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Particle Nature Of Light Mcq
Quiz 8
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Q.1
the energies of an photon of a mass m and electron are same. velocity of electron is v, the ratio of wave lengths associated with them will be
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a)
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b)
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c)
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d)
Explanation
As de-broglie wavelength is same mc=me v me=mc / v --eq(1) wave length of electron λe=h/ √(2me ) from eq(1) we get As energies are same Wave length of photo λ=hc/E by taking ratios of wavelength we get Answer: (c)
Q.2
Two particles of same mass m1 and m2 respectively are identically charged and are accelerated by same potential. If de-Broglie wavelength associated with them are λ1 and λ2 then
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a) λ1 / λ2=m2 / m1
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b) λ1 / λ2=√( m2 / m1)
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c)λ1 / λ2=m1 / m2
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d)λ1 / λ2=√( m1 / m2)
Explanation
energies are same because of same charge and same potential use equation for wavelength in terms of energy Answer: (b)
Q.3
An electron is 2000 times lighter than a proton. An electron and a proton are moving with such a velocity that de-Broglie wave associated with them is 1 Å
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a) 1:2000
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b) 2000 : 1
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c)1:1
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d)1:(4.0106)
Explanation
USe formula for E for both the particles, given wavelength are same Answer: (b)
Q.4
A double slit interference experiment is performed by a beam of electrons of energy 100eV and fring spacing is observed to be β. Now if the electrons energy is increased to 10keV, then the fring spacing
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a) remains the same
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b) becomes 10β
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c)becomes 100β
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d)becomes β/10
Explanation
formula for fringe width β=λD/d Initial electron energy E ∝ 1/λ2 Thus λ ∝ 1/ √ENow energy is 100 times of original thus λ' ∝ 1/√(100E) λ ∝ 1/10√EOr λ'=λ/10∴ β'=λ'D/dβ'=λ'd/(10 ×D)β'=β/10 Answer:(d)
Q.5
An electron beam of energy 10keV is passed through a slit of width 1 mm. the observed phenomenon will be
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a) interference
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b) diffraction
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c) rectilinear propogation
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d) polarisation
Explanation
To observe diffraction λ/d should be equal to one In given problem wavelength of electron will be very small compared to slit width thus we will observe rectilinear propagation Answer: (c)
Q.6
The wave-nature of electron was verified by
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a)photoelectric effect
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b) Compton effect
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c)the incidence of electron on metallic surface
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d)diffraction of electron by crystal
Explanation
Answer: (d)
Q.7
In Davisson -Germaer experiment, electrons are produced by
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a) the photo-electric effect
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b) the thermionic emission
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c)the electric effect
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d)Schottky effect
Explanation
Answer: (b)
Q.8
Wrong statement in connection with Devisson-Germer experiment is
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a) The inter-atomic distance in Nickel crystal is of the order of the de-Broglie wavelength
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b) Electrons of constant energy are obtained by the electron gun
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c)Nickel crystal acts as a three dimensional diffracting grating
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d)Davision-Germer experiment is an interference experiment
Explanation
Answer:(d)
Q.9
In Devision-Germer experiment maximum intensity is observed at
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a) 50° and 54 volts
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b) 54° and 50 volts
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c) 50° and 50 volt
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d) 65° and 50 volt
Explanation
Answer: (a)
Q.10
the interatomic distance between atom in a crystal is 2.8 Å. then if such a crystal is used in Davisson-Germer experiment, the maximum order of diffraction that can be observed for a beam of electrons accelerated by 100V shall be
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a)n=1
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b) n=2
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c)n=10
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d)n=∞
Explanation
From the formula for wavelength in Å, λ=√(150/V)=√(150/100)for the equation for diffraction nλ=d sinθFor maximum n θ=90°n×√(150/100)=2.8 n=2.8 /√1.5 n=2.28n is integer thus n=2 Answer: (b)
Q.11
The angle between the incident and the diffracted electron in the Davisson-Germer experiment is called as
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a) angle of incidence
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b) angle of diffraction
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c)angle of scattering
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d)none of above
Explanation
Answer: (b)
Q.12
The distance between two consecutive atoms of the crystal lattice is 1.227 Å . the maximum order of diffraction of electrons accelerated through 104 volts will be
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a) 10
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b) 1/10
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c)100
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d)1/100
Explanation
Wavelength in Å λ=√(150/V)=√(150/10000)from the formula for diffraction nλ=dsin90n=d/λn=1.227 × √(10000/150)=10 Answer:(a)
Q.13
If the spacing between lattice plane of a crystal be d and wavelength of electron be λ, then for observing the diffraction of electron from the crystal, it is essential that
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a) 2d ≥ λ
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b) λ ≥ 2d
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c) 4d ≥ λ
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d) λ ≥ d
Explanation
Answer: (a)
Q.14
A proton when accelerated through a potential difference of V volt has a wavelength λ associated with it . An alpha particle , in order to have same λ, must be accelerated through a potential difference of
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a)V volt
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b) 4V volt
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c)2V volt
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d)V/8 volt
Explanation
energy of particle E=qV Thus for proton mass of Alpha particle is four times mass of proton and charge on Alpha particle is twice the charge on proton thus Answer: (d)
Q.15
The potential in electron microscope is increased from 3kV to 9kV, if the initial resolving power of microscope is R then, its new resolving power will be
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a) R
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b) 2R
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c)√3 R
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d)√5 R
Explanation
Wavelength λ ∝ 1/√V and resolving power R ∝ λ∴ R ∝ 1/√V New potential is 3V Thus R ∝ 1/√3V and R' ∝1/3V R'/R=√3 R'=√3 RAnswer: (c)
Q.16
Silver has workfunction of 4.7eV. When ultraviolet light of wavelength 100nm is incident upon it, a potential of 7.7 volt is required to stop the photoelectrons from reaching the collector plate. the potential required to stop photo electrons when light of wavelength 200nm is incident upon silver is
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a) 1.5V
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b) 1.85 V
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c)1.95 V
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d)2.37 V
Explanation
use formula put data for first case to determine value of hc/e then substitute data of second case and value of hc/e again in above equation to determine stopping potential Answer:(a)
Q.17
When a light source is placed at a distance of 1m from the emitter, it emits electrons of energy 4eV. IF the distance is changed to 0.5m, then
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a) the number of electrons emitted will be same but their energy will become double
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b) the number of electrons emitter will be same but there energy will become four times
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c)it will emit twice the number of electrons of same energy
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d)it will emit four times the number of electrons in earlier case with same energy
Explanation
since the distance is reduced to half intensity will become four times [ Intensity=energy / area]Number of electrons emitted ∝ intensity Answer:(d)
Q.18
Light of wavelength 5000Å and intensity 3.98 mW/cm2 is incident on a light sensitive surface. If 1% photons of incident light causes emission of the photoelectrons, then how many electrons will be emitted from 1m2 area of the surface
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a) 1016
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b) 1018
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c) 1020
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d) 1022
Explanation
energy per sqm of area=3.98mJ E=nhc/λhere n the number of photon per sqm incident in one second 1% of n=1016Answer: (a)
Q.19
The kinetic energy of the photoelectrons is E. When the incident wavelength is λ/The Kinetic energy become 2E when the incident wavelength is λ/The work function of the metal is
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a)hc/λ
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b) 2hc/λ
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c)3hc/λ
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d)hc/3λ
Explanation
Answer: (a)
Q.20
The electric field associated with a light wave is E=Eo sin [1.57 × 107( x-ct)] where x is in metrre and t is in second. If this light is used to produce photoelectric emission from the surface of metal of workfunction 1.9eV, then the stopping potential will be
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a) 1.2V
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b) 1.5 V
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c)1.75 V
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d)1.9 V
Explanation
from the given equation k=1.57×107 Thus λ=2π/ 1.57×107 substitute above value in following equation to get stopping potentialAnswer: (a)
Q.21
In the diagram, graphs are drawn between stopping potential Vo and frequency ν for the elements K and Ca. According to this to diagram
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a)workfunction of K and Ca are equal
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b) workfunction of K is greater than that of Ca
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c)the workfunction of K is less than that of Ca
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d)no information can be obtained about the workfunction
Explanation
Answer:(c)
Q.22
When light of intensity 1 W/m2 and wavelength 5×10-7 is incident on a surface. It is completely absorbed by the surface. If 100 photons emit one electron and area of surface is 1cm2, then the photoelectric current will be
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a) 2mA
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b) 0.4 µA
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c) 4.0 mA
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d) 4 µA
Explanation
Energy incident on 1sqcm surface=10-4 J IF n are the number of photons striking 1sqcm area then energy E=nhc/λ Only one percent of incident photons emits electrons thus number of electrons emitted N=0.2525 ×1013 Now Current I=Ne/t here t=1 sec I=0.2525×1013=0.4 µA Answer: (b)
Q.23
Electrons of 0.5 eV energy are emitted from a metal surface when photons of wavelength 3000Å are incident. the energy of electrons, when photons of 2000Å are incident ill be
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a)equal to 0.5 eV
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b) higher than 0.5eV
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c)less than 0.5eV
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d)none of the above
Explanation
Answer: (b)
Q.24
Lights of two different frequencies, whose photons have energies 1 and 2.5 eV respectively, successively illuminate a metal whose workfunction is 0.5eV. The ratio of the maximum speed of the emitted electron will be
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a) 1 : 5
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b) 1:4
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c)1:2
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d)1:1
Explanation
use formula ½ mv2=(hν -φ)×e [ as data is in eVAnswer: (c)
Q.25
Photoelectrons of energy 2eV are emitted from a metal surface when photons of energy 5eV are incident on it. What will be the energy of emitted photoelectrons when photons of energy 6eV are incident on it
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a) 3eV
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b) 4eV
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c)2eV
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d)5eV
Explanation
Answer:(a)
Q.26
If 5% of the energy supplied to a bulb is radiated as visible light, how many quanta are emitted per second by a 100 watt lamp? Assume wavelength of visible light as 5.6×10-5 cm
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a) 1.4 × 1019
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b) 2.0 × 10-4
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c) 1.4 × 10-19
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d) 2.0 × 104
Explanation
energy of radiation E=5 J use equation E=nhc/λ Answer: (a)
Q.27
In majority of crystals the value of lattice constant is of order of 3 Å. The proper X-rays with which the crystal structure can be studied are
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a)50Å to 100Å
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b) 10Å to 50Å
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c)5Å to 10Å
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d)0.1Å to 2.7Å
Explanation
Answer: (d)
Q.28
The distance between two success atomic planes of calcite crystal is 0.3Å. the minimum angle for Bragg scattering of 0.3Å X-ray will be
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a) 1.5°
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b) 30°
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c)2.86°
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d)60°
Explanation
According to Bragg equation nλ=2dsinθfor minimum angle of scattering n=10.3=2×0.3sinθ∴ θ=30°Answer: (b)
Q.29
The velocity of electron emitted in photoelectric effect depends on the properties of photosensitive surface an
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a) frequency of incident light
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b) state of polarization of incident light
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c)time for which light is incident
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d) intensity of incident light
Explanation
Answer:(a)
Q.30
De-Broglie wavelength of a particle moving with velocity 2.25×108 m/s is same at the wave length of photon. The ratio of kinetic energy of the particle to the energy of photon is
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a) 1/8
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b) 3/8
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c) 5/8
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d) 7/8
Explanation
wavelength λ is same for both momentm p=h/λ Kinetic energy of particle K=½ mv2=½ pv Kinetic energy of photon E'=hc/λ=pc ∴ K/K'=V/2=3/8 Answer: (b)
0 h : 0 m : 1 s
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