Q.1. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10^{–3} watt will be : (h = 6.6 × 10^{–34 }Js) (2021)
A: 10^{16}
B: 10^{15}
C: 10^{18}
D: 10^{17}
Ans: A
Solution:
Q.2. An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' is mass of photoelectron emitted from the surface has deBroglie wavelength λd, then: (2021)
A:
B:
C:
D:
Ans: A
Solution:
Q.3. Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled? (2020)
A: onefourth
B: zero
C: doubled
D: four times
Ans: B
Solution:
f_{0} < f_{1} = 1.5 f_{0}
∴ f_{2} = 0.75 f_{0}
for given condition
f_{incident }< f_{threshold}
so no photo electron emission
i = 0
Q.4. An electron is accelerated through a potential difference of 10,000 V. Its deBroglie wavelength is, (nearly) : (m_{e} = 9 × 10^{–31} kg) (2019)
A: 12.2 × 10^{–13} m
B: 12.2 × 10^{–12} m
C: 12.2 × 10^{–14} m
D: 12.2 nm
Ans: B
Solution:
For an electron accelerated through a potential V
Q.5. An electron of mass m with an initial velocityenters an electric field E_{0} = constant > 0) at t = 0. If λ_{0} is its deBroglie wavelength initially, then its deBroglie wavelength at time t is : (2018)
A:
C: λ_{0}t
D: λ_{0}
Ans: A
Solution:
Q.6. When the light of frequency 2v_{0} (where v_{0} is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v_{1}. When the frequency of the incident radiation is increased to 5 v_{0}, the maximum velocity of electrons emitted from the same plate is v_{2}. The ratio of v_{1} to v_{2} is : (2018)
A: 1 : 2
B: 1 : 4
C: 4 : 1
D: 2 : 1
Ans: A
Solution:
Q.7. The photoelectric threshold wavelength of silver is 3250 × 10^{–10}m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10–10 m is : (2017)
(Given h = 4.14 × 10^{–15} eVs and c = 3 × 10^{8} ms^{–1})
A: ≈ 0.6 × 10^{6} ms^{–1}
B: ≈ 61 × 10^{3} ms^{–1}
C: ≈ 0.3 × 10^{6} ms^{–1}
D: ≈ 6 × 10^{5} ms^{–1}
Ans: A
Solution:
Q.8. When a metallic surface is illuminated with radiation of wavelength λ the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is : (2016)
A: 3λ
B: 4λ
C: 5λ
D: 5/2λ
Ans: A
Solution:
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V.
Photoelectric equation can be written as,
Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,
From equations (i) and (ii), we get
When a metallic surface is illuminated with radiation of wavelength λ stopping potential is V.
Photoelectric equation can be written as,
Now, when the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. So, photoelectric equation can be written as,
From equations (i) and (ii), we get
Q.9. An electron of mass m and a photon have same energy E. The ratio of deBroglie wavelengths associated with them is: (2016)
A:
B:
C:
D: c(2mE)^{1/2}
Ans: B
Solution:
Given that electron has a mass m.
DeBroglie wavelength for an electron will be given as,
where,
h is the Planck's constant, and
p is the linear momentum of electron
Kinetic energy of electron is given by,
From equation (i) and (ii), we have
Energy of a photon can be given as,
Hence, λ_{P} is the deBroglie wavelength of photon.
Now, dividing equation (iii) by (iv), we get
Q.10. A certain metallic surface is illuminated with monochromatic light of wavelength, λ. The stopping potential for photoelectric current for this light is 3V_{0}. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V_{0}. The threshold wavelength for this surface for photoelectric effect is : (2015)
A: λ/6
B: 6λ
C: 4λ
D: λ/4
Ans: C
Solution:
We have,
where W is the work function and (3V_{0}) is the stopping potential when monochromatic light of wavelength λ is used.
where V_{0} is the stopping potential when monochromatic light of wavelength 2λ is used.
Subtracting equation (2) from equation (1)
We get,
∴
Substituting in equation (2) we get,
∴
The threshold wavelength is therefore 4λ.
Q.11. Which of the following figures represent the variation of particle momentum and the associated deBroglie wavelength? (2015)
A:
B:
C:
D:
Ans: C
Solution:
The deBroglie wavelength is given by
This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.
The deBroglie wavelength is given by
This equation is in the form of yx = c, which is the equation of a rectangular hyperbola.
Q.12. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is : (2014)
A: 1.3 eV
B: 1.5 eV
C: 0.65 eV
D: 1.0 eV
Ans: D
Solution:
Original energy of photon be E_{0}
From equation (i) and (ii)
Q.13. If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de−Broglie wavelength of the particle is : (2014)
A: 60
B: 50
C: 25
D: 75
Ans: D
Solution:
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
134 videos388 docs213 tests
