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Quiz 1
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Q.1
in photoelectric effect work function of any metal is 2.5eV. Emitted electrons are stopped by the potential of 1.5volt then
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a)energy of incident photons is 4eV
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b) energy of incident photons is 1eV
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c)photoelectric current increases when we use photos of high frequency
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d)none of the above
Explanation
Answer: (a)
Q.2
If the wavelength of incident light changes from 4000Å to 3600Å, change in stopping potential will be
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a) +0.35V
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b) -0.35V
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c)+0.4V
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d)-0.4 V
Explanation
Answer: (a)
Q.3
The slope of the graph Kmax Vs f in photo electric effect is
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a) h
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b) h/e
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c)he
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d)e/h
Explanation
K=hf - φ on comparing with standard equation for line we get slope=hAnswer: (a)
Q.4
In photocell, energy conversion is from...
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a) chemical to electrical
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b) mechanical to electrical
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c)optical to electrical
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d)magnetic to electrical
Explanation
Answer:(c)
Q.5
The photoelectric currents at distance r1 and r2 of a light source from a photocell are I1 and I2 respectively. What is the value of I2 / I1 ?
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a) r1 / r2
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b) r2 / r1
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c) (r1 / r2)2
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d) (r2 / r1)2
Explanation
we know that current ∝ Intenisty ∝ 1/r2 Answer: (c)
Q.6
Ratio of momentum of 105eV, X-ray photon (P) with that of 105eV electron (P')
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a) P'/P=1/2
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b) P'/P=16/5
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c) P'/P=1/5
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d) P'/P=5/1
Explanation
Use formula for X-ray P=E/c=eV/c and for electrons P'=√(2eVm), here m is the mass of electron Answer: (b)
Q.7
An atom emits a photon of wavelength 1 Å. The energy of recoil of the atom will be ( mass of atom=1 amu)
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a) 1.304 × 10-20 J
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b) 1.304 eV
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c) 1.532 × 10-19J
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d) 1.532 eV
Explanation
use formula Answer: (a)
Q.8
The de-Broglie wavelength of a proton and alpha particle is same, the ratio of their velocities is
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a)1:4
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b) 1:2
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c)2:1
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d)4:1
Explanation
λ=h/mv and mass of alpha particle is four times mass of protonAnswer: (d)
Q.9
The energy of an electron having de_broglie wavelength λ is
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a) h/2m
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b) h2/2mλ
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c)h2/2λ2
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d)h2/2mλ2
Explanation
Answer: (d)
Q.10
If the momentum of a particle is doubled, then its de-Broglie wavelength will become.. [ AFMC 1997]
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a)unchanged
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b) four times
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c)two times
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d)half times
Explanation
de-Broglie wave length λ=h / p If p is doubled then λ will be halved.Answer: (d)
Q.11
The threshold frequency for a photosensitive metal is 3.3×1014Hz. If light of frequency 8.2×1014Hz 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
V0 is stopping potential ν0 is threshold frequencyeV0=hν -hν0 V0=h/e( ν - ν0)Answer: (a)
Q.12
Light of wave length 5000 Angstrom falls on a sensitive plate with photoelectric work function of 1.9 eV. The maximum kinetic energy of the photo electron emitted will be..[ AFMC 1997]
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a) 1.16 eV
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b) 2.38 eV
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c)0.58 eV
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d)2.98 eV
Explanation
Energy of incident radiation=hc/λ Now according to Einstein's formulaK.E of photo electron=Incident energy - work functionK. E. of photo electron=2.475-1.9=0.58eVAnswer: (c)
Q.13
The work function of a metallic substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from the substance is approximately.. [ AFMC 1998]
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a)220 nm
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b) 310 nm
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c)400 nm
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d)540 nm
Explanation
Work function Φ=hc/λλ=hc/Φgiven Φ=4 × 10-16 J Answer:(b)
Q.14
The work function of aluminum is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminum, then emission of electron will be ..[ AFMC 1999]
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a) depends up on the density of the surface
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b) data is incomplete
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c) not possible
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d) possible
Explanation
photoelectron will be emitted if energy of each photon is equal to or more than work function In problem energy of photon is less than work function hence emission is not possible Answer: (c)
Q.15
The de-Broglie wave length of an electron of energy 600eV is [ AFMC 2000]
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a) 4 Angstrom
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b) 2 Angstrom
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c) 1 Angstrom
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d) 0.5 Angstrom
Explanation
We can use following formula By substituting values we get Above formula can be as follows:- We know that E = p2 / (2m) p = √ (2mE) --eq(1) and λ = h/ p --eq(2) substituting value of p from equation (1) in equation (2) we get above mention formula Answer: (d)
Q.16
The wave length of a particle having a momentum of 2×10-28 kgm/s is ..[ AFMC 2002]
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a) 3.3×10-6m
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b) 3.3×105 m
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c) 3.3×10-4m
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d) 30 m
Explanation
Wave length λ = h/p λ = 6.6×10-34 / 2×10-6 λ = 3.3×10-6 Answer: (a)
Q.17
What is the de-Broglie wavelength of 1 kg mass moving with a velocity of 10 m/s? [ AFMC 2001]
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a) 6.626×10-35 m
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b) 6.626×10-33 m
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c) 6.626×10-34 m
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d) none of these
Explanation
λ = h/p = h/(mv) λ = 6.6×10-34 / (1×10) λ = 6.6×10-35 Answer:(a)
Q.18
The magnitude of saturation photoelectric current depends upon... [ AFMC 2005]
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a) frequency
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b) intensity
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c) work function
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d) stopping potential
Explanation
The magnitude of saturation photoelectric current depends upon the intensity of radiation because higher the intensity of radiation, larger number of electrons coming out giving rise to increased intensity of photoelectric current, but frequency must be more or equal to threshold frequency Answer: (b)
Q.19
According to Einstein's photoelectric equation, the plot of kinetic energy of the emitted photoelectrons from a metal vs frequency of the incident radiation gives a straight line whose slope.. [ AFMC 2004]
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a)depends on the intensity of radiation
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b) depends on the nature of the metal used
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c)depends both on the intensity of the radiation and the metal used
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d)is the same for all metals and independent of the intensity of the radiation
Explanation
Einstein's photoelectric equation, is hν=Φ + EHere hν is energy of incident radiationΦ is work functionE is energy of electron ν=Φ/ h + (1/h) E This line is straight line ,slope of the line is 1/h which is constant So, it depends on metals usedAnswer: (d)
Q.20
For photoelectric emission, tungsten requires light of 2300Å. If light of 1800Å wave length is incident then emission ...[ AFMC 2005]
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a) takes place
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b) doesn't take place
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c)may or may not take
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d)depends on frequency
Explanation
Since light of lower wavelength than threshold wave length is used, emission will take place.Answer: (a)
Q.21
If the minimum energy of photons needed to produce photoelectric effect is 3eV, bombarding the photoelectric material by a number of photons of 2.5eV also one can get...[ AFMC 2006]
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a) photoelectrons of the same kinetic energy
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b) photoelectrons of higher kinetic energy
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c)higher current
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d)none
Explanation
As one photon is necessary to eject one photoelectron, more than one photon are not required for ejection of photoelectron. Hence, the energy of photon should be equal to or greater than the minimum energy needed to produce photoelectron.Since in the given problem, energy of photon is 2.5eV is less than the minimum energy 3eV needed to produce photoelectric effect, hence no photoelectrons will be produced. Answer:(d)
Q.22
What is incorrect about photon? ... [ AFMC 2009]
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a) Its rest mass is zero
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b) its energy is hν
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c) Its momentum is (hν)/c
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d) It does not exert pressure
Explanation
A photon exerts pressure as it has momentum and energy Answer: (d)
Q.23
The momentum of pfoton of an electromagnetic radiation is 3.3×10-29 kgms-What is the frequaency of the associated waves? [ CBSE-PMT 1990] h=6.6×10-34Js; c=3×108 ms-1
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a)1.5 × 1013 Hz
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b) 7.5 × 1012 Hz
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c)6.0×103
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d)3.0×103
Explanation
As λ=h/p and λ=c/ν soν=cP/hby substituting the values we gettν=1.5× 1013 HzAnswer: (a)
Q.24
Number of ejected photoelectron increases with increase ... [ CBSE-PMt 1993]
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a) in intensity of light
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b) in wave length of light
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c)in frequency of light
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d)never
Explanation
Answer: (a)
Q.25
Which of the following moving particles ( moving with same velocity) has largest wave length of matter waves [ CBSE-PMT 2002]
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a) Electron
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b) α-particle
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c)Proton
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d)Neutron
Explanation
de-Broglie wave length λ=h/ mvFor same velocity, λ ∝ 1/m Out of given particles, the mass of electron is minimum, so the associated de-Broglie wave length is maximum for electron Answer:(a)
Q.26
Monochromatic light of frequaency 6.0×1014 Hz is produced by a laser. the power emitted is 2×10-3 W. The number of photons emitted, on the average, by the sources per second is ... [ CBSE-PMT 2007]
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a) 5×1016
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b) 5×1017
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c) 5×1014
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d) 5×1015
Explanation
Since power p=nhν here n is number of photons per second ∴ n=p/ hν Answer: (d)
Q.27
Which of the following statement is correct? [ CBSE-PME 1997]
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a) Photo-current increases with intensity of light
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b) Photo-current is proportional to the applied voltage
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c)Current in photocell increases with increasing frequency
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d)Stopping potential increases with increase of incident light
Explanation
Answer: (a)
Q.28
Which of the following statement is correct? [ CBSE-PME 1997]
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a) Photo-current increases with intensity of light
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b) Photo-current is proportional to the applied voltage
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c)Current in photocell increases with increasing frequency
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d)Stopping potential increases with increase of incident light
Explanation
Answer: (a)
Q.29
The photoelectric work function for a metal surface is 4.125eV. The cut off wavelength for this surface is . [ CBSE-PMT 1999]
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a) 4125 Å
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b) 3000 Å
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c)6000 Å
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d)2062.5 Å
Explanation
let λ0 be cut off wave lengthWork function Φ=hc/ λ0 ∴ λ0=hc/ Φ here Φ=4.125×1.6×10-19h=6.26×10-34c=3×108 Substituting values in above equation on solving we getλ=3000Å Answer:(b)
Q.30
The momentum of photon of energy 1 MeV in kg m/s, will be [ CBSE-PMT 2006]
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a) 7×10-24
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b) 10-22
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c) 5×10-22
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d) 0.33×106
Explanation
1MeV=106×1.6×10-19 Momentu of photon=E/c Momentum of photon=1.6×10-13 / 3×108 Momentum of photon=5×10-22 kg m/sec Answer: (c)
0 h : 0 m : 1 s
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