then exacts a promise from Kemp that he will not be betrayed in his sleep and finally locks the door, barring Kemp from his own room.

Kemp retires to his dining room to speculate upon the strange events. There he sees the day’s newspaper, which he had ignored earlier. He reads it eagerly, but assigns the more terrifying elements of the stores to “fabrication.” In the morning, he sends his housekeeper for all available papers and reads those as well. The papers contain stories of the previous evening’s events at the Cricketers along with a rather badly written account of Marvel’s experience. Marvel doesn’t tell how he came upon the money in his pockets, nor does he mention the location of the three books. Kemp becomes alarmed at the possibilities of what Griffin could do and writes a note to Colonel Adye at Port Burdock.

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Notes – Kemp experiences his first apprehension because of what his own intelligence reveals to him rather than from the hysterical reports in the papers. He is motivated, however, from personal interest. When he recalls the behavior of Marvel, he realizes that Marvel-a mere tramp-was being pursued by Griffin. He suddenly realizes that Griffin is insane to the point of being homicidal.

CHAPTER 19 Certain First Principles

Griffin explains how he became invisible. He had been a medical student, but had dropped medicine and taken up physics. He discovered a formula of pigments that lowers the refractive index of a substance, allowing light to pass through it rather than being reflected or refracted. After experimenting with pigments for three years, he came upon the secret whereby animal tissue could be rendered transparent. He was continuously trying to hide his work from another professor. He was finally brought to a halt in his experimenting by a lack of funds, a problem he solved by robbing his own father. Because the money did not belong to him, his father shot himself.

Notes – From this chapter through XXIII, the point of view changes as Griffin tells his own story. He explains how he became invisible and tells the story up to the time when he had first entered the Coach & Horses. He explains his use of and contempt for Marvel, justifying his own behavior as necessary to his survival.

CHAPTER 20 Doctor Kemp’s Visitor

Griffin explains how he had found lodging in a boarding house on Great Portland Street. After his father’s funeral, he went to his apartment to continue with his experiments. He successfully made a piece of cloth disappear, then he tried his process on a stray cat. The cat was not entirely successful, as the animal’s eyes and claws never completely disappeared.

Later the next day he had a minor altercation with the landlord who brought reports of Griffin tormenting a cat in the night. The landlord wanted to know what Griffin was doing in the room and what all the paraphernalia was for. The two argued and Griffin shoved the landlord out of the room. Griffin knew he would have to act quickly, so he made arrangements to have his belongings stored, then he drank some of his own potion. In the evening the landlord returned with an ejection notice, but was too terrified at the stone white face of Griffin to serve it. In spite of extreme illness and pain, Griffin finished his treatment and watched himself gradually disappear.

In the morning, the landlord, his stepsons and the elderly neighbor lady who had complained about the cat enter Griffin’s apartment and are astonished to see no one. A day later, afraid, lest his equipment reveal too much information, Griffin smashes the items and sets fire to the house. Believing that he has covered his tracks with impunity, he begins to

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imagine all sorts of “wild and wonderful” things he will be able to do under the cover of invisibility.

Notes – Griffin’s explanations are completely absent of any sense of humanity or conscience. His intentions suggest anarchy or lawlessness resulting from an absence of social restriction. Killing his own father seems to have killed his conscience, and the novelty of invisibility highlights his immaturity and seems to divorce him from a normal sense of responsibility.

CHAPTER 21 In Oxford, Street

Griffin continues to explain his experiences with invisibility. He soon discovered that being invisible had as many drawbacks as advantages. People ran into him and stepped on him. He had to be continually on guard as to the movements and positions of others in order to avoid accidental contact. To make matters worse, although people could not see him, dogs could detect him with their keen sense of smell. As he had to remain naked, he was soon uncomfortable. Also, he could not eat, as food was visible until it was fully assimilated into his system.

At one point, he had run up the steps of a house in order to avoid a unit of a marching Salvation Army band. While he waited, two youngsters spotted the prints of his bare feet in the mud. Soon a crowd of people had gathered to look at the “ghost prints.” He leapt over the railing and ran through a bunch of back roads to avoid the press. Fortunately for him, his escape at that time was aided with the distraction created by conflagration engulfing his former dwelling.

Notes – Griffin’s initial error was that he became so obsessed with a single scientific notion that he failed to take consequences into consideration. No doubt, he was not concerned about people reacting to him as though he were some kind of mutation or monster. As an albino human, he was already a marginalized individual who did not fit into ordinary society. College was the perfect place for him, but he was so concerned about the possibility of any one getting credit for his discovery that he failed to take advantage of collaboration and more mature knowledge that he might have had access to.

CHAPTER 22 In the Emporium

Griffin explains his first attempts to get clothing and render his situation more tolerable. He had gone into the Omniums, a large apartment type store where one could buy everything from groceries to clothing. He made his way to an area of bedsteads and mattresses, hoping that once the store closed for the night, he would be able to sleep on the mattresses and steal some clothes with which to mask his condition.

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In the night, he procured a complete set of clothes for himself, helped himself to food in a refreshment department, and then slept in a pile of down quilts. He failed to awaken before the morning crew had entered, however, and was unable to escape as long as they could see him. Thus, he was forced to shed the clothing and run, naked, back out into the cold.

Notes – Griffin was preoccupied with getting his food and clothes by illicit means. His plans are continually evil even as the reactions of other people are consistently behaviors of suspicion and rejection. At no point, does he consider trying to get anyone to understand his situation. His imagination drives him only toward evil, as if the grotesque and the evil are natural partners.

CHAPTER 23 In Drury Lane

Griffin’s peril increased daily. He had no clothes or shelter and dared not eat. Also, he soon realized that walking through the streets of London was going to result in an accumulation of dirt on his skin- which would make him visible in a grotesque way.

He made his way into a costume shop, hoping to make way with some clothes and dark glasses after the proprietor had gone to bed. In the shopkeeper’s room, he had to stand and watch the man eat his breakfast. Furthermore, the man had exceptionally acute hearing and nearly discovered Griffin several times. When evening came, he was finally able to explore the house and found a pile of old clothes. In his excitement, he forgot about the noise he was making and was nearly caught when the shopkeeper investigated the noise. Unable to see the source, but positive someone was in the house, the proprietor went about locking all the doors in the house and pocketing the keys. In desperation, Griffin struck the old man on the head, then gagged and tied him with a sheet. Then he put together a costume of old clothes, stole all the money he could find and went out into the street.

Believing his troubles were over, Griffin went into a restaurant and ordered a meal, but soon realized he couldn’t eat it without exposing his invisible face. He ordered the lunch and left, telling the proprietor that he would be back in ten minutes.

Griffin went to “another place” (which happens to be the Coach & Horses Inn) and demanded a private room, explaining that he was “badly disfigured.” Thus, he had set himself up at Iping, hoping to find a way to reverse the process of invisibility. Here he was finally discovered.

Notes – This chapter brings us current with events in the first chapter of the book.

CHAPTER 24 In Oxford, Street

Griffin tells how his original plan, after being discovered by the people of Iping, had been to get his books and get out of the country, but that plan had changed upon meeting

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Kemp. He thinks that Kemp can work with him. Together they can set up a “reign of terror” to take full advantage of the Invisibility. Griffin does not realize that Kemp has already betrayed him and is only trying to keep him talking until the police arrive. Kemp stands in front of the window to keep Griffin from seeing the police, but Griffin soon hears them on the stairs and realizes he has been deceived.

Griffin quickly begins to disrobe even as Kemp springs to the door and attempts to lock him in. A dropped key spoils the effort as the now invisible Griffin shoves him aside, then hurls his weight at Colonel Adye, the chief of the Burdock Police who is approaching on the stairs. Griffin escapes past two more policemen in the hall; they hear the front door of the house slam violently.

Notes – In assuming that he can make demands and others will simply capitulate to him, Griffin has misjudged Kemp. Kemp is self-centered, but is not a murderer. As for Griffin himself, he appears to have abandoned any intention of searching for an antidote and is only interested in trying to terrorize as much of the country as he can. He wants to set himself up as a vindictive god with Kemp as his personal henchman.

CHAPTER 25 The Hunting of the Invisible man

Kemp explains the situation to the police, informing them of Griffin’s intentions to cause general mayhem. They talk of using dogs to sniff him out and of putting powdered glass in the streets.

Notes – The narrator tells us that if he had used his time more wisely, Griffin may have been able to escape during the 24 hours it took the countryside to organized. He slept instead, however, and by the time he had awakened there was no escape possible.

CHAPTER 26 The Wicksteed Murder

By 2:00 in the afternoon, the entire countryside around Burdock has been mobilized. Men set out with guns, clubs and dogs, and the police warn the village people to lock their doors and stay inside. Griffin manages to evade his pursuers for a 24-hour period except for one encounter with a middle-aged man who had apparently cornered him. Griffin kills the man by beating him with an iron rod.

CHAPTER 27 The Siege of Kemp’s House

Kemp receives a letter telling him that the Reign of Terror is beginning and that Kemp himself will be the first execution for the sake of an example. Kemp decides that he himself will be the bait and that Griffin will be caught because he will have gone too far. A

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knock at the door turns out to be Adye with news that Kemp’s housekeeper-who was carrying notes for the police-had been attacked and the notes taken from her.

Griffin makes his presence known by smashing windows in Kemp’s house. During the battle that follows, Adye is shot. Griffin gets inside the house and tries to tell the police to “stand away” as he is after only Kemp. He swings an ax at them, but one of them manages to strike him with an iron poker. By this time Kemp has followed his housekeeper through a window and is nowhere to be found.

Notes – The police express contempt for Kemp, believing he has run off and left them to face Griffin alone. The truth is, he has, because he knows Griffin will follow through on his threats. However, even though Kemp tries to escape, he does not forget his earlier idea of using himself as bait. It is ironic that he runs the same course he watched Marvel run just a couple days earlier. He, too, is white faced and terrified, but keeps his wits; whenever he finds a bit of uneven ground or a patch that is scattered with broken glass, he takes it, knowing it will slow down the invisible, barefooted Griffin.

CHAPTER 28 The Hunter Hunted

Griffin chases Kemp through the town. People begin to join in the chase. When Kemp realizes that the people are chasing Griffin, he stops running, which allows the Invisible Man to catch him. Even though people cannot see him, they are able to grab hold of him and keep him down. The effort is not needed for long as Griffin has been fatally injured and seems to have lost a lot of blood. As the town people watch, the effect of invisibility is gradually reversed, and soon, Griffin, now dead, is visible.

Notes – When Griffin becomes visible, his albino condition is also revealed. It is interesting that the people are not horrified or even surprised. Nor is there any speculation about how this bizarre incident could have happened. The people watch as his broken, battered body slowly becomes visible from his extremities to the center of his being. It is only when his white face and hair and staring garnet eyes are revealed that someone calls for them to “cover that face” before the children in the town can see it.

EPILOGUE

Mr. Marvel, formerly the tramp, has become the landlord of the little inn near Port Stowe and the “owner” of all the information about Griffin. He has been able to keep all the money Griffin stole because lawyers could not identify the sources accurately. The books seem to have disappeared entirely; at least whenever anyone asks Marvel about them, he denies knowing anything. However, when the inn is closed and he is alone, he takes the books out of their hiding place and tries to study the “wonderful secrets.”

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Notes – The epilogue implies that the people, represented by the tramp-turned-innkeeper, not only have learned very little from the experience of the invisible man, but that they would not be above trying the invisibility themselves if only they knew how to do it. Regardless of the horrors perpetrated by Griffin, it seems to be part of human nature to want to be able to cause chaos and commit obscenities with impunity. While Marvel says that he would not do the same things Griffin did, there is little doubt that anyone, given such advantage over others, would resist the temptation to dabble in behaviors that are unacceptable in normal civilized society.

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January 16, 2019
NCERT Exemplar Problems Solutions for Class 12 Physics
Physics NCERT Exemplar Solution pdf download, get all Physics NCERT Exemplar problem solution here with Step by step problem solution of class 12th CBSE.

NCERT Exemplar Problems Class 12 Physics
Also check: NCERT Exemplar Problems Solutions for Class 12 Maths
March 20, 2018
Communication Systems NCERT Exemplar Problems Solutions Physics

NCERT Exemplar Problems Class 12 Physics Chapter 15 Communication Systems

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?
(a) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave
Solution: (b)
Key concept: The radio waves emitted from a transmitter antenna can reach the receiver antenna by the following mode of operation.
• Ground wave propagation
• Sky wave propagation
• Space wave propagation
Mode of communication frequency range:
• Ground wave propagation— 500 kHz to 1710 kHz
• Sky wave propagation — 2 MHz to 40 MHz
• Space wave propagation— 54 MHz to 42 GHz
So, A is transmitted via ground wave, B via sky wave and C via space wave.

Question 2. A loom long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with λ.
(a) ~400m (b) -25 m (c) -150 m (d) -2400 m
Solution: (a) Length of the building (l) is
l = 500 m
and length of antenna = 100 m
and we know, wavelength of the wave which can be transmitted by
L =λ/4. So, λ~ 4l= 4 x 100 = 400 m
Wavelength (λ) is nearly equal to 400 m.

Question 3. A1 kW signal is transmitted using a communication channel which provides attenuation at the rate of -2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in d_B =10 log₁₀(p₀/p_i)]
(a) 900 W (b) 100 W (c) 990 W (d) 1010 W
Solution:

Question 4. A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be
(a) 1.003 MHz and 0.997 MHz (b) 3001 kHz and 2997 kHz
(c) 1003 kHz and 1000 kHz (d) 1 MHz and 0.997 MHz
Solution: (a)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged.
Side band frequencies: The AM wave contains three frequencies f_c, (f_c + fm) and (f_c -f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c +f_m): Upper side band (USB) frequency
(f_c -f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
According to the problem, frequency of carrier signal is f_c = 1 MHz and frequency of speech signal = 3 kHz
= 3 x 10^-3 MHz
= 0.003 MHz
We know that, Frequencies of side bands = (f_c ± f_m) = (1 + 0.003) and (1 – 0.003)
So, side band frequencies are 1.003 MHz and 0.997 MHz.

Question 5. A message signal of frequency ω_m is superposed on a carrier wave of frequency ω_c to get an Amplitude Modulated Wave (AM). The frequency of the AM wave will be

Solution: (b)
Key concept: The process of changing the amplitude of a carrier wave in accordance with the amplitude of the audio frequency (AF) signal is known as amplitude modulation (AM).
In AM, frequency of the carrier wave remains unchanged or we can say that the frequency of modulated wave is equal to the frequency of carrier wave. Now, according to the problem, frequency of carrier wave is f_c.
Thus the amplitude modulated wave also has frequency f_c.

Question 6. I-V Characteristics of 4 devices are shown in figure.

Solution: Key concept: A square law modulator is the device which can produce modulated waves by the application of the message signal and the carrier wave.
Square law modulator is used for modulation purpose. Characteristics shown by (i) and (iii) correspond to linear devices.
And by (ii) corresponds to square law device which shows non-linear relations. Some part of (iv) also follow square law.
Hence, (ii) and (iv) can be used for modulation.

Question 7. A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0 < m <1)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission.
Solution: (b) In this problem, the frequency of modulated signal received becomes more, due to improper selection of bandwidth.
This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal.
But, the frequency of male voice is less than that of a female.

Question 8. A basic communication system consists of
A. transmitter.
B. information source.
C. user of information.
D. channel.
E. receiver.
Choose the correct sequence in which these are arranged in a basic communication system.
(a) ABCDE (b) BADEC (c) BDACE (d) BEADC
Solution: (b) A basic communication system consists of an information source, a transmitter, a link (channel) and a receiver or a communication system is the set-up used in the transmission and reception of information from one place to another.
The whole system consist of several elements in a sequence. It can be represented as the diagram given below:

Question 9. Identify the mathematical expression for amplitude modulated wave,

Solution:

One or More Than One Correct Answer Type

Question 10. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal cannot be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km
Solution: (a, b, d)
Key concept: Size of the antenna or aerial. For transmitting a signal, we need an antenna or an aerial. This antenna should have a size comparable to the wavelength of the signal (at least 1/4 in dimension) so that the antenna properly senses the time variation of the signal. For an electromagnetic wave of frequency 20 kHz, the wavelength λ is 15 km. Obviously, such a long antenna is not possible to construct and operate. Hence direct transmission of such baseband signals is not practical. We can obtain transmission with reasonable antenna s if transmission frequency is high (for example,
if n is 1 MHz, then λ is 300 m). Therefore, there is a need of translating the information contained in our original low frequency baseband signal into high or radio frequencies before transmission.
Effective power radiated by an antenna: A theoretical study of radiation from a linear antenna (length l) shows that the power radiated is proportional to (1/λ)² . This implies that for the same antenna length, the power radiated increases with decreasing λ, i.e., increasing frequency. Hence, the effective power radiated by a long wavelength baseband signal would be small. For a good transmission,we need high powers and hence this also points out to the need of using high frequency transmission.

Question 11. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?
(a) The side band frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6 kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The side band frequencies are 1503 kHz and 1497 kHz
Solution: (b, d)

Also,bandwidth =2 f_m=2 x 3=6 KHz

Question 12. A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be (6.4 x 10⁶ m)
(a) 100 km (b) 24 km (c) 55 km (d) 50 km
Solution: (b, c, d)

Therefore, the range of 55.4 km covers the distance 24 km, 55 km and 50 km.

Question 13. The frequency response curve (figure) for the filter circuit used for production of AM wave should be

Solution:(a, b, c)
Key concept:
(i) Side band frequencies-. The AM wave contains three frequencies f_c ,(f_c +f_m) and (f_c-f_m),f_c is called carrier frequency, (f_c +f_m) and (f_c -f_m) are called side band frequencies.
(f_c + f_m)- Upper side band (USB) frequency
(f_c – f_m): Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency,
(ii) Bandwidth: The two side bands lie on either side of the carrier frequency at equal frequency interval ω_m.
So, bandwidth = {(ω_c + ω_m) – (ω_c – ω_m)} = 2ω_m

To produce an amplitude modulated wave, bandwidth is given by the difference between upper side band frequency and lower side band frequency. Bandwidth = ω_USB – ω_LSB = (ω_c + ω_m) – (ω_c – ω_m)

Question 14. In amplitude modulation, the modulation index m is kept less than or equal to 1 because
(a) m> 1, will result in interference between carrier frequency and message frequency, resulting into distortion.
(b) m > 1, will result in overlapping of both side bands resulting into loss of information
(c) m > 1, will result in change in phase between carrier signal and message signal.
(d) m > 1, indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion.
Solution: (b, d)
Key concept: Modulation index: The ratio of change of amplitude of carrier wave to the amplitude of original carrier wave Is called the modulation factor or degree of modulation or modulation index (m_a).

Very Short Answer Type Questions

Question 15. Which of the following would produce analog signals and which would produce digital signals?
(a) A vibrating tuning fork
(b) Musical sound due to a vibrating sitar string
(c) Light pulse
(d) Output of NAND gate
Solution: Analog and digital signals are the gateway of information or we can say that they are used to transmit information through electric signals. In both these signals, the information such as any audio or video is transformed into electric signals.
The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes. So, output of a NAND gate and a light pulse produces a digital signal.
Thus, (a) and (b) would produce analog signal and (c) and (d) would produce digital signals.

Question 16. Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?
Solution: A signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.
But, here the frequency of TV signals are 60 MHz which is beyond the required range (frequency range: there is a maximum frequency of EM waves called critical frequency, above which wave cannot reflect back).
So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.
Important point: Sky wave propagation: These are the waves which are reflected back to the earth by ionosphere.
Ionosphere is a layer of atmosphere having charged particles, ions and electrons and extended above 80 km – 300 km from the earth’s surface.

Question 17. Two waves A and B of frequencies 2 MHz and 3 MHz, respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?
Solution: We know that refractive index p of a layer is

The refractive index of wave B is more than refractive index of wave A because frequency of wave B is more than wave A (as refractive index increases with frequency increases).
Sin i / sin r = µ (lesser the value of r larger the value of µ )
For higher frequency wave (i.e., higher refractive index) the angle of refraction is less, i.e., bending is less. So, wave B travels longer distance in the ionosphere before suffering total internal reflection.
Importance point: Refractive index of a medium is that characteristic which decides speed of light in it.
Dependence of Refractive index:
(i) Nature of the media of incidence and refraction.
(ii) Colour of light or wavelength of light.
(iii) Temperature of the media: Refractive index decreases with the increase in temperature.
Total internal reflection: When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90°. This angle of incidence is called critical angle (C).
When angle of incidence exceeds the critical angle then light ray comes back into the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

Question 18. The maximum amplitude of an AM wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index?
Solution:

Question 19. Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.
Solution:

Question 20. Why is an AM signal likely to be more noisy than a FM signal upon transmission through a channel?
Solution: An AM signal likely to be more noisy than FM signal through a channel because in case of AM, the instantaneous voltage of carrier waver waves is varied by the modulating wave voltage So, during the transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal. In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal transmitting in channel. So, noise does not affect FM signal or simply we can say that noise signals are difficult to filter out in AM reception whereas FM receivers easily filter out noise.
Important point: In frequency modulation m_f (frequency modulation index) is inversely proportional to modulating frequency f_m. While in PM it does not vary with modulating frequency. Moreover, FM is more noise immune.

Short Answer Type Questions

Question 21. Figure shows a communication system. What is the output power when input signal is of 1.01 mW? [Gain in d_B = 10 log₁₀ (P₀ / P₁)]

Solution:

Question 22. A TV transmission tower antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to case (i).
Solution:

Question 23. If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earth’s radius.?
Solution:
Key concept: Distance or range of transmission tower, d_T =√2Rh_T
where, R is the radius of the earth (approximately 6400 km). h_T is the height of transmission tower, .
d_T is also called the radio horizon of the transmitting antenna.
Let us consider the figure given below to solve this problem.
Assume the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is h₁, i.e. h_T = h_R and radius of earth is R. If d_M is the line-of-sight distance between the transmission and receiving antennas, then maximum distance

Question 24. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max= 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F₁ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere. Estimate the maximum electron densities of the F₁ and F₂ layers on that day.
Solution:

Question 25. On radiating (sending out) and AM modulated signal, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c + ω_m). Suggest ways to minimise cost of radiation without compromising on information.
Solution:
Key concept: Side band frequencies. The AM wave contains three frequencies ω_c, (ω_c + ω_m) and (ω_c – ω_m), ω_c is called carrier frequency, (ω_c + ω_m) and ( ω_c – ω_m) are called side band frequencies.
(ω_c + ω_m) = Upper side band (USB) frequency
(ω_c – ω_m) =Lower side band (LSB) frequency
Side band frequencies are generally close to the carrier frequency.
Only side band frequencies contain information in amplitude modulated signal, [only (ω_c+ ω_m) and (ω_c + ω_m)].
Here, the total radiated power is due to energy carried by ω_c, (ω_c – ω_m) and (ω_c+ω_m)
For reduction of cost of radiation without compromising on information ω_c can be left and transmitting the frequencies (ω_c + ω_m), (ω_c – ω_m) or both (ω_c + ω_m) and (ω_c – ω_m).

Long Answer Type Questions

Question 26. The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I₀e^-ax, where I₀ is the intensity at x = 0 and α is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of (In4/α).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 log₁₀(I/I₀).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
Solution:

Question 27. A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above Earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
Solution: Let the receiver is at point A and source is at B.

Question 28. An amplitude modulated wave is as shown in figure. Calculate
(i) the percentage modulation,
(ii) peak carrier voltage and
(iii) peak value of information voltage

Solution:

Question 29. (i) Draw the plot of amplitude versus ω for an amplitude modulated wave whose carrier wave (ω > ω_c) is carrying two modulating signals, ω1 and ω₂ (ω₂> ω₁).
(ii) Is the plot symmetrical about ω_c? Comment especially about plot in region (ω < ω_c ).
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
Solution:

(ii) In the plotted graph shown, we note that frequency spectrum is not symmetrical about ω_c. Crowding of spectrum is present for ω < ω_c.
(iii) If more modulating signals are present then there will be more crowding in the modulation signal in the region ω <ω_c. That will result more chances of mixing of signal.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves ω_c. This shows that large carrier frequency
enables to carry more information (i.e., more ω_m) and the same will in turn increase bandwidth.

Question 30. An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
(i) R = 1 kΩ, C= 0.01 µF?
(ii) R= 10 kΩ, C=0.01 µF?
(iii) R = 10 kΩ, C = 0.1 µF?
Solution:

March 20, 2018
Semiconductor Electronics: Materials, Devices and Simple Circuits NCERT Exemplar Problems Solutions Physics
NCERT Exemplar Problems Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. The conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free current carries increases
(b) relaxation time increases
(c) both number density of carries and relaxation time increase
(d) number density of carries increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density .
Solution: (d)

Question 2.

Solution: (b)
Key concept: P-N Junction Diode:
– When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting arrangement is called P-N junction or P-N junction diode.

(1) Depletion region: On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region.
Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions, appears near the junction in N-crystal. This layer is called depletion layer.

(2) Potential barrier: The potential difference created across the P-N junction due to the diffusion of electron and holes is called potential barrier.

Height of potential barrier is decreases when p-n junction is forward biased, it opposes the potential barrier junction, when p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

Question 3. In figure given on next page, assuming the diodes to be ideal
(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice-versa
(c) D₁ and D₂ are both forward biased and hence current flows from A to B
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice-versa

Solution:

Question 4. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor?

Solution: (d)
Key concept: Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called half wave rectifier.

Question 5. Hole is
(a) an anti-particle of electron
(b) a vacancy created when an electron leaves a covalent bond
(c) absence of free electrons
(d) an artificially created particle
Solution: (b) Concept of holes in the semiconductor:
1. .When an electron is removed from a covalent bond, it leaves a vacancy behind. An electron from a neighbouring atom can move into this vacancy, leaving the neighbour with a vacancy. In this way the vacancy formed is called a hole (or cotter), and can travel through the material and serve as an additional current carriers.
2. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron.
3. Holes acts as a virtual charge, although there is no physical charge on it.
4. Effective mass of hole is more than an electron.
5. Mobility of hole is less than an electron.
Question 6. The output of the given circuit in figure is given below.

(a) would be zero at all times
(b) would be like a half wave rectifier with positive cycles in output
(c) would be like a half wave rectifier with negative cycles in output
(d) would be like that of a full wave rectifier
Solution: (c)

When the diode is forward biased during positive half cycle of input AC voltage, the resistance of p-n junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero output voltage across p-n junction.
And when the diode is reverse biased during negative half cycle of AC voltage, the p-n junction is reverse biased. The resistance of p-n junction becomes high which will be more than resistance in series. That is why, there will be voltage across p-n junction with negative cycle in output, hence option (c) is correct.

Question 7. In the circuit shown in figure given below, if the diode forward voltage between A and B is

(a) 1.3 V (b) 2.3 V (c) 0 (d) 0.5 V
Solution:

Question 8. Truth table for the given circuit is

Solution:

One or More Than One Correct Answer Type

Question 9. When an electric field is applied across a semiconductor
(a) electrons move from lower energy level to higher energy level in the conduction band
(b) electrons move from higher energy level to lower energy level in the conduction band
(c) holes in the valence band move from higher energy level to lower energy level
(d) holes in the valence band move from lower energy level to higher energy level
Solution: (a, c)
In valence band electrons are not capable of gaining energy from external electric field. While in conduction band the electrons can gain energy from external electric field.
When electric field is applied across a semiconductor, the electrons in the conduction band (which is partially filled with electrons) get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Question 10. Consider an n-p-n transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
(a) Electrons crossover from emitter to collector
(b) Holes move from base to collector
(c) Electrons move from emitter to base
(d) Electrons from emitter move out of base without going to the collector.
Solution: (a, c)

In normal operation base-emitter is forward biased, i.e., the positive pole of emitter base battery is connected to base and its negative pole is connected to the emitter. And collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base. Thus, electron moves from emitter to base and crossover from emitter to collector.

Question 11.

Solution: (b, c, d) According to above graph transfer characteristics of a base biased common emitter transistor, we note that .
(a) when V_i= 0.4 V, output voltage remain same,there is no collection current. So, transistor circuit is not in active state.
(b) when V_i = 1 V (This is in between 0.6 V to 2 V), the transistor circuit is in active state and when input is increasing output is decreasing because when CE is used as an amplifier input and output voltages are 180° out of phase. Then it is used as an amplifier.
(c) when V_i = 0.5 V, there is no collector current. The transistor is in cut off state. The transistor circuit can be used as a switch to be turned off.
(d) when V_i = 2.5 V, the collector current becomes maximum and transistor is in a saturation state and can used as switch turned on state.

Question 12. In a n-p-n transistor circuit, the collector current is 10 mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
(a) The emitter current will be 8 mA
(b) The emitter current will be 10.53 mA
(c) The base current will be 0.53 mA
(d) The base current will be 2 mA
Solution: (b, c) According to the problem, the collector current is 95% of electrons reaching the collector after emission. And collector current, I_C = 10 mA

Question 13. In the depletion region of a diode
(a) there are no mobile charges
(b) equal number of holes and elections exist, making the region neutral
(c) recombination of holes and electrons has taken place
(d) immobile charged ions exist
Solution: (a, b, d) On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-rcgion diffuse through the junction into P-region and the hole from P-region diffuse into N-region.

Due to diffusion, neutrality of both N-and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This layer is called depletion layer.
The thickness of depletion layer is 1 micron = 10^-6 m.
Width of depletion layer ∞ 1/Dopping
Depletion is directly proportional to temperature.
Important point: The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a dielectric.

Question 14. What happens during regulation action of a Zener diode?
(a) The current and voltage across the Zener remains fixed
(b) The current through the series Resistance (R_s) changes
(c) The Zener resistance is constant
(d) The resistance offered by the Zener changes

Solution: (b, d) Symbolically zener diode represents like this:
In the forward bias, the zener diode acts as an ordinary diode. It can be used as a voltage regulator.

A zener diode when reverse biases offers constant voltage drop across in terminals as unregulated voltage is applied across circuit to regulate. Then during regulation action of a Zener diode, the current through the series resistance R_s changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

Question 15. To reduce the ripples in rectifier circuit with capacitor filter
(a) R_L should be increased
(b) input frequency should be decreased .
(c) input frequency should be increased
(d) capacitors with high capacitance should be used
Solution: (a, c, d)
Ripple factor may be defined as the ratio of r.m.s. value of the ripple voltage to the absolute value of the DC component of the output voltage, usually expressed as a percentage. However ripple voltage is also commonly expressed as the peak-to-peak value. Ripple factor (r) of a full wave rectifier using capacitor filter is given by

Ripple factor is inversely proportional to R_L, C and v.
Thus to reduce r, R_L should be increased, input frequency v should be increased and capacitance C should be increased.

Question 16. The breakdown in a reverse biased p-n junction is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strongelectricfieldinadepletionregionifthedopingconcentrationissmall
(d) strong electric field in the depletion region if the doping concentration is large
Solution: (a, d)
Reverse biasing: Positive terminal of the battery is connected to the N-crystal and negative terminal of the battery is connected to P-crystal.

(i) In reverse biasing width of depletion layer increases
(ii) In reverse biasing resistance offered R_Reverse = 10⁵ Ω
(iii) Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority carriers.
(A very small reverse current may exist in the circuit due to the drifting of minority carriers across the junction)
(iv) Break down voltage: Reverse voltage at which break down of semiconductor occurs. For Ge it is 25 V and for Si it is 35 V.

So, we conclude that in reverse biasing, ionization takes place because the minority charge carriers will be accelerated due to reverse biasing and striking with atoms which in turn cause secondary electrons and thus more number of charge carriers.
When doping concentration is large, there will be a large number of ions in the depletion region, which will give rise to a strong electric field.

Very Short Answer Type Questions

Question 17. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Solution: When pure semiconductor material is mixed with small amounts of certain specific impurities with valency different from that of the parent material, the number of mobile electrons/holes drastically changes. The process of addition of impurity is called doping. The size of the dopant atom should be compatible such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Silicon or Germanium atoms, which are provided by group XIII or group XV elements.

Question 18. Sn, C and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?
Solution: The conduction level of any element depends on the energy gap between its conduction band and valence band.
In conductors, there is no energy gap between conduction band and valence band. For insulator, the energy gap is large and for semiconductor the energy gap is moderate.
The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV and for Ge is 0.7 eV related to their atomic size. Therefore Sn is a conductor, C is an insulator, and Ge and Si are semiconductors

Question 19. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Solution: We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of voltmeter is very high as compared to the junction resistance. Potential of potential barrier for Ge is V_B = 0.3 V and for silicon is V_B = 0.7 V.
On the average the potential barrier in P-N junction is ~0.5 V.

Question 20. Draw the output waveform across the resistor in the given figure.

Solution: The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased. So the output is obtained only when positive input is given,so the output waveform is

Question 21. The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10,20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value).
(i) if DC supply voltage is 10 V?
(ii) if DC supply voltage is 5 V?
Solution: Total voltage amplification is defined as the ratio of output signal voltage and input signal voltage.

Question 22. In a CE transistor amplifier, there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?
Solution:

The power gain is very high in CE transistor amplifier. In this circuit, the extra power required for amplified output is obtained from DC source. Thus, the circuit used does not violate the law of conservation.

Short Answer Type Questions

Question 23. (i) Name the type of a diode whose characteristics are shown in figure (a) and (b).
(ii) What does the point P in fig. (a) represent?
(iii) What does the points P and Q in fig. (b) represent?

Solution:
(i) Fig. (a) represents the characteristics of Zener diode and curve (b) is of solar cell.
(ii) In fig. (a), point P represents Zener breakdown voltage.
(iii) In fig. (b), the point Q represents zero voltage and negative current. Which means the light falling on solar cell with atleast minimum threshold frequency gives the current in opposite direction to that due to a battery connected to solar cell. But for the point Q the battery is short circuited. Hence it represents the short circuit current.
And the point Pin fig. (b) represents some open circuit, voltage on solar cell with zero current through solar cell.
It means, there is a battery connected to a solar cell which gives rise to the equal and opposite current to that in solar cell by virtue of light falling on it.

Question 24. Three photo diodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000 Å ?
Solution:Key concept: In Photo diodes electron and hole pairs are created by junction photoelectric effect. That is the covalent bonds are broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals.

The incident radiation which is detected by the photodiode D₂ because energy of incident radiation is greater than the band-gap.

Question 25. If the resistance R₁ is increased (see figure), how will the readings of the ammeter and voltmeter change?

Solution: Let us redrawn the circuit diagram to find the change in reading of ammeter and voltmeter.

So, R₁ is increased, I_B is decreased.
Now, the current in ammeter is collector current I_C.
I_C =βI_B as I_B is decreased, I_C is also decreased and the reading of voltmeter and ammeter also decreased.

Question 26. Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Draw a circuit that resembles this situation using diodes for this situation.
Solution: As car enters in either of the garages or both, the common gate opened automatically.
This means that if any one input is high, output will high otherwise low.
The device is shown like this:

Question 27. How would you set up a circuit to obtain NOT gate using a transistor?
Solution:
(1)It has only one input and only one output.
(2)Boolean expression is Y = Ᾱ and is read as “y equals not A” .
Logical symbol of NOT gate.

(3)Realization of NOT gate: The transistor is so biased that the collector voltage V_CC = V (Voltage corresponding to 1 state)
The resistors R and R_B are so chosen that if the input is low, i.e. 0, the transistor is in the cut off and hence the voltage appearing at the output will be the same as applied V = 5 V. Hence Y= V(or state I)
If the input is high, the transistor current is in saturation and the net voltage at the output Y is 0 (in state 0).

Question 28. Explain why elemental semiconductor cannot be used to make visible LEDs.
Solution:

Question 29. Write the truth table for the circuit shown in figure given below. Name the gate that the circuit resembles.

Solution:

Question 30.

Solution:

Long Answer Type Questions

Question 31. If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents I₁, I₂, I₃ and I₄?

Solution:

Question 32. In the circuit shown in figure, when the input voltage of the base resistance is 10 V, V_BE is zero and V_CE is also zero. Find the values of I_B, I_C and β .

Solution:

Question 33. Draw the output signals C₁ and C₂ in the given combination of gates.

Solution:

Question 34. Consider the circuit arrangement shown in figure for studying input and output characteristics of n-p-n transistor in.CE configuration.
Select the values of R_B and R_C for a transistor whose V_BE = 0.7 V so that the transistor is operating at point Q as shown in the characteristics (see figure).

Solution:

Question 35. Assuming the ideal diode, draw the output waveform for the circuit given in figure, explain the waveform.

Solution:
Key concept: An ideal diode is a diode in which it has a very large resistance in reverse biased and very, low resistance in forward biased. So, it acts like a perfect conductor when voltage is applied forward biased and like a perfect insulator when voltage is applied reverse biased.
In reverse biased when the input voltage is equal to or less than 5 V diode,then it will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.
If input voltage is greater than +5 V, diode is in conducting state, then it will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.
When input voltage is negative, there will be opposition to 5 V battery in p-n junction input voltage becomes more than -5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals.
The output waveform will be like this (as shown below).

Question 36. Suppose a n-type wafer is‘created by doping Si crystal having 5 x 10²⁸ atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm boron is added to create ‘p’ region in this wafer. Considering n_i = 1.5 x 10¹⁶ m^-3,
(i) Calculate the densities of the charge carriers in the n and p regions,
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
Solution: n-type wafer is created when As is implanted in Si crystal. The number of majority carriers electrons due to doping of As is

Question 37. An XOR gate has the following truth table.

It is represented by following logic relation Y = Ᾱ . B + A . B’ . Build this gate using AND, OR and NOT gates.
Solution: XOR gate can be realized by the combination of two NOT gates, two AND gates and one OR gate. According to the problem, the logic relation for the . given truth table is

Question 38. Consider a box with three terminals on top of it as shown in figure.

Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit as shown in figure.
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit. The graphs are
(i) when A is positive and B is negative .

Solution: The V-I characteristics of these graph is discussed in points:
(a)In V-I graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n-side of p-n junction I and B is connected top-side of p-n junction I with a resistance in series.
(b)In V-I graph of condition (ii), a forward characteristics is shown in fig. (d), where 0.7 V is the knee voltage of p-n junction I. 1/slope = (1/1000) Ω.
It means A is connected to n-side of p-n junction I and B is connected to p-side of p-n junction I and resistance R is in series of p-n junction I between A and B.
(c)In V-I graph of condition (iii), a forward characteristics is shown in figure (e) , where 0.7 V is the knee voltage. In this case p-side of p-n junction II is connected to C and n-side of p-n junction II to B.
(d)In V-I graphs of conditions (iv), (v), (vi) also concludes the above connection of p-n junctions I and II along with a resistance R.
Thus, the arrangement of p-n I, p-n II and resistance R between A, B and C will be as shown in the figure.

Question 39. For the transistor circuit shown in figure, evaluate V_E, R_B, R_E, given I_C= 1 mA, V_CE= 3, V_BE = 0.5 V and V_CC= 12 V, β= 100.

Solution: Let us redraw the circuit diagram given here to solve this problem.

Question 40. In the circuit shown in figure, find the value of R_{c .}

.

Solution:
March 20, 2018
Nuclei NCERT Exemplar Problems Solutions Physics
NCERT Exemplar Problems Class 12 Physics Chapter 13 Nuclei

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year,
(a) all the containers will have 5000 atoms of the material
(b) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000
(c) the containers will in general have different numbers of the atoms of the material but their average will be close to 5000
(d) none of the containers can have more than 5000 atoms
Solution: (c)
Key concept: Half life ( T_1/2):
Radioactivity is a process due to which a radioactive material spontaneously decays. Time interval in which the mass of a radioactive substance or the number of its atom reduces to half of its initial value is called the half life of the substance.

In half-life (t= 1 yr) of the material on the average half the number of atoms will decay. Therefore, the containers will in general have different number of atoms of the material, but their average will be approx 5000.

Question 2. The gravitational force between a H-atom and another particle of mass m will

Solution: (b)
Key concept: The gravitational force between a H-atom and another
particle of mass m will be given by Newton’s law, F = G M.m/r²
Here M is the effective mass of Hydrogen atom.
Let us learn how to find the effective mass of a Hydrogen atom.
Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just m_proton + m_electron.
Now let the proton and electron fall towards each other under their mutual electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they are moving with a high speed. Note that we haven’t added or removed any energy, so the mass/energy of the system is still m_proton + m_electron.
The trouble is that this would not form a hydrogen atom because the proton and electron will just speed past each other and fly away again. To form a hydrogen atom we have to take the kinetic energy of the electron and proton out of the system so we can bring them to a stop. Let’s call the kinetic energy E_k. This energy has a mass given by Einstein’s famous equation E = mc², so die mass of our atom is the mass we started with less the energy we’ve taken out:

Question 3. When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom
(a) do not change for any type of radioactivity
(b) change for α and β-radioactivity but not for γ-radioactivity
(c) change for α -radioactivity but not for others
(d) change for β-radioactivity but not for others
Solution: (b)
Key concept:

A /3-particle carries one unit of negative charge (-e), an α-particle carries 2 units of positive charge (+ 2e ) and γ (particle) carries no charge. Hence electronic energy levels of the atom charges for α and β decay, but not for γ-decay.

Question 4. M_x and M_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β^–decay is Q₁ and that for a β⁺decay is Q₂. If me denotes the mass of an electron, then which of the following statements is correct?

Solution: (a)
Key concept: Q value or energy of nuclear reaction: The energy absorbed or released during a nuclear reaction is known as Q-value of nuclear reaction.
Q-value = (Mass of reactants – mass of products)c² Joules
= (Mass of reactants – mass of products) amu
If Q < 0, the nuclear reaction is known as endothermic. (The energy is absorbed in the reaction)
If Q > 0, the nuclear reaction is known as exothermic. (The energy is released in the reaction)

Question 5. Tritium is an isotope of hydrogen whose nucleus triton contains 2 neutrons and 1 proton. Free neutrons decay into p + e + n . If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because
(a) Triton energy is less than that of a He³nucleus
(b) The electron created in the beta decay process cannot remain in the nucleus
(c) Both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He³ nucleus.
(d) Free neutrons decay due to external perturbations which is absent in triton nucleus .
Solution: (a)

Question 6. Heavy stable nuclei have more neutrons than protons. This is because of the fact that .
(a) neutrons are heavier than protons
(b) electrostatic foree between protons are repulsive
(c) neutrons decay into protons through beta decay
(d) nuclear forces between neutrons are weaker than that between protons
Solution: (b)

Question 7. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose, because
(a) they will break up
(b) elastic collision of neutrons with heavy nuclei will not slow them down
(c) the net weight of the reactor would be unbearably high
(d) substances with heavy nuclei do not occur in liquid or gaseous state at room temperature
Solution: (b)
Key concept: A moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission. By slowing the neutrons down the probability of a neutron interacting with Uranium-235 nuclei is greatly increased thereby maintaining the chain reaction. Moderators are made from materials with light nuclei which do not absorb the neutrons but rather slow them down by a series of collisions.
The moderator only slows neutrons down in order to increase the interaction with Uranium nuclei. They do not give any protection if the reaction goes out of control. 1 fa chain reaction is heading out of control the reactors needs to be able to reduce the concentration of neutrons. For this the reactor uses control rods. Control rods are matte from material with the ability to absorb neutrons. Cadmium and Boron are examples of suitable materials. By inserting.control rods between the fuel rods the chain reaction can be slowed dowp-or shut down. Withdrawing the control rods can restart or speed up the reaction.
In our given question, the moderator used have light nuclei (like proton). When protons undergo perfectly elastic collision with the neutron emitted their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons.
Heavy nuclei will not serve the purpose because elastic collisions of neutrons with heavy nuclei will not slow them down.

One or More Than One Correct Answer Type

Question 8. Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact
(a) nuclear forces have short range
(b) nuclei are positively charged
(c) the original nuclei must be completely ionized before fusion can take place
(d) the original nuclei must first break up before combining with each other
Solution: (a, b)
Key coneept:
Nuclear Fusion: In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of a single nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in the release of tremendous amount of energy To achieve fusion, you need to create special conditions to overcome this tendency.
Here are the conditions that make fusion possible:
High Temperature: The high temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons.
• Fusion requires temperatures about 100 million Kelvin (approximately six times hotter titan the sun’s core).
• At these temperatures, hydrogen is a plasma, not a gas. Plasma is a high-energy state of matter in which all the electrons are stripped from atoms and move freely about.
• The sun achieves these temperatures by its large mass and the force of gravity compressing this mass in the core. We must use energy from microwaves, lasers and ion particles to achieve these temperatures.
High pressure: Pressure squeezes the hydrogen atoms together. They must be within 1 x 10^-15 metres of each other to fuse.
• The sun uses its mass and the force of gravity to squeeze hydrogen atoms together in its core.
• We must squeeze hydrogen atoms together by using intense magnetic fields, powerful lasers or ion beams.
Fusion processes are impossible at ordinary temperatures and pressures. The reason is that nuclei are positively charged and nuclear forces are short range strongest forces. In order to force two hydrogen nuclei together, we need to have a very high pressure, or a very high temperature, or both. A high pressure helps because it causes all the hydrogen nuclei in the sun to squeeze into a smaller space. Then there is more chance of one hydrogen bumping into another. A high temperature helps because it makes the hydrogen nuclei move faster. They need this extra speed so that they can get close together and join. It is as if the nucleus has to break through a barrier, and so the faster it is moving, the greater chance it has.
So, at the “normal” temperature and pressure on earth, a hydrogen nucleus has basically no chance of ever joining with another hydrogen nucleus.
Important point: We know that in the middle of the sun, where the temperature is about 16 million degrees, and the pressure is 250 billion atmospheres, hydrogen nuclei will sometimes have enough energy to join together. (An atmosphere is the “normal”, pressure of the air here on earth. A pressure of 250 billion atmospheres is like having a large mountain piled on top of you!)

Question 9. Samples of two radioactive nuclides A and B are taken. λ_A and λ_B are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?
(a) Initial rate of decay of A is twice the initial rate of decay of B and λ_A = λ_B
(b) Initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B
(c) Initial rate of decay of B is twice the initial rate of decay of A and λ_A> λ_B
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λ_A< λ_B
Solution: (b, d)
Key concept:
Law of radioactive disintegration : According to Rutherford and Soddy law for radioactive decay is as follows:
“At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant.” i.e.
dN/dt ∞ N => dN/dt = -λN
it can be proved that N=N₀e^-λ1
In terms of mass M— M₀e^-λ1
where N = Number of atoms remains undecayed after time t,
N₀ = Number of atoms present initially (i.e., at t = 0),
M = Mass of radioactive nuclei at time t,
M₀ = Mass ofradioactive nuclei at time t = 0,
N₀-N= Number of disintegrated nucleus in time t,
dN/dt= rate of decay, λ = Decay constant or disintegration constant or radioactivity constant or Rutherford Soddy’s constant or the probability of decay per unit time of a nucleus.
The samples of the two radioactive nuclides A and B can simultaneously have the same decay rate at any time if initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B.
Also, when initial rate of decay of B is the same as rate of decay of A at t = 2h and λ_B < λ_A.

Question 10.

Solution:

Hence at point P, rate of decay for both A and B is the same.

Very Short Answer Type Questions

Question 11. He₂³ and He₁³ nuclei have the same mass number. Do they have the same binding energy?
Solution: The nuclei He₂³ and He₁³ have the same mass number. He₂³ has two protons and one neutron. He₂³ has one proton and two neutrons. As He3 has only one proton hence the repulsive force between protons is missing in ₁He³, so the binding energy of ₁He³ is greater than that of ₂He³.

Question 12. Draw a graph showing the variation of decay rate with number of active nuclei.
Solution:

Question 13. Which sample AoxB shown in figure has shorter mean-life?

Solution:
Key concept:
Mean (or average) life (ґ) : The time for which a radioactive material remains active is defined as mean (average) life of that material.
• It is defined as the sum of lives of all atoms divided by the total number of atoms.

Question 14. Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.
Solution:
Key concept: The energy of internal motion of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typically of the order of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by emission of one or more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. This proceks is called gamma (γ) decay.
Excited electron cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV ( mega electron volt), y-radiations have energy of the order of MeV.

Question 15. In pair annihilation, an electron and a positron destroy each other to produce gamma radiations. How is the momentum conserved?
Solution: In pair annihilation, an electron and a positron destroy each other to produce 2yphotons which move in.opposite directions to conserve linear momentum. The annihilation is shown below:

Short Answer Type Questions

Question 16. Why do stable nuclei never have more protons than neutrons?
Solution: The reason is that protons, being charged particles, repel each other. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.
Important point: As you get to heavier elements, with each new proton you add, there is a larger repulsive force. The nuclear force is attractive and stronger than the electrostatic force, but it has a finite range. So you need to add extra neutrons, which do not repel each other, to add extra attractive force. You eventually reach a point where the nucleus is just too big, and tends to decay via alpha decay or spontaneous fission.
To view this in quantum mechanical terms, the proton potential well is not as deep as the neutron well due to the electrostatic repulsion. [Due to the Pauli exclusion principle, you only get two particles per level (spin up and spin down)]. If one well is filled higher than the other, you tend to get a beta decay to even them out. As the nuclei get larger, the neutron well gels deeper as compared to the proton well and you get more neutrons than protons.

Question 17. Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence:
A -> B -> C
Here B is an intermediate nuclei which is also radioactive. Considering that there are N₀ atoms of A initially, plot the graph showing the variation of number of atoms of A and B versus time.
Solution: Consider radioactive nucleus A have N₀ atoms of A initially; or at t = 0, N_A = N₀ (maximum) whole N_B = 0. As time increases, N_A decreases exponentially and the number of atoms of B increases. After some time N_B becomes maximum. As B is an intermediate nuclei which is also radioactive, it also start decaying and finally drop to zero exponentially by radioactive decay law. We can represent the situation as shown in the graph.

Question 18. A piece of wood from the ruins of an ancient building was found to have a ¹⁴C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ¹⁴C is 5760 yr.
Solution:
Key concept:
Carbon dating: Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method tor determining the age of an object containing organic material by using the properties of radiocarbon -14(¹⁴C), a radioactive isotope of carbon.
Radiocarbon, or carbon 14, is an isotope of the element carbon that is unstable and weakly radioactive. The stable isotopes are carbon 12 and carbon 13.
Carbon 14 is continually being formed in the upper atmosphere by the effect of cosmic ray neutrons on nitrogen 14 atoms. It is rapidly oxidized in air to form carbon dioxide and enters the global carbon cycle.
Plants and animals assimilate carbon 14 from carbon dioxide throughout their lifetimes. When they die, they stop exchanging carbon with the biosphere and their carbon 14 content then starts to decrease at a rate determined by the law of radioactive decay.
Radiocarbon dating is essentially a method designed to measure residual radioactivity. By knowing how much carbon 14 is left in a sample, the age of the organism when it died can be known. It must be noted though that radiocarbon dating results indicate when the organism was alive but not when a material from that organism was used.

Question 19. Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10^-15 m.
Solution:
Key concept: A nucleon is one of the particles that makes up the atomic nucleus. Each atomic nucleus consists of one or more nucleons, and each atom in turn consists of a cluster of nucleons surrounded by one of more electrons. There are two known kinds of nucleon: the neutron and the proton. The mass number of a given atomic isotope is identical to its number of nucleons. Thus the term nucleon number may be used in place of the more common terms mass number or atomic mass number.
For resolving two objects separated by distance d, the wavelength A of the proving signal must be less than d. Therefore, to detect separate parts inside a nucleon, the electron must have a wavelength less than 10^-15 m.

Important point: Until the 1960s, nucleons were thought to be elementary particles, each of which would not then have been made up of smaller parts. Now they are known to be composite particles, made of three quarks bound together by the so-called strong interaction. The interaction between two or more nucleons is called intemucleon interactions or nuclear force, which is also ultimately caused by the strong interaction. (Before the discovery of quarks, the term “strong interaction” referred to just intemucleon interactions.)

Question 20. A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z₁=N₂ and Z₂ = N₁ (a) What nuclide is a mirror isobar of ₁₁²³Na ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?
Solution:
Key concept: Mirror nuclei are nuclei where the number of protons of element one (Z₁) equals the number of neutrons of element two (N₂), the number of protons of element two (Z₂) equal the number of neutrons in element one (N₁) and the mass number is the same.
Pairs of mirror nuclei have the same spin and parity. If we constrain to odd number of nuclcons(A), then w find mirror nuclei that differ one another by exchanging a proton by a neutron. Interesting to observe is their binding energy which is mainly due to the strong interaction and also due to Coulomb interaction. Since the strong interaction is invariant to protons and neutrons one can expect these mirror nuclei to have very similar binding energies.

Long Answer Type Questions

Question 21. Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is

Assume that we start with 1000 ³⁸S nuclei at time t = 0. The number of ³⁸Cl is of count zero at t = 0 and will again be zero at t =∞, At what value of t, would the number of counts be a maximum?
Solution:

Question 22. Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen.
Solution: Given the binding energy of a deuteron, B = 2.2 MeV Let kinetic energy and momentum of neutron and proton be K_n, K_P and p_n, p_p respectively.
From conservation of energy,

Question 23. The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in a deuteron as given in the form a coulomb potential but with an effective charge e’.

Solution:

Question 24. Before the neutrino hypothesis, the beta decay process was thought to be the transition.
n —> p + e
If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.
Solution:

Question 25. The activity R of an unknown radioactive nuclide is measured at hourly intervals. The result found are tabulated as follows:

Solution:

Question 26. Nuclei with magic number of proton Z = 2, 8, 20, 28, 50, 52 and magic number of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable.
(i) Verify this by calculating the proton, separation energy S_P for ¹²⁰Sn (Z= 50) and ¹²¹Sb(Z= 51).
The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by

Solution:

Important point: “Magic Numbers” in Nuclear Structure Careful observation of the nuclear properties of elements, showed certain patterns that seemed to change abruptly at specific key elements. Mayer noticed that magic numbers applied whether one counts the number of neutrons (A). the atomic number (Z), or the sum of the two, known as mass number (A). I-xamples are Helium Z = 2, Lead Z = 82, Helium A = 2, Oxygen N = 8, Lead A = 126, Neon A = 20, Silicon A = 28.
Magic numbers in the nuclear structure have been coming up during all this time, but no plausible explanation for their existence has ever been given. Interestingly, there are peaks and dips for binding energy, repeating every fourth nucleon. This periodicity is one clear indication of a geometrical structure within the nucleus. In particular, those nuclei that can be thought of as containing an exact number of alpha particles (2P + 2A), are more tightly bound than their neighbours. This effect is more pronounced for the lightest nuclei, but is still perceptible up to A – 28. For those nuclei with A > 20, the number of neutrons exceeds the number of protons, so some sort of distortion occurs wi thin the cluster.
It is found that nuclei with even numbers of protons and neutrons are more stable than those with odd numbers. This comes from the fact that the physical structure must have an even number of vertices. A type of regular polyhedron would satisfy this condition, since no regular polyhedron exists with an odd number of vertices. These specific “magic numbers” of neutrons or protons which seem to be particularly favoured in terms of nuclear stability are:
2, 8. 20. 28, 50, 82, 126
. Note that the structure must apply to both protons and neutrons individually, so that we can speak of “magic nuclei” where any one nucleon type, or their sum, is at a magic number. .
The existence of these magic numbers suggests closed shell configurations, like the shells in atomic structure. They represent one line of reasoning which led to the development of a shell model of the nucleus. Other forms – of evidence suggesting shell structure include the following.
1. Enhanced abundance of those elements for which Z or N is a magic number.
2. The stable elements at the end of the naturally occurring radioactive series all have a “magic number” of neutrons or protons.
3. The neutron absorption cross-sections for isotopes where N = magic number are much lower than surrounding isotopes.
4. The binding energy for the last neutron is a maximum for a magic neutron number and drops sharply for the next neutron added.
5. Electric quadrupole moments are near zero for magic number nuclei.
6. The excitation energy from the ground nuclear state to the first excited state is greater for closed shells.
Visualizing the densely packed nucleus in terms of orbits and shells seems much less plausible than the corresponding shell model for atomic electrons. You can easily believe that an atomic electron can complete many orbits without running into anything, but you expect protons and neutrons in a nucleus to be in a continuous process of collision with each other. But dense-gas type models of nuclei with multiple collisions between particles didn’t fit the data, and remarkable patterns like the “magic numbers” in the stability of nuclei suggested the seemingly improbable shell structure.
March 20, 2018
Atoms NCERT Exemplar Problems Solutions Physics

NCERT Exemplar Problems Class 12 Physics Chapter 12 Atoms

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Taking the Bohr radius as a₀= 53 pm, the radius of Li⁺⁺ion in its ground state, on the basis of Bohr’s model, will be about
(a) 53 pm (b) 27 pm (c) 18 pm (d) 13 pm
Solution: (c)

Question 2. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is

Solution: (c) In a hydrogen atom, electron revolving around a fixed proton nucleus have some centripetal acceleration. Therefore its frame of reference is non- inertial. If the frame of reference, where the electron is at rest, the given expression is not true as it forms the non-inertial frame of reference.

Question 3. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with man electrons. This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law
Solution: (a) The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons because when we derive the formula for radius/energy levels etc, we make the assumption that centripetal force is provided only by electrostatic force of attraction by the nucleus. Hence, this will only work for single electron atoms. In multi-electron atoms, there will also be repulsion due to other electrons. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons.

Question 4. For the ground state, the electron in the H-atom has an angular .momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) because Bohr model gives incorrect values of angular momentum
(b) because only one of these Would Have a minimum energy
(c) angular momentum must be in the direction of spin of electron
(d) because electrons go around only in horizontal orbits
Solution: (a)

Bohr’s model gives only the magnitude of angular momentum. The angular momentum is a vector quantity. Hence we cannot express angular momentum completely by Bohr model. Hence it does not give correct values of angular momentum of revolving electron.

Question 5. 02 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short-ranged
(b) is as important as electrostatic force for binding the two atoms
(c) cancels the repulsive electrostatic force between the nuclei
(d) is not important because oxygen nucleus have equal number of neutrons and protons
Solution: (a)
Key concept: Forces that keep the nucleons bound in the nucleus are called nuclear forces.
• Nuclear forces are short range forces. These do not exist at large distances greater than 10~15 m.
• Nuclear forces are the strongest forces in nature.
• These are attractive force and causes stability of the nucleus.
• These forces are charge independent.
• Nuclear forces arc non-central force.
The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. .
In O₂molecule which consists of two oxygen atoms molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged and act inside the nucleus only.

Question 6. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(a) 10.20 eV (b) 20.40eV (c) 13.6 eV
Solution: (a)

Question 7. A set of atoms in an excited state decays
(a) in general to any of the states with lower energy
(b) into a lower state only when excited by an external electric field
(c) all together simultaneously into a lower state
(d) to emit photons only when they collide .
Solution: (a)

One or More Than One Correct Answer Type

Question 8. An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,
(a) the electron would not move in circular orbits
(b) the energy would be (2)⁴times that of a H-atom
(c) the electrons, orbit would go around the protons
(d) the molecule will soon decay in a proton and a H-atom
Solution: (a, c) In a hydrogen atom, electron revolves around a fixed proton nucleus in circular path. This can be explained by Bohr model. But in case of ionised H-molecule which consists of two protons in nucleus and where protons are separated by a small distance of the order of angstrom, cannot be explained by Bohr model. Hence in this case the ground state the electron would not move in circular orbits, the electrons orbit would go around the protons.

Question 9. Consider aiming a beam of free electrons towards free protons. When-they scatter, an electron and a proton cannot combine to produce a H-atom.
(a) Because of energy conservation
(b) Without simultaneously releasing energy in the form of radiation
(c) Because of momentum conservation
(d) Because of angular momentum conservation
Solution: (a, b) A moving electron and proton cannot combine to produce a H-atom because of energy conservation and without simultaneously releasing energy in the form of radiation.

Question 10. The Bohr model for the spectra of a H-atom
(a) will not be applicable to hydrogen in the molecular form
(b) will not be applicable as it is for a He-atom
(c) is valid only at room temperature
(d) predicts continuous as well as discrete spectral lines
Solution: (a, b) Bohr proposed a-model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom, e.g.: H, He⁺, Li⁺², Na⁺¹ etc). It is not applicable to hydrogen in the molecular form and also, it will not be applicable as it is for a He-atom.

Question 11. The Balmer series for the H-atom can be observed
(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) if we measure the frequencies of light emitted due to transitions between
excited states and the first excited state
(c) in any transition in a H-atom .
(d) as a sequence of frequencies with the higher frequencies getting closely
Solution: (b, d)

From above discussion we can say Balmer series for the H-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the’higher frequencies getting closely packed.

Question 12.

Solution: (b, d) Let E₂ and E₁be the energy corresponding to n = 2 and n = 1 respectively. If radiation of energy ∆E = (E₂ – E₁) = hf incident on a sample where all the H-atoms are in the ground state, according to Bohr model some of the atoms will move to the first excited state. As this energy is not sufficient for transition from n = 1 to n =3, hence no atoms will make a transition to the n = 3 state.

Question 13. The simple Bohr model is not applicable to He⁴ atom because
(a) He⁴ is an inert gas
(b) He⁴ has neutrons in the nucleus
(c) He⁴ has one more electron
(d) electrons are not subject to central forces
Solution: (c, d)
Key concept: Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom). It is valid only for one electron atoms, e.g. : H, He⁺, Li⁺², Na⁺¹ etc.
The Bohr model is not applicable to He⁴, as it has one more electron and electrons are not subjected to central forces.

Very Short Answer Type Questions

Question 14. The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this?
Solution:

Question 15. Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.
Solution: Bohr model is applicable for hydrogen atom-and some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge Ze (called hydrogen like atom). If we remove one electron from He⁴ and He³, atoms contain one electron and becomes hydrogen like atoms. Now we can apply Bohr model to define the energy levels.

Question 16. When an electron falls from a higher energy to a lower energy level, the
difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? .
Solution: The electrons are charged particles. When an electron falls from a higher energy to a lower energy level, it accelerates. We know accelerating charged particle radiates energy in the form of electromagnetic radiation.

Question 17. Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+ 4/3)e and electron a charge (-3/4)e, where e = 1.6 x 10^-19C. Give reasons for your answer.
Solution: According to Bohr, for an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force.

Hence the magnitude of electrostatic force F ∞ q₁x q₂
If proton had a charge (+4/3)e and electron a charge (-3/4)e, then the Bohr formula for the H-atom remains same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

Question 18. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?
Solution: Bohr postulated that an electron in an atom can move around the nucleus in a certain circular stable orbits without emitting radiations.
Bohr found that the magnitude of the electron’s angular momentum is

Short Answer Type Questions

Question 19. Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?
Solution:
Key concept: Positronium (Ps) is a system consisting of an electron and its anti-particle a positron, bound together into an exotic atom, specifically anonium. The system is unstable: the two particles annihilate each other to predominantly produce two or three gamma-rays, depending on the relative spin states. The orbit and energy levels of the two particles arc similar to that of the hydrogen atom (which is a bound slate of a proton and an electron). I lowever, because of the reduced mass, the frequencies of the spectral lines are less than half of the corresponding hydrogen lines.

Question 20. Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
Solution:

Question 21. Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.
Solution:

Question 22. Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than n, are approximate harmonics (i.e., in the ratio 1:2:3 …) when n >> 1.
Solution:

Question 23. What is the minimum energy that must be given to a H-atom in ground state so that it can emit an H_ґ Line in Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such H_ґ photon?
Solution:

Long Answer Type Questions

Question 24. The first four spectral in the Lyman series of a H-atom are λ= 1218 Å, 1028 Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider deuterium, calculate the shift in the wavelength of these lines.
Solution: Let µ_H and µ_D are the reduced masses of electron for hydrogen and deuterium respectively.

Question 25. Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ²H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre bf mass.
Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ=m_eMI(m_e + M), wfiere M is the nuclear mass and m_e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in ‘H and ²H
(Mass of ¹H nucleus is 1.6725 x 10^-27 kg, mass of ²H nucleus is 3.3374 x 10^-27 kg, Mass of electron = 9.109 x 10^-31 kg.)
Solution:

Question 26. If a proton had a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R = 0.1 Å and (ii) R = 10 Å .
Solution: In a H-atom in ground state, electron revolves round the point-size proton in a circular orbit of radius r_B (Bohr’s radius).

Question 27. In the Auger process, an atom maizes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n – 2 to n = 1 transition.
Solution:
Key concept:
Auger Effect: The Auger effect is a process by which electrons with characteristic energies are ejected from atoms in response to a downward transition by another electron in the atom. In Auger spectroscopy, the vacancy is produced by bombardment with high energy electrons, but the Auger effect can occur if the vacancy is produced by other interactions. It is observed as one of the methods of electron rearrangement after electron capture into the nucleus.
If an inner shell electron is removed from an atom, an electron from a higher level will quickly make the transition downward to fill the vacancy. Sometimes this transition will be accompanied by an emitted photon whose quantum energy matches the energy gap between the upper and lower level. Since for heavy atoms this quantum energy will be in the x-ray region, it is commonly called x.-ray fluorescence. This emission process for lighter atoms and outer electrons gives rise to line spectra.
In other cases, the energy released by the downward transition is given to one of the outer electrons instead of to a photon, and this electron is then ejected from the atom with an energy equal to. the energy lost by the electron which made the downward transition minus the binding energy of the electron that is ejected from the atom. Though more involved in interpretation than optical spectra, the analysis of the energy spectrum of these emitted electrons does give information about die atomic energy levels. The Auger effect bears some resemblance to internal conversion of the nucleus, which also ejects an electron
Sometimes an upper election drops to fill the vacancy, emitting a photon.

As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model.

Question 28.

Solution:

Question 29. The Bohr model for the H-atom relies on the Coulomb ’s law of electrostatics. Coulomb’s law has not directly been verified for very
short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge +q₁, – q₂ modified to

Solution:

March 20, 2018
Dual Nature of Radiation and Matter NCERT Exemplar Problems Solutions Physics

NCERT Exemplar Problems Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to (a) H (b) H^1/2 (c) H⁰ (d) H^-1/2
Solution: (d)
Key concept: According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
The wave associated with a moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de-Broglie wave is given by

Question 2. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10^-3nm
(c) 1.2 x 10^-6 nm (d). 1.2 x 10 nm
Solution: (b)
Key concept: According to Einstein’s quantum theory light propagates in the bundles (packets or quanta) of energy, each bundle being called a photon and possessing energy.
Energy of photon is given by

Question 3. Consider a beam of electrons (each electron with energy E₀) incident on a metal surface kept in an evacuated chamber. Then,
(a) no electrons will be emitted as only photons can emit electrons
(b) electrons can be emitted but all with an energy, E₀
(c) electrons can be emitted with any energy, with a maximum of E₀ – ɸ (ɸ is the work function)
(d) electron can be omitted with energy ,with a maximum of E₀
Solution: (d) If a beam of electrons of having energy E₀ is incident on a metal surface kept in an evacuated chamber.

The electrons can be emitted with maximum energy E₀ (due to elastic collision) and With any energy less than E₀, when part of incident energy of electron is used in liberating the electrons from the surface of metal.

Question 4. Consider the figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that (a) will be larger than the earlier value
(b) will be the same as the earlier value
(c) will be less than the earlier value
(d) will depend on the target

Solution: (c)
Key concept: Davision and Germer Experiment:
1. It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by an . electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

2. The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it .about the point of incidence. The energy of the incideni beam of electrons can also be varied by changing the applied voltage to the electron gun.
According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering. It is maximum for diffracting angle 50° at 54 volt potential difference.
3. If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula 2d sinθ = nλ, we can determine the wavelength of these waves.
The de-Broglie wavelength associated with electron is where V is the applied voltage.
Using the Bragg’s formula we can determine the wavelength of these waves. If there is a maxima of the, diffracted electrons at an angle θ, then
2d sin θ = A (ii)
From Eq. (i), we note that if V is inversely proportional to the wavelength λ. i.e., V will increase with the decrease, in λ.
From Eq. (ii), we note that wavelength λ is directly proportional to sinθ and hence θ.
So, with the decrease in λ , θ will also decrease.
Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that will be less than the earlier value.

4. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as

Question 5. A proton, a neutron, an electron and an a-particle have same energy. Then, their de-Broglie wavelengths compare as (a) λ_p = λ_n > λ_e > λ_α (b) λ_α < λ_p = λ_n > λ_e
(c) λ_e< λ_p = λn> λ_α (d) λ_e = λ_p = λn = λ_α
Solution: (b)
Key concept:
• Matter Waves (de-Broglie Waves)
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.

Question 6. An electron is moving with an initial velocity v = v₀i and is in a magnetic field B = B₀j. Then, its de-Broglie wavelength
(a) remains constant
(b) increases with time
(c) decreases with time
(d) increases and decreases periodically
Solution: (a)
Key concept: If a particle is carrying a positive charge q and moving with a velocity v enters a magnetic field .5 then it experiences a force F which is given by the expression
F = q(v x B)=$ F = qvB sin θ. As this force is perpendicular to v and B , so the magnitude of v will not change, i.e. momentum (p = mv) will remain constant in magnitude. Hence,

Question 7. An electron (mass m) with an initial velocity v = v₀i(v₀ > 0) is in an electric field E =E₀î (E₀ = constant > 0). Its de-Broglie wavelength at time t is given by

Solution: (a)
Key concept: The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity. According to de-Broglie theory, the wavelength of de- Broglie wave is given by

Question 8. An electron (mass m) with an initial velocity v =v₀î is in an electric field

Solution: (c) According to the problem de-Broglie wavelength of electron at time t=0.

One or More Than One Correct Answer Type

Question 9. Relativistic corrections become necessary when the expression for the kinetic
energy 1/2 mv² , becomes comparable with mc². where m is the mass of the
particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?
(a) A=10nm (b) A =10^-1 nm (c) A=10^-4nm (d) A=10^-6nm
Solution: (c, d)
Key concept: De-Brogile or matter wave is independent of die charge on the material particle. It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
The de-Broglie wavelength at which relativistic corrections become important that the phase velocity of the matter waves can be greater than the speed of the light (3 x 10⁸ m/s).
The wavelength of de-Broglie wave is given by
λ = h/p = h/mv
Here, h = 6.6 x 10^-34 Js
and for electron, m = 9 x 10^-31 kg
To approach these types of problem we use hit and trial method by picking up each option one by one.

Question 10. Two particles A₁ and A₂ of masses m₁ , m₂ ( m₁> m₂) have the same de-Broglie
wavelength. Then,
(a) their momenta are the same (b) their energies are the same
(c) energy of A₁ is less than the energy of A₂
(d) energy of A₁ is more than the energy of A₂
Solution: (a. c)

Question 11. De-Broglie wavelength associated with uncharged particles: For Neutron efe-Broglie wavelength is given as v_e=c/100.Then

Solution:(b,c)

Question 12. Photons absorbed in a matter are converted to heat. A source emitting v photon/sec of frequency v is used to convert 1 kg of ice at 0°C to water at 0°C. Then, the time T taken for the conversion
(a) decreases with increasing n. with v fixed
(b) decreases with n fixed, v increasing
(c) remains constant with n and v changing such that nv = constant
(d) increases when the product nv increases
Solution: (a, b. c).

Question 13. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-BrOglie wavelength of the particle varies cyclically between two values λ₁, λ₂with λ₁> λ₂ Which of the following statements are true?
(a) The particle could be moving in a circular orbit with origin as centre
(b) The particle could be moving in an elliptic orbit with origin as its focus
(c) When the de-Broglie wavetength is λ₁ the particle is nearer the origin than when its value is λ₂
(d) When the de-Broglie wavelength is λ₂ the particle is nearer the origin than when its value is λ₁
Solution:

Very Short Answer Type Questions

Question 14. Aproton and an a-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λ_pand λ_α related to each other?
Solution:

Question 15. (i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy E_max of the emitted electron as E_max = hv – ɸ₀
where ɸ₀ is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
Solution:

Question 16. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength?
Solution: In the first case, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is low when it has a longer wavelength or in short we can say that energy given out is less than the energy supplied.
But in second case, the energy of the incident photon is low for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light Of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.
But this is not possible for a stable substances.

Question 17. Do all the electrons that absorb a photon come out as photo electrons?
Solution:
Key concept: Photo-Electric Effect:
The photo-electtic effect is the emission of electrons (called photo-electrons when light strikes a surface. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface.
The photoelectric effect is based on the principle of conservation of energy.
1. Two conducting electrodes, the anode (Q) and cathode (P) are enclosed in an evacuated glass tube as shown on next page.
2. The battery or other source of potential difference creates an electric field in the direction from anode to cathode.
3. Light of certain wavelength or frequency falling on the surface of cathode causes a current in the external circuit called photoelectric current.
4. As potential difference increases, photoelectric current also increases till saturation is reached.

Question 18. There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.
Solution:
Key concept: X-Rays:
1. X-rays were discovered by scientist Roentgen that is why they are also called Roentgen rays.
2. Roentgen discovered that when pressure inside a discharge tube is kept 10“3 mm of Hg and potential difference is kept 25 kV, then some unknown radiations (X-rays) are emitted by anode.
3. There are three essential requirements for the production of X-rays.
(i) A source of electron
(ii) An arrangement to accelerate the electrons
(iii) A target of suitable material of high atomic weight and high melting point on which these high speed electrons strike.

Question 19. Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Solution:The momentum of incident photon is transferred to the metal ,during photo electric emission.
At microscopic level ,atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.The excited electron is emitted.Therefore,the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electron.

Question 20.Consider a metal exposed to light the wavelength of 600 nm. The maximum energy of electron doubles when light of wavelength 400 nm is used.Find the work function in eV?
Solution:

Question 21. Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg uncertainty principle (∆x x∆p=h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p = ∆p, find the energy of the electron in electron volts.
Solution:

Question 22. Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?
Solution:

Question 23. Two particles A and B of de-Broglie wavelengths A, and combine to form a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one-dimensional)
Solution:

Question 24. A neutron beam of energy E scatters from atoms on a surface with a spacing d = 0.1 nm. The first maximum intensity in the reflected beam occurs at θ= 30°. What is the kinetic energy E of the beam in eV?
Solution:

Long Answer Type Questions

Question 25. Consider a thin target (10^-2cm square, 10^-3 m thickness) of sodium, which produces a photocurrent of 100 µA when a light of intensity 100 W/m² (λ = 660 nm) falls on it. Find the probability that a photoelectron is produced when a photon strikes a sodium atom. [Take density of Na = 0.97 kg/m³]
Solution:

Question 26. Consider an electron in front of metallic surface of a distance d.Assume the force of attraction by the plate is given as . Calculate work in taking the to an infinite distance from the plate .Taking d=0.1 nm. find the work done in electron volts?
Solution:

Question 27. A student performs an experiment on photoelectric effect, using two materials A and B. A plot of V_stop versus v is given in figure.

Solution:

Question 28. A particle A with a mass m_A is moving with a velocity v and hits a particle B (mass m_B) at rest (one dimensional motion). Find the change in the de-Broglie wavelength of the particle A. Treat the collision as elastic.
Solution:

Question 29. Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2 m. Assume that the metal surface has work function of 2 eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å.
(i) Estimate number of photons emitted by the bulb per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the atomic disk to receive energy equal to work function (2 eV)?
(iv) How many photons would atomic disk receive within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was observed instantaneously?
Solution:

March 20, 2018
Wave Optics NCERT Exemplar Problems Solutions Physics

NCERT Exemplar Problems Class 12 Physics Chapter 10 Wave Optics

Multiple Choice Questions (MCQs)

Single Correct Answer Type
Question 1. Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in figure.

A Polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.
(a) For a particular orientation, there shall be darkness as observed through the polaroid.
(b) The intensity of light as seen through the polaroid shall be independent of the rotation.
(c) The intensity of light as seen through the polaroid shall go minimum but not zero for two orientations of the polaroid.
(d) The intensity of light as seen through the polaroid shall go minimum for four orientations of the polaroid.
Solution: (c)
Key concept:
Brewster’s law: Brewster discovered that when a beam of unpolarised light is reflected from a transparent medium (refractive index = µ), the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarisation θ_p).
From figure, it is clear that θ_p+ θ_r = 90°
Also n = tan θ_p (Brewster’s law)
(i) For i < θ_p or i > θ_p
Both reflected and refracted rays becomes partially polarised.
(ii) For glass θ_p = 51° for water θ_p = 53°

If a light beam is incident on a glass slab at Brewster’s angle, the transmitted beaiji is unpolarised and reflected beam is polarised.
In the given figure, the light beam is incident from air to the glass slab at Brewster’s angle (i_p). The incident ray is unpolarised and is represented by dot (•). The reflected light is plane polarised represented by arrows.
As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

Important points:
Polarised light: The phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of light wave is called polarization of light.
(i) The plane in which oscillation occurs in the polarised light is called plane of oscillation.
(ii) The plane perpendicular to the plane of oscillation is called plane of polarisation.
(iii) Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.
Polaroids: It is a device used to produce the plane polarised light. It is based on the principle of selective absorption and is more effective than the tourmaline crystal, or it is a thin film of ultramicroscopic crystals of quinine iodosulphate with their optic axis parallel to each other.

Question 2. Consider sunlight incident on a slit of width 104 Å. The image seen through the slit shall
(a) be a fine sharp slit white in colour at the centre
(b) a bright slit white at the centre diffusing to zero intensities at the edges
(c) a bright slit white at the centre diffusing to regions of different colours
(d) only be diffused slit white in colour.
Solution: (a)
Key concept: –
Diffraction of Light is the phenomenon of bending of light around the comers of an obstacle/aperture of the size of the wavelength of light.

In figure (A), no diffraction phenomenon is observed as the size of slit is weary large compared to wavelength. But in figure(B), there will be diffraction of light as size of slit is compared to the wavelength of light incident.
Here in the question it is given, width of the slit
b = 104 Å = 104 x 10^-10m = 10^-6m = 1 pm
Wavelength of (visible) sunlight varies from 4000 Å to 8000 Å.
Hence the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear, i.e., mixing of colours form white patch at the centre.

Question 3. Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is

Solution: (a)
Key concept: If slab of a glass is placed in air, the wave reflected from the upper surface (from a denser medium) sutlers a sudden phase change of π, while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change.
It is useful to draw an analogy between reflected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary:
Figure (a) shows that ray reflecting from a medium of higher refractive index undergoes a 180° phase change.

Question 4. In a Young’s double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case,
(a) there shall be alternate interference patterns of red and blue
(b) there shall be an interference pattern for red distinct from that for blue
(c) there shall be no interference fringes
(d) there shall be an interference pattern for red mixing with one for blue
Solution: (c)
Key concept:
Condition for Observing Interference
The initial phase difference between the interfering waves must remain constant. Otherwise the interference will not be sustained.
The frequency and wavelengths of two waves should be equal. If not the phase difference will not remain constant and so the interference will not be sustained.
The light must be monochromatic. This eliminates overlapping of patterns as each wavelength corresponds to one interference pattern.
Here in this problem of Young’s double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filtration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.
The wave front emitted by a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent sources so obtained are imaginary.

Question 5. Figure shows a standard two slit arrangement with slits S₁, S₂, P₁, P₂ are the two minima points on either side of P (figure).

At P₂ on the screen, there is a hole and behind P₂.is a second 2-slit arrangement with slits S₃, S₄ and a second screen behind them.
(a) There would be no interference pattern on the second screen but it would be lighted
(b) The second screen would be totally dark
(c) There would be a single bright point on the second screen
(d) There would be a regular two slit pattern on the second screen
Solution: (d)
Key concept:
Wave front
Every point on the given wave front acts as a source of new disturbance called secondary wavelets which travel in all directions with the .velocity of light in the medium.
A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wave front at that instant. This is called secondary wave front.In the given question, there is a hole at point which is a maxima point. From Huygen’s principle, wave will propagate from the sources S₁ and S₂. Each point on the screen will act as secondary sources of wavelets.

One or More Than One Correct Answer Type

Question 6. Two sources S₁ and S₂ of intensity I₁ and I₂are placed in front of a screen .
(a) The pattern of intensity distribution seen in the central portion is given by Fig. (b).
In this case, which of the following statements are true?

Solution: Key concept:

Question 7. Consider sunlight incident on a pinhole of width 103 Å. The image of the pinhole seen on a screen shall be
(a) a sharp white ring
(b) different from a geometrical image
(c) a diffused central spot, white in colour
(d) diffused coloured region around a sharp central white spot
Solution: (b, d)
Key concept: Diffraction of Light can be observed only if the size of obstacle/aperture is less than the wavelength of light.
Given, width of pinhole = 103 Å = 1000 Å
We know that wavelength of sunlight ranges from 4000 Å to 8000 Å. Clearly, wavelength X < width of the slit.
Hence, light is diffracted from the hole. Due to diffraction from the slit the image formed on the screen will be different from the geometrical image.

Question 8. Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
(a) the size decreases (b) the intensity increases
(c) the size increases (d) the intensity decreases
Solution: (a,b)

The central bright disc is known as Airy’s disc.
As the size of the hole is increased, AO will decrease and size of Airy’s disc will decrease.
As the size of the hole is increased, the width of the central maximum of the diffraction pattern of hole decreases. Since the same amount of light is now distributed over a small area, as intensity ∞ 1/area, the area is decreasing so area intensity increases.

Question 9. For light diverging from a point source,
(a) the wavefront is spherical
(b) the intensity decreases in proportion to the distance squared
(c) the wavefront is parabolic
(d) the intensity at the wavefront does not depend on the distance
Solution: (a, b)

Question 10. Is Huygen’s principle valid for longitudinal sound waves?
Solution: Yes it can. Huygen’s principle basically states that every point wave front can be considered as a secondary source of tiny wavelets that spread out in the forward direction of the wave itself. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of wavefront is in accordance with Huygen’s principle.
So, Huygen’s principle’is valid for longitudinal sound.waves also.

Question 11. Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
Solution: Orientation of wave front is perpendicular to ray. The ray diagram of the situation is as shown in figure.

Parallel rays incident on lens L₁ forms the image I₂ at the focal point of the lens. This image acts as object for the lens L₂ Now, due to the converging lens L₂, let final image formed is I which is point image. Hence the wavefront for this image will be of spherical symmetry.

Question 12. What is the shape of the wavefront on earth for sunlight?
Solution: The sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. We can treat it like a point object as seen from the surface of earth.
Because of large distance, the radius of wavefront can be considered as Large (infinity) and hence, wavefront is almost plane.

Question 13. Why is the diffraction of Sound waves more evident in daily experience than that of light wave?
Solution: The frequencies of sound waves lie between 20 Hz to 20 kHz, their wavelength ranges between 15 m to 15 mm. The diffraction occurs if the wavelength of waves is nearly equal to slit width.
The wavelength of light waves is 7000 x 10^-10 m to 4000 x 10^-10 m. For observing diffraction of light we need very narrow slit width. In daily life experience we observe the slit width very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

Question 14. The human eye has an approximate angular resolution of ɸ = 5.8 x 10^-4 rad and a typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.
Solution: It is given, angular resolution of human eye ɸ = 5.8 x 10^-4 rad and printer prints 300 dots per inch.

Question 15. A polaroid (I) is placed in front,of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.
Solution: A polaroid (I) is placed in front of a monochromatic source. Another polariod (II) is placed in front of this polaroid (I) when the pass axis of (II) is parallel to (I), light passes through polaroid-II is unaffected.

Now polariod (II) is rotated till no light passes. In this situation the pass axis of polariod (II) is perpendicular to polariod (I), then (I) and (II) are set in crossed positions. No light passes through polaroid-II.
Now third polaroid (III) is now placed in between (I) and (II). Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Short Answer Type Questions

Question 16. Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
Solution: If angle of incidence is equal to Brewster’s angle, the transmitted light is slightly polarised and reflected light is plane polarised.

Question 17. For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance.
Solution:
Key concept:
• Resolving power is the ability of an imaging device to separate (i.e., to see as distinct) points of an object that are located at a small angular distance or it is the power of an optical instrument to separate far – away objects, that are close together, into individual images. The term resolution or minimum resolvable distance is the minimum distance between distinguishable objects in an image, although the term is loosely used by many users of microscopes and telescopes to describe resolving power. In scientific analysis, in general, the term “resolution” is used to describe the precision with which any instrument measures Ratio of the least separation,
For electrons accelerated through 100 V, the de-Broglie wavelength, 12.27

Question 18. Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is. half the distance between the slits. Obtain the value of D in terms of X such that the first minima on the screen falls at a distance D from the centre O.

Solution:

Long Answer Type Questions

Question 19.

Figure shown has a two sift arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I₀ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
Solution: The resultant amplitude of wave reaching on screen will be the sum of amplitude of either wave in perpendicular and parallel polarisation.

Question 20.

A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

Solution:

Question 21.

Solution:

Question 22.

Solution:

March 20, 2018