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Particle Nature Of Light Mcq
Quiz 2
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Q.1
When photons of energy hν fall on an aluminium plate ( of work function Φ), photoelectrons of maximum kinetic energy of K are ejected. If the frequency of the radiation is doubled then ejected photoelectrons will be.. [ CBSE-PMT 2006]
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a)2K
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b)K
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c)K + hν
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d)K + Φ
Explanation
Applying Einstein's formula fro photo-electricityhν=Φ + K; here K is kinetic energy of electronIf we use 2ν frequency then let kinetic energy becomes K' soh.2ν=Φ + K'From above two equationsK'=hν + K Answer: (c)
Q.2
The X-rays cannot be diffracted by means of an ordinary grating because of .. [ CBSE-PMT 1997]
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a) high speed
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b) short wave length
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c)large wave length
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d)none of these
Explanation
We know that the X-rays are of short wave length as compared to grating constant of optical grating. As a result of this, it makes difficult to observe X-rays diffraction with ordinary gratingAnswer: (b)
Q.3
The wave length associated with an electron accelerated through a potential difference of 100V, is of the order of .. [ CBSE-PMT 1996]
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a) 1000 Å
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b) 100 Å
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c)10.5 Å
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d)1.2 Å
Explanation
Potential difference V=100V. Kinetic energy=P.E Kinetic energy=qV Kinetic energy=1.6×10-19 ×100=1.6×10-17 J now energy E=hc/ λ ∴ λ=hc / E Answer: (d)
Q.4
In the Davisson and Germar experiment, the velocity of electrons emitted from the electron gun can be increased by[ CBSE-PMT 2011]
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a) increasing the potential difference between the anode and filament
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b) increasing the filament current
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c) decreasing the filament current
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d) decreasing the potential difference between the anode and filament
Explanation
In Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and filament. Answer: (a)
Q.5
The number of photo electrons emitted for light of frequency ν ( higher than the threshold frequency ν0 is proportional to [ CBSE - PMT 2009]
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a) Threshold frequency ( ν0
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b) Intensity of light
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c)Frequency of light ν
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d)ν - ν0
Explanation
The number of photoelectrons emitted is proportional to the intensity of incident light. Saturation Current ∝ IntensityAnswer: (b)
Q.6
Light of two different frequencies whose photos have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5eV successively. Ratio of maximum speeds of emissions will be [ CBSE-PMT 2011]
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a)1:4
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b) 1:2
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c)1:1
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d)1:5
Explanation
The mximum kinetic energy of emitted electrons is given by E=Φ - Φ0E1=1eV - 0.5eV=0.5eVE2=2.5eV - 0.5eV=2eVNow E=½ ( m v2)Thus E ∝ v2 ∴ E1 / E2=v12 / v22 ∴ 1/4=v12 / v22 ∴ v1 / v2=1/2 Answer:(b)
Q.7
A beam of cathode rays is sunjected to crossed Electric field(E) and Magnetic field (B). The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by [ CBSE-PMT 2010]V is the potentia; difference between cathode and anode
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a)
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b)
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c)
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d)
Explanation
For no deflection of beam eE=evB ∴ v2=E2/B2 also ½ ( m v2)=eV ∴ v2=2eV / m ∴ E2/B2=2eV / m ∴ e/m=E2 / 2VB2 Answer: (d)
Q.8
The work function of a surface of a photosensitive material is 6.2 eV. The wave length of incident radiation for which the stopping potential is 5V lies in the [ CBSE-PMT 2008]
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a)Ultraviolet region
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b) Visible region
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c)Inferred region
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d)X-ray region
Explanation
Work functio Φ=6.2eVStopping potential V0=5VAs eV0=hc/ λ - Φ Or Thus the wave length of the incident radiation lies in the ultraviolet regionAnswer: (a)
Q.9
A photo-cell employ photoelectric effect to convert [ CBSE-PMT 2006]
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a) Change in the intensity of illumination into a change in photoelectric current
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b) Change in the intensity of illumination into a change in the work function of the photocathode
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c)change in the frequency of light into a change in the electric current
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d)change in the frequency of light into a change in electric voltage
Explanation
A photo-cell employs photoelectric effect to convert light energy into photoelectric currentAnswer: (a)
Q.10
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function 5.01 eV, when ultraviolet light of 200nm falls on it, must be..[ CBSE-PMT 2010]
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a) 2.4V
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b) -1.2V
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c)-2.4V
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d)1.2V
Explanation
Answer:(d)
Q.11
Photoelectric work function of a metal is 1eV. Light of wave length λ=3000 Å falls on it. The photo electron come out with velocity [ CBSE-PMT 1991]
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a) 10 m/sec
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b) 102 m/sec
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c) 104 m/sec
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d) 106
Explanation
We know that hν=Φ + ½ ( m v2) Φ=1eV=1× 1.6×10-19 J Answer: (d)
Q.12
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of 5000 Å its work function is [ CBSE-PMT 1988]
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a)4 ×10-19 J
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b) 1 J
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c)2× 10-19 J
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d)3 ×10-19 J
Explanation
Answer: (a)
Q.13
Electrons used in an electron microscope are accelerated by a voltage of 25kV. If the voltage is increased by 100kV then the de-Broglie wavelength associated with electrons would.. [ CBSE-PMT 2011]
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a) increase by 2 times
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b) decrease by 2 times
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c)decrease by 4 times
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d)increase by 4 times
Explanation
λ ∝ 1/ √ V ∴ Answer: (b)
Q.14
A particle of mass 1mg has the same wave length as an electron moving with velocity of 3×106 ms-The velocity of the particle is [ CBSE-PMT 2008] mass of electron=9.1×10-31 kg
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a)2.7×10-18 m/s
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b) 9×10-2 m/s
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c)3×10-31m/s
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d)2.7×10-21m/s
Explanation
let m1 be the mass of particle, and velocity be v1me be mass of electron , and ve be the velocity of electronSince wave length is same thus h/p1=h/ pe ∴ m1 v1=me ve v1=me ve / m1on substitution we get Answer:(d)
Q.15
Monochromatic light of wave length 667 nm is produced by a helium neon laser. The power emitted is 9mW. The number of photons arriving per second on the average at a target irradiated by this beam is [ CBSE-PMT 2009]
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a) 3×1016
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b) 9×1015
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c) 3×1019
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d) 9×1017
Explanation
if n is number of photons per second then p=nh (c/λ) n=(pλ) / (hc) Answer: (a)
Q.16
In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of [ CBSE-PMT 2008]
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a)excitation of electons in the atoms
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b) Collision between the atoms of the gas
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c)Collision between the charges particles emitted from the cathode and the atoms of the gas
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d)Collision between different electrons of atoms of the gas
Explanation
The coloured glow in the tube appears as a result of excitations of electrons in atoms.Answer: (a)
Q.17
In discharge tube ionization of enclosed gas is produced due to collisions between [ CBSE-PMT 2006]
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a) negative electrons and neutral atoms / molecules
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b) photons and neutral atoms/ molecule
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c)neutral gas atoms / molecules
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d)positive ions and neutral atoms/molecule
Explanation
When electrons emitted from cathode collide with gas molecules or atoms, they knock out outer electrons and produce positively charged ions. They become part of positive raysAnswer: (a)
Q.18
Photoelectric emission occurs only when the incident light has more than a certain minimum [ CBSE-PMT 2011]
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a) power
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b) wavelength
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c)intensity
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d)frequency
Explanation
Answer:(d)
Q.19
When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If a second beam causes photoelectrons to be ejected, it may consist of [ CBSE-PMT 2002]
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a) infra-red waves
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b) X-rays
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c) Visible light rays
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d) radio waves
Explanation
Energy of photon of X-rays is more than energy of photon of ultraviolet rays. Because frequency of X rays is more than ultraviolet rays. Answer: (b)
Q.20
A source S1 is producing, 1015photons per second of wavelength 5000Å . Another source S2 is producing 1.02×1015 photon per second of wave length 5100Å, then ratio of power of Source S2 and power of Source S1 [ CBSE-PMT 2010]
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a)1.00
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b) 1.02
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c)1.04
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d)0.98
Explanation
Energy emitted per second by S1Energy emitted per second by S2Answer: (a)
Q.21
A photoelectric cell is illuminated by a point source of light 1m away. When the source is shifted to 2m then [ CBSE-PMT 2003]
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a) number of electrons emitted is a quarter of initial number
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b) each emitted electron carries one quarter of the initial energy
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c)number of electrons emitted is half the initial number
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d)each emitted electron carries half the initial energy
Explanation
Power P ∝ Number of electrons emitted (N)and p ∝ 1/ r2 ∴ N ∝ 1/ r2 since distance is doubled intensity is one quarter of the initialAnswer: (a)
Q.22
When light of wave length 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of 600nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters? [ CBSE-PMT 1993]
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a) 1:2
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b) 2:1
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c)4:1
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d)1:4
Explanation
Work function Φ=hc/ λ ∴ Φ ∝ 1 / λ Answer:(b)
Q.23
A radio transmitter operates at a frequency 880kHz and a power of 10kW. the number of photons emitted per second is .. [ CBSE-PMT 1990]
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a) 1.72×1031
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b) 1.327×1025
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c) 1.327×1037
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d) 1.327×1045
Explanation
Number of photons emitted per sec n=power / Energy of Photon n=P/nν Answer: (a)
Q.24
The wave length of 1keV photon is 1.24×10-9m. What is the frequency of 1MeV photon?
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a)1.24×1015
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b) 2.4×1020
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c)1.24×1018
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d)2.4×1023
Explanation
Here hν=103 eV and λ=1.24×10-9m.Now ν=c/λ ν=3×108 / 1.24×10-9ν=2.4×1017∴ frequency of 106 will be Thus frequency of 106 is 2.4×1020 Answer: (b)
Q.25
An electron of mass m and charged e is accelerated from rest through a potential difference of V volt in vacuum. Its final speed will be.. [ CBSE-PMT 1996]
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a)
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b)
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c)
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d)
Explanation
Kinetic energy of electron accelerated through a potential V is eVthus kinetic energy=Potential energy½ ( m v2)=eVAnswer: (c)
Q.26
The de-Broglie wavelength associated with a proton changes by 0.25% if its momentum is changed by Po. The initial momentum of proton is
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a)100Po
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b) 400Po
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c)Po/400
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d)Po
Explanation
Answer: (b)
Q.27
Work function for metals A, B, and C are respectively 1.92eV, 2.0eV and 5eV According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength 4100Å
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a)none
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b) A only
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c)A and B only
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d)All three metals
Explanation
E=hc/ λ So, only metals having work function less than 3eV can emit photoelectrons from the incident radiation of wave length 4100 Å Answer:(c)
Q.28
A 5 watt source emits monochromatic light of wavelength 5000Å. When placed 0.5m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0m, the number of photoelectrons liberated will be reduced by a factor of [ CBSE-PMT 2007]
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a) 8
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b) 16
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c) 2
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d) 4
Explanation
Number of electrons emitted ∝ 1 / (distance)2 Since the distance is doubled number of electrons emitted will be 1/4 of initial Answer: (d)
Q.29
The threshold frequency for a photosensetive metal is 3.3×1014 Hz. If light of frequency 8.2×1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly [ CBSE-PMT 2011]
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a)2V
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b) 3V
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c)5V
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d)1V
Explanation
K.E.=hν - hν0=eV0Answer: (a)
Q.30
The cathode of photoelectric cell is changed such that the work function changes from W1 to w2 ( W1 > w2). If the current before and after changes are I1 and Iall other conditions remain unchanged, then ( assuming hν > W2) [ CBSE-PMT 1992]
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a) I1=I2
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b) I1 < I2
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c)I1 > I2
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d)I1< I2<2I1
Explanation
The work function has no effect on photoelectric current so long as hν > W0. The photoelectric current is proportional to the intensity of incident light. Since there is no change in the intensity of light, hence I1=I2Answer: (a)
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