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Physics NEET MCQ
Particle Nature Of Light Mcq
Quiz 4
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Q.1
Figure shows graph of kinetic energy of electrons emitted verses ν for a material exhibiting photoelectric effect. The work function of the material is ..
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a) 1eV
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b) 1.5eV
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c)2eV
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d)3eV
Explanation
The intercept on the negative Y-axis give value of stopping potential Answer:(b)
Q.2
If in a photoelectric experiment, the wavelength of incident radiation is reduced from 6000Å to 4000Å, then [ MPPMT 1999]
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a) Stopping potential will decrease
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b) Stopping potential will increase
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c) Kinetic energy of emitted electrons will decrease
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d) The value of work-function will decrease
Explanation
Wavelength decreased thus energy of incident photon increased. Kinetic energy of emitted photoelectrons will increase. Stopping potential will increase Answer: (b)
Q.3
the maximum velocity of an electron emitted by light of wavelength λ incident on the surface of metal of workfunction Φ is [ h=plank constant, m=mass of electron, and c=speed of light] [ MPPMT 1998]
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.4
Maximum kinetic energy E of photo-electron varies with the frequency (ν) of the incident radiation as one of the following graphs:[ MPPMT 1994]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (d)
Q.5
A gold leaf electroscope is negatively charged and the leaves are observed to diverged by a certain amount. A beam of X-ray is allowed to fall upon the electroscope for a short period. The effect is as follows
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a)The leaves will diverge further
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b) There will be no chnage in the divergence of leaves
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c) The leaves will collapse
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d)The gold leaves will melt
Explanation
Due to X ray electrons will leave the surface and Positive change will develop. Will nullify negative change on gold leaf Answer:(c)
Q.6
According to Einstein's interpretation of the photoelectric effect, the maximum K.E. of photoelectrons depends on ( hν -W) where ν is the frequency of incident radiation and W is the work function. For three different metals graph plotted between maximum K/E. and the frequency of incident radiations. The three graphs obtained:
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a) Have different slopes but a constant intercept on y-axis
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b) Are all parallel to y-axis but at different heights
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c) same slope but different intercept on y-axis
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d) none of these
Explanation
Answer: (c)
Q.7
According to the modern theory for nature of light, the light has..[ MPPMT 1998]
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a)Wave nature only
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b) Particle nature only
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c)Both wave and particle ( dual) nature
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d)neither particle nature nor wave nature
Explanation
Answer: (c)
Q.8
The energy of photon is 10eV. The momentum of a photon is
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a) 5.33×10-25 kg×m /s
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b) 5.33×10-27 kg×m /s
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c)5.33×10-29 kg×m /s
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d)5.33×10-23 kg×m /s
Explanation
Energy E=mc2 momentum p=mc Thus E=pc or p=E/cp=10×1.6×10-19 / 3×108 p=5.33×10-27 kg×m / secAnswer: (b)
Q.9
When the light of wavelength 2537Å is made incident over the surface of copper slab, the stopping voltage 0.24 volt obtained. The threshold frequency for copper will be .. [ Raj.PMT 1996]
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a) 1.124×1015 Hz
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b) 1.414×1014 Hz
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c)2.248×1015 Hz
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d)None of the above
Explanation
incident energy=work function + maximum kinetic energy incident energy=hc/λ Incident energy=6.6×10-34 ×3×108 / 2537×10-10 Incident energy=0.0078×10-16maximum kinetic energy=eV=1.6×10-19×0.24=0.384×10-19Now Work function=Incident energy - maximum kinetic energyWork function=0.0078×10-16 - 0.384×10-19Work function=0.007416×10-16Work function=h×νo h×νo=0.007416×10-16νo=0.007416×10-16 / 6.6×10-34νo=1.124×1015 Answer:(a)
Q.10
Given h=6.6×10-34 joules×sec. the momentum of each photon in a given radiation is 3.3×10-29 kg×metre / sec .. [ CBSE 1990]
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a) 3×10-2 hz
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b) 6×1010 hz
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c) 7.5×1012 hz
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d) 1.5×1013hz
Explanation
λ=h/p and λ=c/ν c/ν=h/p ν=pc/λ ν=3.3×10-29 ×3×108 / 6.6 ×10-34 ν=1.5×1013 hz Answer: (d)
Q.11
According to de Broglie, the waves are associated with
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a)Moving charged particles only
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b) Moving neutral particles only
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c)Electrons only
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d)All moving particles
Explanation
Answer: (d)
Q.12
The duration of laser pulse is 10-8 sec. The uncertainty in its energy will be (ΔE.Δt ≥ h)
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a) 6.6 × 10-26 J
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b) 6.6 × 10-34 J
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c)6.6 × 10-42 J
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d)1 /(6.6×1026) J
Explanation
Answer: (a)
Q.13
Matter waves
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a) Are electromagnetic waves
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b) are transverse waves
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c)are longitudinal waves
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d)exhibit diffraction
Explanation
Answer:(d)
Q.14
If the kinetic energy of moving particle is E, then de Broglie wavelength is
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a)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.15
de Broglie wave length associated with a moving particle of mass m, velocity v is given by ( c=speed of light)
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a)λ hmv
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b) λ=h/mv
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c)λ=v/mh
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d)
Explanation
Answer: (b)
Q.16
Neglecting variation of mass with energy the wavelength associated with an electron having a kinetic energy E is proportional to
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a) √E
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b) E
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c)E-1/2
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d)E-2
Explanation
Answer: (a)
Q.17
Of the following, moving with the same velocity, the one which has largest wavelength λ is
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a) an electron
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b) a proton
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c)an alpha particle
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d)all have same de Broglie wave length
Explanation
Answer:(a)
Q.18
The energy of photon is 10eV. The momentum of a photon is
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a) 5.33 × 10-25 kg×m/s
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b) 5.33 × 10-27 kg×m/s
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c) 5.33 × 10-29 kg×m/s
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d) 5.33 × 10-23 kg×m/s
Explanation
Answer: (b)
Q.19
The energy of photon corresponding to the visible light of a maximum wavelength is approximately [ MPPMT 1985]
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a)1 eV
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b) 1.6 eV
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c)3.2 eV
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d)7.0 eV
Explanation
E=hc/ λ Answer: (b)
Q.20
If position uncertainty of an electron is 2 Å, the uncertainty in energy will be [ raj. PET 1996]
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a) 0.1 eV
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b) 1.0 eV
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c)9.0 eV
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d)94.0 eV
Explanation
Answer: (b)
Q.21
The period of laser is 10-8 sec. The uncertainty in energy
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a) 1.05 × 10-26 J
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b) 1.5 × 10-25 J
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c)6.62 × 10-26 J
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d)zero
Explanation
Answer:(a)
Q.22
To reduce the de Broglie wave length of an electron from 100pm to 50pm, the required increase in energy will be [ Raj PET 1997]
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a) 600 eV
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b) 450 eV
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c) 300 eV
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d) 150 eV
Explanation
Answer: (b)
Q.23
The ratio of de Broglie wavelength for a proton and an α-particle of same energy is [ raj-PET 1996]
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a)2:1
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b) 1:2
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c)4:1
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d)1:4
Explanation
λ ∝ (1/√m)λp /λalpha=√malpha/ √mp λp /λalpha=2:1Answer: (a)
Q.24
If a photon has velocity c and frequency ν, then its wave length is equal to [ CBSE 1996]
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a) hc/E
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b) hν/c
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c)hν/ c2
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d)hν
Explanation
Answer: (a)
Q.25
An electron and α particle are accelerated by potential V. The masses are me and malpha then ratio of their momentums will be [ raj. PET 1997]
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a) me / malpha
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b)√(me / malpha)
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c)√(me / 2malpha)
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d)√(2me / malpha)
Explanation
Answer:(c)
Q.26
A proton and an α particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths ( λp / λα) is
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a) 1
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b) 2
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c) √8
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d) 1 / √8
Explanation
Charge on alpha particle is 2 times charge on proton and mass of alpha particle is 4 times mass of proton Answer: (c)
Q.27
If Plank's constant is larger than its present value. The de-Broglie wavelength associated with material particles would have been
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a)unchanged
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b) larger
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c)smaller
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d)larger for some particles and smaller for others
Explanation
Answer: (b)
Q.28
Given Plank's constant h=6.6×10-34 J-sec. The momentum of each photon in a given radiation is 3.3 ×10-29 kg-m/sec. The wave-length of radiation is
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a) 3 × 10-3 m
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b) 6 × 10-10 m
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c)7.5 × 10-2 m
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d)2 × 10-5 m
Explanation
Use formula λ=h/pAnswer: (d)
Q.29
When electrons are accelerated through potential difference of V volts, the de Broglie wavelength associated is given by
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a) √(150/V) Å
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b) √(150/V) m
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c)(150/ √V) Å
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d)(√150/V) Å
Explanation
λ=h / √(2meV)on substituting values of h, m and e we get λ=12.3×10-10 / √V metreλ=12.3/ √V ÅOr λ=√(150/V) Å Answer:(a)
Q.30
A material particle with a rest mass mo is moving with speed of light c. The de-Broglie wavelength associated is given by
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a) h/moc
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b) moc / h
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c) 0
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d) ∞
Explanation
When particle moves with velocity nearer to velocity light its mass is given by Now particle is moving at velocity of light its mass m=∞λ=h/p=h/∞=0 Answer: (c)
0 h : 0 m : 1 s
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