Q: 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?

Zero. Initially \nu = 1.5\nu_0. When halved, \nu' = 0.75\nu_0 < \nu_0 — below the threshold frequency. No photoelectric emission takes place regardless of intensity, so the photoelectric current is zero.

Q: The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photon emission from the substance is (approximately)

310 nm. \phi = \frac{hc}{\lambda} \lambda = \frac{hc}{\phi} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.6 \times 10^{-19}} = 3.1 \times 10^{-7} m \approx 310 nm

Q: When the light of frequency 2\nu_0 (where \nu_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\nu_0, the maximum velocity of electrons emitted from the same plate is v_2. The ratio of v_1 to v_2 is

1 : 2. By Einstein's equation, \frac{1}{2}mv^2 = h\nu - h\nu_0 For 2\nu_0: \frac{1}{2}mv_1^2 = h\nu_0; for 5\nu_0: \frac{1}{2}mv_2^2 = 4h\nu_0 \frac{v_1^2}{v_2^2} = \frac{1}{4} \Rightarrow v_1 : v_2 = 1 : 2

Q: The photoelectric threshold wavelength of silver is 3250 \times 10^{-10} m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 \times 10^{-10} m is (Given, h = 4.14 \times 10^{-15} eV-s and c = 3 \times 10^8 ms⁻¹)

\approx 6 \times 10^5 ms⁻¹. \frac{1}{2}m_e v^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) v = \sqrt{\frac{2hc}{m_e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)} \approx 6 \times 10^5 ms⁻¹ (Note: options (a) and (b) are numerically the same value — 0.6 \times 10^6 = 6 \times 10^5; the official key accepted both.)

Q: When a metallic surface is illuminated with radiation of wavelength \lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2\lambda, the stopping potential is \frac{V}{4}. The threshold wavelength for the metallic surface is

3\lambda. Case 1: eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} Case 2: \frac{eV}{4} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}, i.e. eV = \frac{4hc}{2\lambda} - \frac{4hc}{\lambda_0} Equating: \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{4}{\lambda_0} \Rightarrow \lambda_0 = 3\lambda

Q: Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

−3 V. Work function: \phi = 5 - 2 = 3 eV With 6 eV photons, K_{max} = 6 - 3 = 3 eV. To stop these electrons the anode must be 3 V negative relative to the cathode: V_A - V_C = -3 V

Q: A certain metallic surface is illuminated with monochromatic light of wavelength \lambda. The stopping potential for photoelectric current for this light is 3V_0. If the same surface is illuminated with light of wavelength 2\lambda, the stopping potential is V_0. The threshold wavelength for this surface for photoelectric effect is

4\lambda. \frac{hc}{\lambda} = W + 3eV_0 and \frac{hc}{2\lambda} = W + eV_0 (so \frac{hc}{\lambda} = 2W + 2eV_0) Subtracting: W = eV_0, hence \frac{hc}{\lambda} = 4eV_0 \lambda_{th} = \frac{hc}{W} = \frac{4eV_0\lambda}{eV_0} = 4\lambda

Q: A photoelectric surface is illuminated successively by monochromatic light of wavelength \lambda and \frac{\lambda}{2}. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck's constant, c = speed of light)

\frac{hc}{2\lambda}. K_{max} = \frac{hc}{\lambda} - \phi and 3K_{max} = \frac{2hc}{\lambda} - \phi Multiplying the first by 3 and equating: \frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi \phi = \frac{hc}{2\lambda}

Q: Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is

\geq 2.8 \times 10^{-9} m. Photon energy E = \frac{hc}{\lambda} = 2.48 eV, so K_{max} = 2.48 - 2.28 = 0.2 eV Minimum de-Broglie wavelength: \lambda_e = \frac{12.27}{\sqrt{0.2}} Å \approx 2.8 \times 10^{-9} m Slower electrons have longer wavelengths, so \lambda \geq 2.8 \times 10^{-9} m.

Q: 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

1.0 eV. 0.5 = E - \phi_0 and 0.8 = 1.2E - \phi_0 Subtracting: 0.3 = 0.2E \Rightarrow E = 1.5 eV \phi_0 = 1.5 - 0.5 = 1.0 eV

Question 1

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?

Accepted Answer

Zero. Initially 
u = 1.5
u_0. When halved, 
u' = 0.75
u_0 < 
u_0 — below the threshold frequency.

No photoelectric emission takes place regardless of intensity, so the photoelectric current is zero.

Question 2

The work function of a photosensitive material is 4.0 eV. The longest wavelength of light that can cause photon emission from the substance is (approximately)

Accepted Answer

310 nm. \phi = \frac{hc}{\lambda}

\lambda = \frac{hc}{\phi} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{4 	imes 1.6 	imes 10^{-19}} = 3.1 	imes 10^{-7} m \approx 310 nm

Question 3

When the light of frequency 2
u_0 (where 
u_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
u_0, the maximum velocity of electrons emitted from the same plate is v_2. The ratio of v_1 to v_2 is

Accepted Answer

1 : 2. By Einstein's equation, \frac{1}{2}mv^2 = h
u - h
u_0

For 2
u_0: \frac{1}{2}mv_1^2 = h
u_0; for 5
u_0: \frac{1}{2}mv_2^2 = 4h
u_0

\frac{v_1^2}{v_2^2} = \frac{1}{4} \Rightarrow v_1 : v_2 = 1 : 2

Question 4

The photoelectric threshold wavelength of silver is 3250 	imes 10^{-10} m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 	imes 10^{-10} m is (Given, h = 4.14 	imes 10^{-15} eV-s and c = 3 	imes 10^8 ms⁻¹)

Accepted Answer

\approx 6 	imes 10^5 ms⁻¹. \frac{1}{2}m_e v^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}ight)

v = \sqrt{\frac{2hc}{m_e}\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}ight)} \approx 6 	imes 10^5 ms⁻¹

(Note: options (a) and (b) are numerically the same value — 0.6 	imes 10^6 = 6 	imes 10^5; the official key accepted both.)

Question 5

When a metallic surface is illuminated with radiation of wavelength \lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2\lambda, the stopping potential is \frac{V}{4}. The threshold wavelength for the metallic surface is

Accepted Answer

3\lambda. Case 1: eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}

Case 2: \frac{eV}{4} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}, i.e. eV = \frac{4hc}{2\lambda} - \frac{4hc}{\lambda_0}

Equating: \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{2}{\lambda} - \frac{4}{\lambda_0} \Rightarrow \lambda_0 = 3\lambda

Question 6

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

Accepted Answer

−3 V. Work function: \phi = 5 - 2 = 3 eV

With 6 eV photons, K_{max} = 6 - 3 = 3 eV.

To stop these electrons the anode must be 3 V negative relative to the cathode: V_A - V_C = -3 V

Question 7

A certain metallic surface is illuminated with monochromatic light of wavelength \lambda. The stopping potential for photoelectric current for this light is 3V_0. If the same surface is illuminated with light of wavelength 2\lambda, the stopping potential is V_0. The threshold wavelength for this surface for photoelectric effect is

Accepted Answer

4\lambda. \frac{hc}{\lambda} = W + 3eV_0 and \frac{hc}{2\lambda} = W + eV_0 (so \frac{hc}{\lambda} = 2W + 2eV_0)

Subtracting: W = eV_0, hence \frac{hc}{\lambda} = 4eV_0

\lambda_{th} = \frac{hc}{W} = \frac{4eV_0\lambda}{eV_0} = 4\lambda

Question 8

A photoelectric surface is illuminated successively by monochromatic light of wavelength \lambda and \frac{\lambda}{2}. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is (h = Planck's constant, c = speed of light)

Accepted Answer

\frac{hc}{2\lambda}. K_{max} = \frac{hc}{\lambda} - \phi and 3K_{max} = \frac{2hc}{\lambda} - \phi

Multiplying the first by 3 and equating: \frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi

\phi = \frac{hc}{2\lambda}

Question 9

Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de-Broglie wavelength of the emitted electron is

Accepted Answer

\geq 2.8 	imes 10^{-9} m. Photon energy E = \frac{hc}{\lambda} = 2.48 eV, so K_{max} = 2.48 - 2.28 = 0.2 eV

Minimum de-Broglie wavelength: \lambda_e = \frac{12.27}{\sqrt{0.2}} Å \approx 2.8 	imes 10^{-9} m

Slower electrons have longer wavelengths, so \lambda \geq 2.8 	imes 10^{-9} m.

Question 10

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

Accepted Answer

1.0 eV. 0.5 = E - \phi_0 and 0.8 = 1.2E - \phi_0

Subtracting: 0.3 = 0.2E \Rightarrow E = 1.5 eV

\phi_0 = 1.5 - 0.5 = 1.0 eV

Question 11

For photoelectric emission from certain metal, the cut-off frequency is 
u. If radiation of frequency 2
u impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

Accepted Answer

\sqrt{\frac{2h
u}{m}}. \frac{1}{2}mv_{max}^2 = h(2
u) - h
u = h
u

v_{max} = \sqrt{\frac{2h
u}{m}}

Question 12

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be

Accepted Answer

1 : 2. KE_1 = 1 - 0.5 = 0.5 eV; KE_2 = 2.5 - 0.5 = 2 eV

\frac{v_1^2}{v_2^2} = \frac{KE_1}{KE_2} = \frac{1}{4} \Rightarrow \frac{v_1}{v_2} = \frac{1}{2}

Question 13

Photoelectric emission occurs only when the incident light has more than a certain minimum

Accepted Answer

frequency. Photoelectric emission requires 
u \geq 
u_0 (threshold frequency). The incident light must have more than a certain minimum frequency, regardless of intensity.

Question 14

In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is

Accepted Answer

0.5 V. The stopping potential corresponds to the maximum kinetic energy:

eV_0 = K_{max} = 0.5 eV \Rightarrow V_0 = 0.5 V

Question 15

The number of photoelectrons emitted for light of a frequency 
u (higher than the threshold frequency 
u_0) is proportional to

Accepted Answer

intensity of light. The number of photoelectrons emitted depends only on the intensity of the incident light (one photon ejects one electron). Frequency decides their maximum kinetic energy, not their number.

Question 16

In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of

Accepted Answer

collisions between the charged particles emitted from the cathode and the atoms of the gas. In electric discharge through gases at low pressure, charged particles emitted from the cathode collide with the atoms of the gas, producing the coloured glow in the tube.

Question 17

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the

Accepted Answer

X-ray region. h
u = \phi + eV_0 = 6.2 + 5 = 11.2 eV

A photon of 11.2 eV corresponds to \lambda \approx 1100 Å, which lies in the X-ray region.

Question 18

A 5 W source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of

Accepted Answer

4. Intensity \propto \frac{1}{d^2}. Doubling the distance reduces intensity to \frac{1}{4}.

The number of photoelectrons is proportional to intensity, so it is reduced by a factor of 4.

Question 19

A photocell employs photoelectric effect to convert

Accepted Answer

change in the intensity of illumination into a change in photoelectric current. For a fixed frequency, the photoelectric current increases linearly with the intensity of the incident light. A photocell therefore converts a change in the intensity of illumination into a change in photoelectric current.

Question 20

A photosensitive metallic surface has work function h
u_0. If photons of energy 2h
u_0 fall on this surface, the electrons come out with a maximum velocity of 4 	imes 10^6 m/s. When the photon energy is increased to 5h
u_0, then maximum velocity of photoelectrons will be

Accepted Answer

8 	imes 10^6 m/s. \frac{1}{2}m(4 	imes 10^6)^2 = 2h
u_0 - h
u_0 = h
u_0

\frac{1}{2}mv^2 = 5h
u_0 - h
u_0 = 4h
u_0

Dividing: v^2 = 4 	imes (4 	imes 10^6)^2 \Rightarrow v = 8 	imes 10^6 m/s

Photoelectric Effect — NEET Physics PYQ MCQs with Solutions

Questions with solutions