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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
Light of wavelength `2475 Å` is incident on barium. Photoelectrons emitted describe a circle of radius `100 cm` by a magnetic field of flux density `(1)/(sqrt(17)) xx 10^(-5) Tesla`. Work function of the barium is (Given `(e )/(m) = 1.7 xx 10^(11)`) |
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Answer» Radius of circular path described by a charged particle in a magnetic field is given by `r=sqrt((2mk)/(qB)) , rArr k=(q^2B^2 r^2)/"2m"=(e/m)(eB^2r^2)/2` `=1/2xx1.7xx10^11 xx 1.6 xx 10^(-19)xx(1/sqrt17 xx 10^(-5))^2xx(1)^2` `=8xx10^(-20) J ` =0.5 eV `E=phi+K_"max"` `phi=E-K_"max"` `=(-12375/2475)` eV-0.5 eV=4.5 eV |
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| 252. |
The wavelength of a photon is 1.4 Å. It collides with an electron. The energy of the scattered electron is `4.26xx10^(-16)` J. Find the wavelength of photon after collision. |
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Answer» Given h=`6.63xx10^(-34)` Js Initial wavelength of photon `lambda_1` Final wavelength of photon `lambda_2` `4.26xx10^(-16)="hc"/lambda_1="hc"/lambda_2` `"hc"/lambda_2="hc"/lambda_1-4.26xx10^(-16)` `(6.63xx10^(-34)xx3xx10^8)/lambda_2=14.21xx10^(-16)-4.26xx10^(-16)` `lambda_2=2xx10^(-10) ` m =2.0 Å. |
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| 253. |
Light of wavelength `2475 Å` is incident on barium. Photoelectrons emitted describe a circle of radius `100 cm` by a magnetic field of flux density `(1)/(sqrt(17)) xx 10^(-5) Tesla`. Work function of the barium is (Given `(e )/(m) = 1.7 xx 10^(11)`) A. `1.8 eV`B. `2.1 eV`C. `4.5 eV`D. `3.3 eV` |
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Answer» Correct Answer - C Radius of circular path described by a charged particle in a magnetic field is given by `r = (sqrt(2 mK))/(qB)`, where `K` is Kinetic energy of electron `rArr K = (q^(2) B^(2) r^(2))/( 2m) = ((e )/(m)) (eB^(2) r^(2))/(2)` `= (1)/(2) xx 1.7 xx 10^(11) xx 1.6 xx 10^(-19) xx ((1)/(sqrt(17)) xx 10^(-5))^(2) xx (1)^(2)` ` = 8 xx 10^(-20) J = 0.5 eV` By using `E = W_(0) + K_(max)` `rArr W_(0) = E - K_(max) ((12375)/(2475))V - 0.5 V = 4.5 V` |
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| 254. |
Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified ? |
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Answer» V = 10 MV = `10^(7) V , v = sqrt(e/m xx2v)= sqrt(1.76 xx10^(11) xx 2 xx10^(7)) :. v = 1.8762 xx10^(9) ` m/s This speed is greater than speed of light , which is not possible . As `v` approaches to c, then mass m = ` (m_(0))/(sqrt(1-(v^(2))/(c^(2))))` |
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| 255. |
Photon is not a..........but it is a.......... . |
| Answer» Correct Answer - material body ; packet of energy | |
| 256. |
Assuming photoemission to take place , the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times , isA. 4B. `1/4`C. `2`D. `1/2` |
| Answer» Correct Answer - 4 | |
| 257. |
Which of the follwing statements about photon is incorrect ?A. Photons exert no pressure.B. Momentum of photon is `(hv)/(c )` .C. Rest mass of photon is zero.D. Energy of photon is hv. |
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Answer» Correct Answer - A (a) : Photons move with velocity of light and have energy hv. Therefore, they also exert pressure. |
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| 258. |
Which of the following figure represents the variation of particle momentum and the associated de - Broglie wavelength ?A. B. C. D. |
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Answer» Correct Answer - D (d) : de Broglie wavelength, `lamda=(h)/(p)or lamdaprop(1)/(p)` Hence, curve (d) is the correct option. |
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| 259. |
When a point source of light is at a distance of one metre from a photo cell , the cut off voltage is found to be `V`. If the same source is placed at `2 m` distance from photo cell , the cut off voltage will beA. When a point source is at a distance of one metre from a photo cell , the cut off voltage is found to be `V`. If the same source is placed at `2 m` distance from photo cell , the cut off voltage will beB. `V`C. `V//2`D. `V//4` |
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Answer» Correct Answer - A By changing distance of source , photoelectric current changes . But there is no change in stopping potential. |
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| 260. |
An X - ray tube produces a continous spectrum of radiation with its short wavelength end at 0.45 `Å` . What is the maximum energy of a photon in the radiation? |
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Answer» Given , ` lambda = 0.45 Å = 0.45 xx 10^(-10) m, E = (hc)/lambda = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(0.45 xx 10^(-10) xx 1.6xx 10^(-19) ) eV ` ` :. E = 27.6 xx 10^(3) eV = 27.6 KeV` |
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| 261. |
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength ) for yellow - green colour and about 760 nm for red colour . (a ) What are the energies of photons in (ev) at the (i) violet end, (ii) average wavelength , yellow - green colour , and (ii) red end of the visible spectrum ? (Take h = `6.63 xx 10^(-34)` J s and 1 eV= ` 1.6 xx 10^(-19)` J ) , |
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Answer» Energy of the incident photon, `E= hv = hc//lambda` ` E = (6.63 xx10^(-34) Js) (3 xx 10^(8) m//s)//lambda` ` = ((1.989xx10^(-25))Jm)/lambda ` (i) For violet light , `lambda _(1) = 390 ` nm (lower wavelength end) Incident photon energy , ` E_(1) = (1.989 xx 10^(-25)Jm)/(390xx10^(-9)m)` ` = 5.10xx10^(-19) J = (5.10xx10^(-19)J)/(1.6 xx 10^(-19)J//eV)` = 3.19 eV (ii) For yellow - green light , `lambda_(2) = 550 ` nm (average wavelength ) Incident photon energy , ` E_(2) = (1.989 xx 10^(-25)Jm)/(550 xx10^(-9)m)` ` = 3.62 xx 10^(-19) J = 2.26 eV` (iii) For red light , ` lambda _(3) = 760` nm( higher wavelength end) Incident photon energy ` E_(3) = (1.989 xx 10^(-25)Jm)/(760 xx 10^(-9) m)` ` = 2.62 xx 10^(-19) J = 1.64 eV` |
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| 262. |
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength ) for yellow - green colour and about 760 nm for red colour . From which of the photosensitive materials with work functions listed in table and using the results of (i) , (ii) and (iii) of (a) can you build a photoelectric device that operates with visible light ? |
| Answer» For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function `phi_(0)` of the material . Thus , the photoelectric device material Na ( with `phi_(0) = 2.75` eV) , K ( with `phi_(0) = 2.30` eV) and Cs ( with `phi_(0) = 2.14` eV) . It will also operate with yellow - green light (with E = 2.26 eV) for Cs (with `phi_(0) = 2.14` eV) only . However, it will not operate with red light ( with E = 1.64 eV) for any of these photosensitive materials . | |
| 263. |
Energy of an quanta of frequency `10^(15) Hz` and `h = 6.6 xx 10^(-34) J - sec` will beA. `6.6 xx 10^(-19) J`B. `6.6 xx 10^(-12) J`C. `6.6 xx 10^(-49) J`D. `6.6 xx 10^(-41) J` |
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Answer» Correct Answer - A `E = hv = 6.6 xx 10^(-34) xx 10^(15) = 6.6 xx 10^(-19) J` |
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| 264. |
The electric field associated with a light wave is given by `E=E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]`. Find the stopping potential when this light is used in an experiment on a photoelectric effect with the emitter having work function 2.1eV. `h=6.62xx10^(-34)Js`. |
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Answer» Correct Answer - `1.0V` Here, `E=-E_(0)sin[(1.57xx10^(7)m^(-1))(ct-x)]` Comparing it with the equation of harmonic wave in electric field , `E=E_(0)sin(2pi)/lambda(vt-x)` we have , `(2pi)/lambda=1.57xx10^(7) or lambda=(2pi)/(1.57xx10^(7))m`, `V_(0) =(hc)/(lambdae)-(phi_(0))/e` `=(6.63xx10^(-34)xx3xx10^(8)xx1.57xx10^(7))/(2xx3.142xx1.6xx10^(-19))-2.1` `=1.0V` |
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| 265. |
When an inert gas is filled in the place vacuum in a photo cell , thenA. Photo - electric current is decreasedB. Photo - electric current is increasedC. Photo - electric current remains the sameD. Decrease or increase in photo - electric in photo - electric current does not depend upon the gas filled |
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Answer» Correct Answer - B In the pressure of inert gas photoelectrons emitted by cathode ionize the gas by collision and hence the current increases. |
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| 266. |
Threshold wavelength for photoelectric effect on sodium is `5000 Å` . Its work function isA. `15 J`B. `16 xx 10^(-14) J`C. `4 xx 10^(-19) J`D. `4 xx 10^(-81) J` |
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Answer» Correct Answer - C `W_(0) = (hc)/(lambda_(0)) = (6.625 xx 10^(-34) xx 3 xx 10^(8))/( 5000 xx 10^(-10)) J = 4 xx 10^(-19) J` |
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| 267. |
If the threshold wavelength for sodium is `5420 Å`, then the work function of sodium isA. `4.58 eV`B. `2.28 eV`C. `1.14 eV`D. `0.23 eV` |
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Answer» Correct Answer - B `W_(0) = (12375)/(lambda_(0)) = (12375)/(5420) = 2.28 eV` |
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| 268. |
From the figure describing photoelectric effect we may infer correctly that A. `Na` and `A1` both have the same threshold frequencyB. Maximum kinetic energy for both the metals depend linearly on the frequencyC. The stoppinf potentials are different for `Na` and `A1` for the same change in frequencyD. `A1` ia a better photo sensitive material than `Na` |
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Answer» Correct Answer - B Stopping potential equals to maximum kinetic energy. Since stopping potential is varying linearly with the frequency. The `max.KE` for both the metals also vary linearly with frequency. |
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| 269. |
Choose the correct alternativeA. Ratio of de-Broglie wavelengths of proton and `alpha`-particle of same kinetic energy is 2L1B. if neutron, `alpha`-particle and `beta`-particle all are moving with same kinetic energy , `beta`-particle has maximum de-Broglie wavelengthC. de-Broglie hypothesis treats particles as wavesD. de-Broglie hypothesis treats waves as made of particles |
| Answer» Correct Answer - 1,2,3 | |
| 270. |
From the figure describing photoelectric effect we may infer correctly that A. Na and Al both have the same threshold frequencyB. Maximum kinetic energy for both the metals depend linearly on the frequencyC. The stopping potentials are different for Na and Al for the same change in frequencyD. Al is better photosensitive material than Na |
| Answer» Correct Answer - 2 | |
| 271. |
Choose the correct statements regarding the phenomena of photoelectric effectA. Photoelectric effect was explained by Einstein considering wave nature of lightB. If intensity of radiations incident on photosensitive material is increases, stopping potential increasesC. Work function of gold is about 5.1 eVD. Slope of a graph between incident frequency (greater than threshold frequency ) verses stopping potentials is independent of metal |
| Answer» Correct Answer - 3,4 | |
| 272. |
The photoelectric effect can be understood on the basis ofA. The principle of superpositionB. The electromagnetic theory of lightC. The special theory of relativityD. Line spectrum of the atom |
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Answer» Correct Answer - D Photoelectric effect can be explained on the basis of spectrum of an atom. |
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| 273. |
The photoelectric effect can be explained on the basis ofA. Corpuscular theoryB. Wave theoryC. electromagnetic theoryD. quantum theory |
| Answer» Correct Answer - d | |
| 274. |
Which of the following has minimum stopping potential?A. BlueB. YellowC. VioletD. Red |
| Answer» Correct Answer - d | |
| 275. |
The maximum value of stopping potential in the following diagram is A. `-4 V`B. `-1 V`C. `-3 V`D. `-2 V` |
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Answer» Correct Answer - A `| - 4V| gt | - 2V|` |
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| 276. |
The frequency of a photon , having energy `100 eV is ( h = 6.610^(-34) J - sec)`A. `2.42 xx 10^(26) Hz`B. `2.42 xx 10^(16) Hz`C. `2.42 xx 10^(12) Hz`D. `2.42 xx 10^(9) Hz` |
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Answer» Correct Answer - B `E = hv rArr 100 xx 1.6 xx 10^(-19) = 6.6 xx 10^(-34) xx v` |
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| 277. |
A photon of wavelength `4400 Å` is passing through vaccum. The effective mass and momentum of the photon are respectivelyA. `5 xx 10^(-36) kg, 1.5 xx 10^(-27) kg - m//s`B. `5 xx 10^(-35) kg, 1.5 xx 10^(-26) kg - m//s`C. Zero , `1.5 xx 10^(-26) kg - m//s`D. `5 xx 10^(-36) kg , 1.67 xx 10^(-43) kg - m//s` |
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Answer» Correct Answer - A `p = (h)/(lambda) = (6.6 xx 10^(-34))/(4400 xx 10^(-10)) = 1.5 xx 10^(-27) kg . m//s` and mass `m = (p)/( c ) = (1.5 xx 10^(-27))/(3 xx 10^(8)) = 5 xx 10^(-36) kg` |
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| 278. |
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which A neutron, and would have the same de Broglie wavelength. |
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Answer» Given, `lambda = 589 nm = 589xx10^(-9)m, m_(e )=9.1xx10^(-31)kg`. `m_(n)=1.67xx10^(-27)kg, h = 6.62xx10^(-34)J-s`. `KE_(n)=(h^(2))/(2m_(n)lambda^(2))=((6.63xx10^(-34))^(2))/(2xx(1.66xx10^(-27))xx(589xx10^(-9))^(2)) therefore KE_(n) = 3.81xx10^(-28)J`. |
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| 279. |
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (i) an electron, and (ii) a neutron, would have the same de-Broglie wavelength . |
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Answer» Given, `lambda = 589 nm = 589xx10^(-9)m, m_(e )=9.1xx10^(-31)kg`. `m_(n)=1.67xx10^(-27)kg, h = 6.62xx10^(-34)J-s`. `KE_(e )=(P_(e )^(2))/(2m)=(h^(2))/(2m lambda^(2)) [because p_(e )=(h)/(lambda)] rArr KE_(e )=((6.63xx10^(-34))^(2))/(2xx(589xx10^(-9))^(2)xx9.1xx10^(_31))=6.96xx10^(-25)J` |
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| 280. |
Electrons with energy `80 keV` are incident on the tungsten target of an X - rays tube , k- shell electrons of tungsten have `72.5 keV` energy X- rays emitted by the tube contain onlyA. A continuous `X` - ray spectrum ("Bremsstrahlung") with a minimum wavelength of ~ 0.155 Å`B. `"A continuous" `X` - "ray spectrum (Bremsstrahlung)" with all wavelengthsC. The characteristic `X` - rays spectrum of tungstenD. A continuous `X` - ray spectrum "(Bremsstrahlung) with a minimum wavelength of" `~0.155 Å` and "the characteristic" `X` - ray spectrum of tungsten |
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Answer» Correct Answer - D Minimum wavelength of continuous `X` - ray spectrum is given by `lambda_(min) (i n Å) = (12375)/(E (eV)) = (12375)/(80 xx 10^(3)) ~~ 0.155` Also the energy of the incident electrons `(80 K eV)` is more than the ionization energy of the `K` - shell electrons (i.e., `72.5 K eV`). Therefore characteristic `X` - ray spectrum will also be obtained because energy of incident electron is enough to knock out the electron from `K` or `L` shells. |
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| 281. |
Assertion : The threshold frequency of photoelectric effect supports the particle nature of sunlight . Reason : If frequency of incident light is less than the threshold frequency , electrons are not emitted from metal surface.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not correct explanation of the assertion.C. If assertion is true but the reason is false.D. If both the assertion and reason are false. |
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Answer» Correct Answer - B There is no emission of photoelectrons till the frequency of incident light is less than a minimum frequency , however intense light it may be . In photoelectric effect, it is a single particle collision. Intensity is `hv xx N` , where `hv` is the individual energy of the photon and `N` is the total number of photon . In the wave theory , the intensity is proportional , not only to `v^(2)` but also to the amplitude squared. For the same frequency , increase in intensity only increase the number of photons ( in the quantum theory of Einstein). |
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| 282. |
The photoelectric threshold of a certain metal is `3000A`. If the radiation of `2000A` is incident on the metalA. Electrons will be emittedB. Positrons will be emittedC. Protons will be emittedD. Protons will be emitted |
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Answer» Correct Answer - A For electron emission `lambda_( incident) lt lambda_(0)` |
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| 283. |
A photocell stoops emission if it is maintained at `2 V` negative potential . The energy of most energetic photoelectron isA. `2 eV`B. `2 J`C. `2 kJ`D. `2 keV` |
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Answer» Correct Answer - A `K_(max) = (|V_(0)|) eV = 2 eV`. |
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| 284. |
The de Broglie wavelength `lamda` of an electron accelerated through a potential V in volts isA. `(1.227)/(sqrt(V))nm`B. `(0.1227)/(sqrt(V))nm`C. `(0.01227)/(sqrt(V))nm`D. `(12.27)/(sqrt(V))nm` |
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Answer» Correct Answer - A (a) : Consider an electron of mass m and charge e accelerated from rest through potential V. Then, K=eV `K=(1)/(2)mv^(2)=(p^(2))/(2m):.p=sqrt(2mK)=sqrt(2meV)` The de Broglie wavelength `lamda` of the electron is `lamda=(h)/(p)=(h)/(sqrt(2mK))=(h)/(sqrt(2meV))` Substituting the numerical values of h, m, e, we `lamda=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xxV))` `=(1.227xx10^(-9))/(sqrt(V))m=(1.227)/(sqrt(V))nm` |
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| 285. |
Photoelectric effect experiments are performed using three different metal plates `p,q` and`r` having work function `phi_(p) = 2.0 eV, phi_(e) = 2.5 eV and phi_(r) = 3.0 eV` respectively A light beam containing wavelength of `550nm , 450 nm` and `350nm ` with equal intensities illuminates each of the plates . The correct `I -V` graph for the experiment is [Take hc = 1240 eV nm]A. B. C. D. |
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Answer» Correct Answer - a Energy of wavelength `550nm` is, `E_(1)=(hc)/(lambda_(1))=(1240 eVnm)/(550nm)=2.25 eV` Energy wavelength `450nm` is , `E_(2)=(1240 eVnm)/(450nm)=2.75 eV` Energy wavelength `350nm` is , `E_(3)=(1240 eVnm)/(350nm)=3.54 eV` Since light of all the source wavelength are incident on a each plate with equal intensities, the stopping potential will be linked by the light of wavelength having maximum energy i.e., `3.54 eV`. Plate-p is having least work function `phi_(p)(=2.0 eV)`, will have maximum value of stopping potential as it will emit photoelectrons due to light of all the three wavelength of energies `E_(1),E_(2)` and `E_(3)` Therefore, for plate-p saturation current will be maximum and stopping potential will be maximum. Plate- q is having a work function `phi_(0)(=2.5 eV)`, will emit photoelectrons due to light of energy `E_(2)` and `E_(3)`. the value of saturation current and stopping potential will be less than plate-p. Plate-r is having a work function `phi_(r) (=3.0 eV)`, will emit photoelectrons due to light of energy `E_(3)`. its saturation current will be minimum and stopping potential will be least. Thus option (a) is true. |
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| 286. |
The minimum wavelength of the `X` - rays produced by electrons accelerated through a potential difference of `V "volts"` is directly proportional toA. `sqrt(V)`B. `V^(2)`C. `1//sqrt(V)`D. `1//V` |
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Answer» Correct Answer - D `hv_(max) = eV rArr (hc)/(lambda_(min)) = eV :. Lambda_(min) prop (1)/(V)` |
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| 287. |
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greter energy? An X-ray photon or the electron? (For quantiative comparision, take the wavelength of the probe equal to `1Å`, which is of the order of interatomic spacing in the lattice), `m_(e)=9.11xx10^(-31)kg`. |
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Answer» For producing X-rays, kinetic energy of electron is involved. K.E. of electron, `E=1/2mv^(2) or mv=sqrt(2Em) and lambda=h/(mv)=h/(sqrt(2Em))` `:. E=(h^(2))/(2lambda^(2)m)=((6.63xx10^(-34))^(2))/(2xx10^(-20)xx9.11xx10^(-31))J = ((6.63xx10^(-34))^(2))/(2xx9.11xx10^(-51)xx1.6xx10^(-19)) eV=150.6 eV` For photons, `lambda=1Å=10^(-10)m` `:.` Energy of photon, `E=hv=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(10^(-10)xx1.6xx10^(-19)) eV=12.4xx10^(3) eV` Thus for the same wavelength, a photon has much greater kinetic energy than an electron. |
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| 288. |
Assertion : Light is produced in gases in the process of electric discharge through them at high pressure. Reason : At high pressure electrons of gaseous atoms collide and reach and excited state.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D Light is produced in gases in the process of electric discharge at low pressure. When accelerated electrons collide with atoms of the gas , atoms get excited . The excited atoms return to their normal state and in this process light radiations are emitted. |
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| 289. |
When yellow light is incident on a surface , no electrons are emitted while green light can emit. If red light is incident on the surface , thenA. No electrons are emittedB. Photons are emittedC. Electrons of higher energy are emittedD. Electrons of lower energy are emitted |
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Answer» Correct Answer - A `lambda_(r) gt lambda_(y) gt lambda_(g)` . Here threshold wavelength `lt lambda_(y)`. |
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| 290. |
In the Davisson and Germer experiment , the velocity of electrons emitted from the electron gun can be increased byA. increasing the filament currentB. decreasing the filament currentC. decreasing the potential difference between the anode and filamentD. increasing the potential difference between the anode and filament |
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Answer» Correct Answer - D In the Davisson and Germer experiment , the velocity of electron emitted from the electron gun can be increased by increasing the potential difference between the anode and filament. |
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| 291. |
When a surface is irradiated with light of wavelength `4950 Å`, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical potential is changed to 1.1 volt. if the photoelectrons (after emission form the source) are subjected to a magnetic field of 10 tesla, the two retarding potentials would |
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Answer» Correct Answer - `1.9 eV, 4125Å`; no changes `1/2mv_(1)^(2)=eV_(1)=hv_(1)-phi_(0)` or ` phi_(0)=hv_(1)-eV_(1)=(hc)/(lambda_(1))-eV_(1)` `:. phi_(0)=(6.63xx10^(-34)xx3xx10^(8))/(4950xx10^(-10))-1.6xx10^(-19)xx0.6` `=3.04xx10^(-19)J=1.9 eV` `(hc)/(lambda_(2))=eV_(0)+phi_(0)` `1.6xx10^(-19)xx1.1+3.04xx10^(-19)` `=4.8xx10^(-19)J` `:. lambda_(2)=(hc)/(4.8xx10^(-19))=(6.63xx10^(-34)xx3xx10^(8))/(4.8xx10^(-19))` `=4125Å` As the magnetic field does not change the speed of the electrons hence there will be no change in the stopping potential. |
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| 292. |
A metal surface of work function `1.07 eV` is irradiated with light of wavelength `332 nm`. The retarding potential required to stop the escape of photo - electrons is |
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Answer» Correct Answer - 2.65 volt `E=phi+eV_s` the correct answer is 2.66 |
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| 293. |
Assertion : Soft and hard `X`-rays differ in frequency as well as velocity. Reason : The penetrating power of hard `X`- rays is more than the penetrating power of soft `X`-rays.A. If both assertion and reason are true and the reason is the correct explanation explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Soft and hard `X` - rays differ only in frequency . But both types of `X`-ray travel with speed of light. |
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| 294. |
An electron of mass `m` and a photon have same energy `E`. The ratio of de - Broglie wavelengths associated with them is :A. `(1)/( C) ((E )/( 2 m))^(1//2)`B. `((E )/(2 m))^(1//2)`C. `C ( 2m E)^(1//2)`D. `(1)/(xC) ((2 m)/( E))^(1//2)` |
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Answer» Correct Answer - A For electron `lambda_(e) = (h)/(sqrt(2 mE))` For photon `E = pc rArr lambda_(ph) = (hc)/( E )` `rArr (lambda_(e ))/(lambda_(ph)) = (h)/(sqrt( 2m E)) xx (E )/(hc) = ((E )/(2 m))^(1//2) (1)/( c )` |
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| 295. |
Who won the Nobel prize in physics in the year 1929 for the discovery of the nature of electrons ?A. Erwin SchrodingerB. R.A MillikanC. Louris Victor de BroglieD. Albert Einstein |
| Answer» Correct Answer - C | |
| 296. |
(a) An X-ray tube produces a continuous spectrum of radiation with its shorts wavelength end ate`0.45 Å`. What is the maximum energy of the photon in the radiation? (b) From your answer to (a) , guess what order of accelerating voltage (for electrons) is required in such a tube? |
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Answer» (a) `lambda_(min)=0.45Å =0.45xx10^(-10)m, h=6.6xx10^(-34)Js, c=3xx10^(8) ms^(-1)` We know, maximum energy of X-ray photon is `E_(max)=hv_(max)=(hc)/(lambda_(min))=(6.63xx10^(-34)xx3xx10^(8))/(0.45xx10^(-10)) J=(6.63xx10^(-34)xx3xx10^(8))/(0.45xx10^(-10)xx1.6xx10^(-19)) eV` `=27.6xx10^(3)eV=27.6keV` (b) In x-ray tube, accelerating voltage provides the energy to the electrons which on striking the anticathode, produce X-rays. For getting X-ray photons of 27.6 keV, it is required that the incident electrons must posses kinetic energy of atleast 27.6keV. Therefore, accelerating voltage of the order of 30 keV should be applied across the X-ray tube to get the required X-ray photons. |
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| 297. |
The figure shows a plot of photo current versus anode potential for a photosensitive surface for three different radiations. Which one of the following is a correct statement ? A. Curves `a` and `b` represent incident radiations of different frequencies and different intensitiesB. Curves `a` and `b` represent incident radiations of the same frequency but no different intensitiesC. Curves `b` and `c` represent incident radiations of different frequencies and different intensitiesD. Curve `b` and `c` represent incident radiations of the same frequency having the same intensity |
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Answer» Correct Answer - B From the two graphs we can conclude that for the graph in question curves `a` and `b` represent incident radiations of the same frequency but of different intensities. |
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| 298. |
The de Broglie wavelength is given byA. `p=(2pih)/(lamda)`B. `p=(h)/(2lamda)`C. `p=(2pi)/(hlamda)`D. `p=(2pi)/(lamda)` |
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Answer» Correct Answer - A (a) : de Broglie wavelength is given by `lamda=(h)/(p)`, where p is the momentum or `p=(h)/(lamda)` Since, `h=(h)/(2pi)" "p=(2pih)/(lamda)` |
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| 299. |
Which phenomenon best supports the theory that matter has a wave nature?A. Electron momentumB. Electron diffractionC. Photon momentumD. Photon diffraction |
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Answer» Correct Answer - B (b) : Matter has a wave nature is best supported by the phenomenon of electron diffraction. |
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| 300. |
Why is the wave nature of matter not more apparent to our daily observations? |
| Answer» De-Broglie wavelength associated with a body of mass m, moving with velocity v is given by `lambda=h/(mv)` Since, the mass of of the object hence the de-Broglie wavelength associated with it is quite small hence it is not visible. Hence the wave nature of matter is not more apparent to our daily observations. | |