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

frequency. Photoelectric emission requires \nu \geq \nu_0 (threshold frequency). The incident light must have more than a certain minimum frequency, regardless of intensity.

Q: Einstein's work on photoelectric effect gives support to

E = h\nu. Einstein explained the photoelectric effect by treating light as concentrated packets (photons), each of energy h\nu — supporting the relation E = h\nu.

Question 1

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 2

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 3

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 4

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 5

When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by incidence of

Accepted Answer

X-rays. If ultraviolet light cannot cause emission, the incident wavelength must be even shorter (higher energy). Among the options, only X-rays have a wavelength shorter than ultraviolet, so only X-rays can cause photoelectric emission.

Question 6

Which of the following is not the property of cathode rays?

Accepted Answer

It does not deflect in electric field. Cathode rays are streams of electrons (negatively charged), so they DO deflect in an electric field — towards the positive plate. They produce heating, travel in straight lines (cast shadows), and cause fluorescence.

Question 7

A light source is at a distance d from a photoelectric cell, then the number of photoelectrons emitted from the cell is n. If the distance of light source and cell is reduced to half, then the number of photoelectrons emitted will become

Accepted Answer

4n. Intensity \propto \frac{1}{d^2}. Halving the distance makes the intensity 4 times, so the number of photoelectrons becomes 4n.

Question 8

Einstein's work on photoelectric effect gives support to

Accepted Answer

E = h
u. Einstein explained the photoelectric effect by treating light as concentrated packets (photons), each of energy h
u — supporting the relation E = h
u.

Question 9

When intensity of incident light increases

Accepted Answer

photocurrent increases. More intensity means more incident photons, hence more ejected electrons and a larger photocurrent. The kinetic energy of the electrons depends on frequency, not intensity.

Question 10

Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be

Accepted Answer

0.58 eV. E = \frac{12375}{5000} = 2.48 eV

KE = E - W_0 = 2.48 - 1.9 = 0.58 eV

Question 11

Which of the following is true?

Accepted Answer

The photocurrent increases with increasing intensity of light. Higher intensity means more photons per second, so more electrons are emitted and the photocurrent increases. Stopping potential depends on frequency (not intensity), and photocurrent does not depend on frequency or applied voltage.

Question 12

If the threshold wavelength for a certain metal is 2000 Å, then the work function of the metal is

Accepted Answer

6.2 eV. W_0 = \frac{hc}{\lambda_0} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{2000 	imes 10^{-10}} = 9.9 	imes 10^{-19} J = 6.2 eV

Question 13

In photoelectric effect, the work function of a metal is 3.5 eV. The emitted electrons can be stopped by applying a potential of −1.2 V. Then,

Accepted Answer

the energy of the incident photons is 4.7 eV. The stopping potential gives the maximum kinetic energy: KE = 1.2 eV

h
u = W_0 + KE = 3.5 + 1.2 = 4.7 eV

Question 14

Number of ejected photoelectrons increases with increase

Accepted Answer

in intensity of light. The number of photoelectrons emitted per second is directly proportional to the intensity of the incident radiation.

Question 15

Ultraviolet radiation of 6.2 eV falls on an aluminium surface. KE of fastest electron emitted is (work function = 4.2 eV)

Accepted Answer

3.2 	imes 10^{-19} J. KE = E - W_0 = 6.2 - 4.2 = 2.0 eV = 2 	imes 1.6 	imes 10^{-19} = 3.2 	imes 10^{-19} J

Question 16

The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of 5000 Å. Its work function is

Accepted Answer

4 	imes 10^{-19} J. W_0 = \frac{hc}{\lambda_0} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = 4 	imes 10^{-19} J

Question 17

Which of the following are thermions?

Accepted Answer

Electrons. Thermionic emission is the emission of electrons from a metal surface when it is suitably heated. The emitted electrons are called thermions.

Question 18

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 19

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 20

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

Photoelectric Effect — JEE Main Physics MCQs with Solutions

Questions with solutions