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Particle Nature Of Light Mcq
Quiz 6
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Q.1
Which of the following graph is correct
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a)
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b)
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c)
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d)
Explanation
Answer: (a)
Q.2
de-Broglie wavelength associated with a neutron at T degree Kelvin is
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a) (1.82 / T )Å
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b) (1.82/√T) Å
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c)(30.7/T ) Å
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d)(30.7/√T ) Å
Explanation
Answer: (d)
Q.3
The stopping potential for electron beam of de-Broglie wavelength of 1 Å
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a)12.3Volts
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b) 123 Volts
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c)151 Volts
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d)1505 Volts
Explanation
λ=12.3/ √V V=(12.3/λ)2 V=151.29V Answer:(c)
Q.4
The pulse duration of radar is 0.25µs. The uncertainty in the frequency of photons is
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a) 4.2 × 10-28
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b) 6.6 × 104
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c) 6.4 × 105
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d) 4.2 × 10-4
Explanation
ΔE ×Δt = h/2π and E=hν ∴ ΔE=hΔν ∴ h Δν Δt=h/2π Δν=1/ Δt ×2π=106/0.25×2π Δν=6.4 ×105 Answer: (c)
Q.5
Uncertainty Principle states that the error in measurement is due to
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a)Dual nature of particles
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b) Due to small size of particles
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c)Due to large size particles
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d)Due to error in measurement
Explanation
Answer: (a)
Q.6
Work function of a photoelectric material is 3.3eV. The threshold frequency will be equal to [ CPMT 1986]
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a) 8 × 1010Hz
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b) 4 × 1014 Hz
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c)8×1014 Hz
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d)4 × 1020 Hz
Explanation
USe formula E=hνAnswer: (c)
Q.7
For light of wavelength 5000Å , the photon energy is nearly 2.5 V. For X-rays of wavelength 1Å, the photon energy will be close to [ CPMT 1987]
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a)2.5 ÷ 5000 eV
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b) 2.5 ÷ (5000)2 eV
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c)2.5 × 5000 eV
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d)2.5 × (5000)2 eV
Explanation
Use formula E=hν and take rations of E1 and E2 Answer:(c)
Q.8
In order to increase the kinetic energy of ejected photoelectrons, there should be an increases in [ PMT 1987]
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a) intensity of radiation
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b) wavelength of radiation
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c) frequency of radiation
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d) both the wavelength and intensity of radiation
Explanation
Answer: (c)
Q.9
When a certain metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential for photoelectric current is 3Vo. When the same surface is illuminated with the light of wavelength 2λ, the stopping potential is Vo. The threshold wavelength for the surface for photoelectric effect is [ MPPMT 1987]
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a)4λ/3
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b) 4λ
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c)6λ
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d)8λ
Explanation
Answer: (b)
Q.10
The specific charge for positive rays is much less than that for cathode rays. This is because [ C.P.M.T 1990]
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a) Positive rays are positively charged
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b) charge on positive rays is less
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c)mass of positive rays is much larger
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d)experimental method is wrong
Explanation
Answer: (c)
Q.11
Light of wavelength 4000Å is incident on a metal plate whose work function is 20eV. The maximum kinetic energy of the emitted photoelectrons would be [ CPMT 1990]
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a) 0.5 eV
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b) 1.1 eV
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c)1.5 eV
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d)2.0 eV
Explanation
USe formula convert work function φ in Jules obtain answer in Joules then convert it in eV Answer:(b)
Q.12
A proton is accelerated through one volt the increase in its kinetic energy is approximately [ CPMT 1993]
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a) 1/1870 eV
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b) 1870 eV
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c) 1 eV
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d) 300 eV
Explanation
Answer: (c)
Q.13
An electron enters a region where magnetic induction B and electric field E are mutually perpendicular to one another then..
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a)it will always move in the direction of B
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b) it will always move in the direction of E
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c)it always possessses circular motion
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d)it can go undeflected also
Explanation
Answer: (d)
Q.14
The threshold wavelength for photoelectric effect of a metal is 6500Å. The work function of the metal is approximately
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a) 2 eV
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b) 1 eV
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c)0.1 eV
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d)3 eV
Explanation
Use formula to obtain energy in eVAnswer: (a)
Q.15
The de-Broglie wavelength of an electron moving with speed of 6.6×105 m/s is approximately [ CPMT 1993]
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a) 10-11 m
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b) 10-9 m
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c)10-7 m
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d)10-5 m
Explanation
use formula λ=h/mv h=6.6×10-34 and m=9.1 ×10 -31 kg Answer:(b)
Q.16
Threshold frequency for photoelectric effect on sodium corresponds to a wavelength 5000 Å. Its work function is ..
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a) 15 joule
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b) 10 × 10-10 joule
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c) 4 × 10-19 joule
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d) 4 ×10+19 joule
Explanation
USe formula work function φ=hc/λo Answer: (a)
Q.17
Number of ejected photoelectrons increases with increase
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a)in intensity of light
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b) in wavelength of light
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c)in frequency of light
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d)never
Explanation
Answer: (a)
Q.18
The ratio of momentum of electron and an alpha particle which are accelerated from rest through a potential difference of 100 volt is [ CEE 1994]
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a) 1
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b) √(2me / mα)
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c)√( me / mα)
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d)√(me /2mα)
Explanation
Use formula energy eV=p2/ 2m take ratios Answer: (d)
Q.19
Doubly ionised Helium atoms and Hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the Helium and the Hydrogen ions is
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a)2
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b) √2
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c)1/2
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d)1/√2
Explanation
Charge on Helium atom=2e and charge on hydrogen aion=e mass of Helium is four times mass of Hydrogen atom take ratios of above equations Answer:(d)
Q.20
Gases begin to conduct electricity at low pressure because
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a) the lectrons in atoms can move freely at low pressures
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b) atoms break up into electrons and protons
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c) colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms
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d) at low pressure gases turn into plasma
Explanation
Answer: (c)
Q.21
Photoelectric effect is due to
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a)wave nature of light
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b) particle nature of light
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c)both a and b
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d)none of these
Explanation
Answer: (c)
Q.22
If the frequency of light in a photoelectric experiment is doubled the stopping potential will [ CPMT 1994]
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a) be doubled
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b) be halved
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c)become more than double
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d)become less than double
Explanation
Answer: (c)
Q.23
In photoelectric effect, the photoelectric current
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a) increases when frequency of incident photons increases
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b) decreases when frequency of incident photons increases
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c)does not depend on photon frequency but only on intensity of incident beam
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d)depends both on intensity and frequency of incident beam
Explanation
Answer:(c)
Q.24
Ultra violet light of wavelength 300 nm and intensity 1.0 watt/m2 falls on the surface of a photosensitive material. If one percent of the incident photon produce photoelectron then the number of photoelectrons emitted per second from an area 1.0 cm2 of the surface is nearly [ AMU PMT 1995]
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a) 2.13 × 1011
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b) 1.51 × 1012
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c) 4.12 × 1013
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d) 9.61 ×1014
Explanation
Intenisty per sq.cm=I/104 number of incident photons falling per sq.cm As 1 percent of the incident photons produc photoelectrons, thereforenumber of photoelectrons produced per second=1013 / 6.6n=1.51 ×1012 Answer: (b)
Q.25
If we consider electrons and photons of the same wavelength, then they will have the same
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a)velocity
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b) angular momentum
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c)energy
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d)momentum
Explanation
Answer: (d)
Q.26
An electron of mass m when accelerated through a potential difference V, the de-Broglie wavelength λ. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be [ pre medical/ dental 1995]
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a) λm/M
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b) λ√(m/M)
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c)λM/m
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d)λ√(M/m)
Explanation
Answer: (b)
Q.27
Assuming photo emission to continue to take place, the factor by which the maximum velocity of the emitted photo electron changes approximately when the wavelength of the incident radiation is increased four times is [ Haryana CET 1996]
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a) 1/4
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b) 1/2
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c)2
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d)4
Explanation
Answer:(b)
Q.28
An X-ray tube operates at 10kV. The ratio of X-ray wavelength to that of de-Broglie is [ CPMT 1996]
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a) 10:1
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b) 1:10
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c) 1:100
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d) 100:1
Explanation
De-Broglie wavelength X-ray wavelength By taking ratioAnswer: (a)
Q.29
In de_broglie's equation wave-length 'λ' depends upon mass 'm' and energy 'E' according to the relation represented as [ CPMT 1996]
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a)mE1/2
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b) m-1/2E1/2
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c)m-1/2 E-1/2
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d)m1/2E-1/2
Explanation
hence option 'd' is correctAnswer: (d)
Q.30
The velocity of the most energetic electrons emitted from a metallic surface is doubled when frequency ν of incident radiation is double. The work function of this metal is [ Pb.CET 1997]
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a) zero
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b) hν/3
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c)hν/2
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d)2hν/3
Explanation
Answer: (d)
0 h : 0 m : 1 s
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