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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.
| 51. |
Kinetic energy with which the electrons are emitted from the metal surface due to photoelectric effect isA. Dependent of the intensity of illuminationB. Dependent on the frequency of lightC. Inversely proportional to the intensity of illuminationD. Directly proportional to the intensity of illumination |
| Answer» Correct Answer - B | |
| 52. |
In a photoelectric experiment , the maximum velocity of photoelectric emittedA. depends on intensity of incident radiationB. does not depend on cathode materialC. depends on frequency of incident radiationD. does not depend on wavelength of incident radiation |
| Answer» Correct Answer - C | |
| 53. |
Emission of electrons in photoelectric effect is possible, ifA. metal surface is highly polishedB. the incident light is of sufficiently high intensityC. the light is incident at right angles to the surfaceD. the incident light is of sufficiently low wavelength |
| Answer» Correct Answer - D | |
| 54. |
With the decrease in the wave length of the incident radiation the velocity of the photoelectrons emitted from a given metalA. remains sameB. increasesC. decreasesD. increases first and then decreases |
| Answer» Correct Answer - B | |
| 55. |
A source of light is placed above a sphere of radius `10cm`. How many photoelectrons must be emitted by the sphere before emission of photoelectrons stop? The energy of incident photon is `4.2eV` and the work function of the metal is `1.5eV`.A. `2.08 xx 10^(18)`B. `1.875 xx 10^(8)`C. `2.88 xx 10^(18)`D. `4 xx 10^(19)` |
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Answer» Correct Answer - B Stopping potential energy =`eV_(0)=E-omega` `V_(0)=(E-omega)/e=(9xx10^(9)"ne")/r`,`n`=no of electrons |
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| 56. |
The number of photoelectrons emitted for light of a frequency `v` (higher than the threshold frequency `V_(0)`) is proportional toA. Frequency of lightB. Work functionC. Theshold wavelengthD. Intensity of light |
| Answer» Correct Answer - D | |
| 57. |
The threshold wavelength for emission of photoelectrons from a metal surface is `6xx10^(-7)m`.The work function of the material of the metal surface is.A. `3.3xx10^(-19) J`B. `6.67xx10^(-19) J`C. `1.23xx10^(-19) J`D. `2.37xx10^(-19) J` |
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Answer» Correct Answer - A `omega_(0)=(hc)/lambda_(0)` |
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| 58. |
Ratio of debroglie wavelengths of uncharged particle of mass m at `27^(0)C` to `127^(0)C` is nearlyA. `1.16`B. `0.16`C. `1.33`D. `0.8` |
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Answer» Correct Answer - A `lambda alpha (1)/(sqrt(T))` |
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| 59. |
Two particles A and B of de-broglie wavelength `lambda_(1) and lambda_(2)` combine to from a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one dimensional).A. `lambda_(A)+lambda_(B)`B. `lambda_(A)-lambda_(B)`C. `(lambda_(A)lambda_(B))/(lambda_(A)+lambda_(B))`D. `(lambda_(A)lambda_(B))/(lambda_(A)-lambda_(B))` |
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Answer» Correct Answer - D For one dimensional motion,`vecp_(C)=vecp_(A)+vecp_(B)` If `p_(A),p_(B) gt 0` or `p_(A),p_(B) lt 0`, i.e., `(p_(A)` and `p_(B)` are in same direction). `p_(C)=p_(A)+p_(B)` `h/lambda_(C)=h/lambda_(A)+h/lambda_(B)=h((lambda_(A)+lambda_(B))/(lambda_(A)lambda_(B)))` `lambda_(C)=(lambda_(A)lambda_(B))/(lambda_(A)+lambda_(B))` If `p_(A)gt0,p_(B)lt0` or `p_(A)lt0,p_(B)gt 0` `(p_(A)` and `p_(B)` are in opposite direction) `p_(C)|p_(A)-p_(B)|` `h/lambda_(C)=|h/lambda_(A)-h/lambda_(B)|rArr(h|lambda_(A)-lambda_(B)|)/(lambda_(A)lambda_(B))` `lambda_(C)=(lambda_(A)lambda_(B))/|(lambda_(A)-lambda_(B))|` |
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| 60. |
Consider a thn target (`10^(-2)` m square, `10^(-3)` m thinkness) of sodium, which produces a photocurrent of 100 `muA` when a light of intensity 100 `W//m^(2)(lamda=660 nm)` falls on it. Find the probability that a photoenectron is produced when a photon strikes a sodium atom. `" "["Taken density of Na"=0.97kg//m^(3)]`A. `0.75`B. `7.5xx10^(-2)`C. `7.5xx10^(-13)`D. `7.5xx10^(-21)` |
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Answer» Correct Answer - D Given `d=10^(-3)m,I=100xx10^(-6)A=10^(-4) A` `A=(10^(-2)m)^(2)=10^(-4)m^(2)` Intensity,`I=100 Wm^(-1)` `lambda=660nm=660xx10^(-9)m` Volume of `6.02xx10^(26)` sodium atoms`=(23kg)/(0.97kg//m^(3))=23.7 m^(3)` Volume of target=`(10^(-2))(10^(-2))(10^(-3))=10^(-7) m^(3)` Number of sodium atoms in the target =`(6.02xx10^(26))/23.7xx10^(-7)=2.5410^(18)` Let `n` be the number of photons falling per second on the target. Each of each photon`=(hc)/lambda` Total energy falling per second on target`=(nhc)/lambda=IA` `thereforen=(IAlambda)/(hc)=(100xx10^(-4)xx660xx10^(-9))/(6.62xx10^(-34)xx(3xx10^(8)))=3.3xx10^(16)` Number of electrons emitted per second by all the atoms in the target if one electron is emitted by each atom for one incident photon. `=(2.54xx10^(18))(3.3xx10^(16))=8.4xx10^(34)` Expected photocurrent =`(8.4xx10^(34))(1.6xx10^(-19))=1.34xx10^(16) A` Observed photocurrrent=`100muA` Probability of photo emission by single photon incident on a single atom `P=(100muA)/(1.34xx10^(16) A)=7.5xx10^(-21)` Thus the probability of emission by single photon on a single atom is very much less than `1`, the probability of absorption of two photons by single atoms is negligible. |
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| 61. |
Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts. Photo electric effect in this metallic surface begains at a frequency `6xx10^(14)s^(-1)`. The frequency of the incident light in `s^(-1)` is `[h=6 x10^(-34) J-sec`, charge on the electron =`1.6xx10^(-19)C`]A. `7.5xx10^(13)`B. `13.5xx10^(13)`C. `14xx10^(14)`D. `7.5xx10^(15)` |
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Answer» Correct Answer - C `K.E.=V_(0)e` and `V_(0)e=h[upsilon - upsilon_(0)]` where `V_(0)` is the stopping potential and `upsilon_(0)` is the Threshold frequency |
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| 62. |
The work function for sodium surface is 2.0eV and that for aluminium surface is 4.2eV. The two metals are illuminated with appropriate radiations so as to cause protoemission. ThenA. Both aluminium and sodium will have the same threshold frequencyB. The threshold freqency of aluminium will be more than that of sodiumC. The threshold freqency of aluminium will be less than that of sodiumD. The threshold wavelength of aluminium will be more than that of sodium |
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Answer» Correct Answer - B The threshold frequency for `AI` must be greater as it has higher work function. (`because W=hv_(0)`) |
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| 63. |
An electron moves with a speed of `sqrt3/2c`. Then its mass becomes…. Times its rest mass.A. 2B. 3C. `3/2`D. 4 |
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Answer» Correct Answer - A `m=(m_(0))/sqrt(1-V^(2)/c^(2))` |
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| 64. |
If the uncertainity in the position of an electron is `10^(-10) m`, then the value of uncertainity in its momentum (in `kg-ms^(-1)`) will beA. `3.33xx10^(-24)`B. `0.53xx10^(-24)`C. `6.6xx10^(-24)`D. `6.6xx10^(-20)` |
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Answer» Correct Answer - B `Deltap=h/(Deltax)=(1.034cc10^(-34))/10^(-10)=1.034xx10^(-24)kg -ms^(-1)` |
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| 65. |
The work function of nickle is `5eV`. When light of wavelength `2000A^(0)` falls on it, emits photoelectrons in the circuit. The the potential difference necessary to stop the fastest electrons emitted is (given `h=6.67xx10^(-34) Js`)A. `1.0V`B. `1.75V`C. `1.2V`D. `0.75V` |
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Answer» Correct Answer - C `V_(0)=(E-omega_(0))/e` |
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| 66. |
Figure show the variation of the stopping potential `(V_(0))` with the frequency `(v)` of the incident radiations for two different photosensitive matererial `M_(1)` and `M_(2)` What are the values of work functions for `M_(1)` and `M_(2)` respectively A. `hv_(01),hv_(02)`B. `hv_(02),hv_(01)`C. `hv_(01),hv_(01)`D. `hv_(02),hv_(02)` |
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Answer» Correct Answer - A `W=hv` |
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| 67. |
Two particles of masses `m` and `2m` have equal kinetic energies. Their de Broglie wavelengths area in the ratio of:A. `1:1`B. `1:2`C. `1:sqrt2`D. `sqrt2:1` |
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Answer» Correct Answer - D `because lambda=h/sqrt(2mK) K`.=kinetic energy `rArr lambda_(1)/lambda_(2)=sqrt(m_(2)k_(2))/sqrt(m_(1)k_(1))=sqrt(m_(2)/m_(1))=sqrt2/1` |
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| 68. |
The two lines A and B in fig. show the photo electron of de Broglie wavelength `(lamda)` as a function of `(1)/(sqrtV) `(V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of heavier mass?A. `A`B. `B`C. Both `A` and `B`D. Data insufficient |
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Answer» Correct Answer - A Slope `alpha1/sqrtm` |
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| 69. |
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 electron will be emittedB. Less electrons will be emittedC. More electrons will be emittedD. we can not predict |
| Answer» Correct Answer - A | |
| 70. |
A particle is droped from a height H. The de-broglie wavelength of the particle as a function of height is proportional toA. `H`B. `H^(1//2)`C. `H^(0)`D. `H^(-1//2)` |
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Answer» Correct Answer - D Velocity of a body falling from a height `H` is given by `v=sqrt(2gH)` We know that de-broglie wavelength `lambda=h/(mv)=h/(msqrt(2gH)) rArr =h/(msqrt(2g)sqrtH)` Here,`h/(msqrt2g)` is a constant `phi` say `K` So, `lambda=K1/sqrtH rArr prop 1/sqrtH rArrlambda prop H^(-1//2)` |
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| 71. |
1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, find the current in the cell.A. `0.59 muA`B. `1.16 muA`C. `0.48 muA`D. `0.79 muA` |
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Answer» Correct Answer - C `L^(0)=q/t=(Ne)/t=(plambdae)/(hc)` |
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| 72. |
Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then the ratio of maximum velocity of the photoelectrons emitted in the two cases respectively, isA. `sqrt2:1`B. `sqrt3:1`C. `sqrt3:2`D. `sqrt1:2` |
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Answer» Correct Answer - D `V_(1)/V_(2)=sqrt((E_(1)-omega_(0))/(E_(2)-omega_(0)))` |
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| 73. |
The photoelectric work function for a metal surface is `4.125 eV`. The cut - off wavelength for this surface isA. `4125 A^(@)`B. `2062.5 A^(@)`C. `3006.06 A^(@)`D. `6000 A^(@)` |
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Answer» Correct Answer - C `lambda "in" A^(0)=12400/E "in" eV` |
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| 74. |
If stopping potentials corresponding to wavelengths `4000A` and `4500A` are 1.3 V and 0.9 V, respectively, then the work function of the metal isA. `0.3eV`B. `1.3eV`C. `1.8eV`D. `5eV` |
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Answer» Correct Answer - C `hgamma=w+K.E rArr hgamma =w+V_(0)e` |
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| 75. |
The de - Broglie wavelength `lambda`A. mass of the particleB. size of the particleC. material of the particleD. shape of the particle |
| Answer» Correct Answer - A | |
| 76. |
The number of photons emitted per second by a `62W` source of monochromatic light of wavelength `4800 A^(@)` isA. `1.5xx10^(19)`B. `1.5xx10^(20)`C. `2.5xx10^(20)`D. `4xx10^(20)` |
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Answer» Correct Answer - B `P=(n/t)((hc)/lambda)` |
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| 77. |
Light rays of wavelength `6000A^(@)` and of photon intensity `39.6 watts//m^(2)` is incident on a metal surface. If only one percent of photons incident on the surface of electons emitted per second unit area from the surface will be [Planck constant =`6.64xx10^(-34) J-S`,Velocity of light =`3xx10^(8) ms^(-1)`]A. `12xx10^(18)`B. `10xx10^(18)`C. `12xx10^(17)`D. `12xx10^(15)` |
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Answer» Correct Answer - C `P=(n/t)((hc)/lambda)` |
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| 78. |
The debroglie wavelength associated with a particle of mass m, moving with a velocity v and energy E isgiven byA. `h//mv^(2)`B. `mv//h^(2)`C. `h//sqrt(2mE)`D. `sqrt(2mE)//h` |
| Answer» Correct Answer - C | |
| 79. |
What is a photon? Show that it has zero rest mass or photons can not exist at rest. Explain.A. zeroB. `1.6xx10^(-19)kg`C. `3.1xx10^(-30)kg`D. `9.1xx10^(-31)kg` |
| Answer» Correct Answer - A | |
| 80. |
The mass of a photon in motion is given its frequency = `x`)A. `(hx)/c^(2)`B. `hx^(3)`C. `(hx)^(3)/c^(2)`D. zero |
| Answer» Correct Answer - B | |
| 81. |
Debroglie wavelength of a particle at rest position isA. zeroB. finiteC. infinityD. cannot be calculated |
| Answer» Correct Answer - C | |
| 82. |
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.A. `h/(m_(A)v)[(m_(A)+(m_(B)))/(m_(A)-(m_(B)))-1]`B. `h/(m_(A)v)[(m_(A)-(m_(B)))/(m_(A)+(m_(B)))-1]`C. `h/(m_(A)v)[(m_(A)(m_(B)))/(m_(A)+(m_(B)))-1]`D. `h/(m_(A)v)[(m_(A)(m_(B)))/(m_(A)-(m_(B)))-1]` |
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Answer» Correct Answer - A From the law of conservation of momentum,`m_(A)v=m_(A)v_(A)+m_(B)v_(B)` or `m_(A)(v-v_(A))=m_(B)v_(B)`...(i) (as particle `B` is at rest, its initial velocity is zero and `v_(A)` and `v_(B)` are the velocities of particles `A` and `B` after collision) Since the collision is elastic, kinetic energy is conserved during collision `therefore1/2m_(A)v^(2)=1/2m_(A)v_(A)^(2)+1/2m_(B)v_(B)^(2)` `m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2)`..(ii) Dividing eqn.(ii) by eqn.(i), we obtain `(m_(A)(v^(2)-v_(A)^(2))=m_(B)v_(B)^(2))/(m_(A)(v-v_(A))=m_(B)v_(B)` or `v+v_(A)=v_(B)`...(iii) From eqns.(i) and (iii), `m_(A)(v-v_(A))=m_(B)(v+v_(A))` or `(m_(A)-m_(B))v=(m_(A)-m_(B))v_(A)` or `v/v_(A)=((m_(A)+m_(B))/(m_(A)-m_(B)))`..(iv) Initial wavelength of the particle `B`,i.e., `(lambda_(A))_(f)=h/(m_(A)v_(A))` Change in wavelength, `Delta lambda=(lambda_(A))_(f)-(lambda_(A))_(i)=h/(m_(A)v_(A))-h/(m_(A)v)` or `Delta lambda=h/(m_(A)v)[v/v_(A)-1]`..(v) From eqns. (iv) and (v), `Delta lambda=h/(m_(A)v)[((m_(A)+m_(B))/(m_(A)-m_(B)))-1]` |
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| 83. |
The uncertainity in the position of a particle is equal to the de-Broglie wavelength. The uncertainity in its momentum will beA. `h/lambda`B. `(2h)/(3lambda)`C. `lambda/h`D. `(3lambda)/(2h)` |
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Answer» Correct Answer - A `delta p=h/(Deltax)=h/lambda` |
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| 84. |
A metal of work function `4eV` is cexposed to a radiation of wavelegth `140xx10^(-9)m`. Find the stopping potential. |
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Answer» `E=(hc)/lambdaE=(6.62xx10^(-34)xx3xx10^(8))/(140xx10^(-9)xx1.6xx10^(-19))eV=8.86eV` work function `W_(0)=4eV` `eV_(0)=E-W_(0)=8.86-4=4.76eV` `:.` Stopping potential `V_(0)=4.86V` |
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| 85. |
The work function of a photosensitive element is 2ev. Calualate the velocity of a photoelectron when the element is exposed to a light of wavelength `4xx10^(3) overset(0)A`. |
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Answer» Einstein,s photoelectric equation is `1/2mv^(2)=(hc)/lambda-W_(0)` `1/2mv^(2)=(6.63xx3)/(4xx10^(3)xx10^(-10)) xx 10^(-26)-2xx1.6xx10^(-19)` `v^(2)=(1.765xx2)/9.1xx10^(12)` `v=sqrt((1.765xx2)/(9.1))xx10^(6)=6.228xx10^(5)ms^(-1)` |
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| 86. |
Photons of energy `2.0eV` fall on a metal plate and release photoelectrons with a maximum velocity `V`. By decreasing `lambda` and `25%` the maximum velocity of photoelectrons is doubled. The work function of the metal of the material plate in `eV` is nearlyA. `2.22`B. `1.985`C. `2.35`D. `1.80` |
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Answer» Correct Answer - D `1/2mv_(max)^(2)=hv-w` & `hv=(hc)/lambda` |
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| 87. |
The work function of a metal is `3.0eV`.It is illuminated by a light of wave length `3xx10^(7)m`.Calculate i) threshold frequency, ii)the maximum energy of photoelectrons, iii) the stopping potential. `(h=6.63xx10^(-34)Js` and `c=3xx10^(8)ms^(-1))` |
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Answer» i)`W=hnu_(0)=3.0eV=3xx1.6xx10^(-19)J` Threshold frequency `nu_(0)=W/h=(3xx1.6xx10^(-19))/(6.63xx10^(-34))=0.72xx10^(15)Hz`. ii) Maximum kinetic energy `(k_(max))=h(nu-nu_(0))` `lambda=3xx10^(-7)m,u=c/lambda=(3xx10^(8))/(3xx10^(-7))=1xx10^(15)Hz` `K_(max)=h(nu-nu_(0))=6.63xx10^(-34)(1-0.72)xx10^(15)` `J=1.8xx10^(-19)J`. iii) `K_(max)=eV_(0)` where `V_(0)` is stopping potential in volt and `e` is the charge of electron `V_(0)=K_(max)/r`. Here `K_(max)=1.86xx10^(-19)J` and `e=1.6xx10^(-19)C, V_(0)=(1.86xx10^(-19)J)/(1.6xx10^(-19)C)=1.16V` |
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| 88. |
Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearlyA. `9.61xx10^(14)`B. `4.12xx10^(13)`C. `1.51xx10^(12)`D. `2.13xx10^(11)` |
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Answer» Correct Answer - C `P=(n/t)((hc)/lambda)` |
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| 89. |
Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface. Calculate: a. the value of work function in joule and in eV. b. maximum kinetic energy of the emitted photoelectrons. c. the maximum velocity of the photoelectrons. (Mass of the electrons`=9.11xx10^(-31)`).A. `1.51`B. `2.36`C. `3.85`D. `4.27` |
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Answer» Correct Answer - A `E=W.E.+K.E.` |
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| 90. |
At stopping potential, the photo electric current becomesA. MinimumB. MaximumC. ZeroD. Infinity |
| Answer» Correct Answer - C | |
| 91. |
The necessary condition for photo electric emission isA. `hupsilon le hupsilon_(0)`B. `hupsilon ge hupsilon_(0)`C. `E_(k) gt hupsilon_(0)`D. `E_(k) lt hupsilon_(0)` |
| Answer» Correct Answer - B | |
| 92. |
The threshold wavelength of lithium is `8000 A^(0)` When light of a wavelength `9000 A^(0)` is made to be incident on it, then the photo electronsA. Will not be emittedB. Will be emittedC. Will sometimes be emitted and sometimes notD. Data insufficient |
| Answer» Correct Answer - A | |
| 93. |
The process of photo electric emission depends onA. Temperature of incident lightB. Nature of surfaceC. Speed of emitted photo electronsD. Speed of the incident light |
| Answer» Correct Answer - B | |
| 94. |
In an experiment of photo electric emission for incident light of `4000 A^(0)`,the stopping potentail is `2V`.If the wavelength of incident light is made `300 A^(0)`, then the stopping potential will beA. Less than `2` voltB. More than `2` voltC. 2 voltD. zero |
| Answer» Correct Answer - B | |
| 95. |
An electron of charge `e` and mass `m` is accelerated from rest by a potential difference `V`. the de Broglie wavelength isA. Directly proportional to the square root of potential difference.B. Inversely proportional to the square root of potential difference.C. Directly proportional to the square root of electron massD. Inversely proportional of the cube root of electron mass |
| Answer» Correct Answer - B | |
| 96. |
Electrons are accelerated through a potential difference of `150V`. Calculate the de broglie wavelength. |
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Answer» `V=150V,h=6.62xx10^(-34)Js,m=9.1xx10^(-31)kg,e=1.6xx10^(-19)C` `:.lambda=h/sqrt(2Vem)=(6.62xx10^(-34))/sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx150)=1overset(0)A` |
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| 97. |
Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures `27^circ` and `127^circC`, respectively. |
| Answer» Since,`lambda=h/(mupsilon)=h/sqrt(3mkT),(lambdaH)/(lambdaHe)=sqrt((m_(He)T_(He))/(m_(H)T_(H)))=sqrt(8/3)` | |
| 98. |
With what velocity must an electron travel so that its momentum is equal to that of a photon with a wavelength of `5000overset(0)A` `(h=6.6xx10^(-34)Js,m_(e)=9.1xx10^(-31) Kg)` |
| Answer» `mv=h/lambdarArrv=(6.6xx10^(-34))/(9.1xx10^(-31)xx5000xx10^(-10))=1450m//s` | |
| 99. |
If electron is having a wavelength of `100 A^(0)` then momentum is `(gm cm s^(-1))` unitsA. `6.6xx10^(-32)`B. `6.6xx10^(-29)`C. `6.6xx10^(-25)`D. `6.6xx10^(-21)` |
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Answer» Correct Answer - D `P=h/lambda` |
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| 100. |
The de Broglie wavelength of an electron and the wavelength of a photon are same. The ratio between the energy of the photon and the momentum of the electron isA. `h`B. `c`C. `1//h`D. `1//c` |
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Answer» Correct Answer - B `lambda_(e) = lambda_(ph), (E_(ph))/(m_(e)V) =(E_(ph))/(((h)/(lambda_(e)))) =((hc)/(lambda_(ph)))/((h)/(lambda_(e))) =C` |
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