List of top Physics Questions on Photoelectric Effect

Light is incident on a metallic plate having work function \(110 \times 10^{-20}\,\text{J}\). If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is _________ rad/s. (\(h = 6.63 \times 10^{-34}\,\text{J·s}\))}

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is \(3.2\ \text{V}\). If a second light having wavelength twice of the first light is used, the stopping potential drops to \(0.7\ \text{V}\). The wavelength of the first light is ________ m.

Light of which of the following colours will have the maximum energy in a photon associated with it?

Assertion : Photoelectric effect is a spontaneous phenomenon. Reason (R): According to the wave picture of radiation, an electron would take hours/days to absorb sufficient energy to overcome the work function and come out from a metal surface.

Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

On a surface, if photon of wavelength $\lambda$ is incident, the stopping potential is $3.2$ V. If the wavelength incident is $2\lambda$, stopping potential is $0.7$ V. Find $\lambda$.

If the electric field of an EM wave is given by \[ 60\,[\sin(3\times10^{14}t) + \sin(12\times10^{14}t)] \] at \( x = 0 \) and it falls on a photosensitive material having work function \( 2.8\,\text{eV} \), find the maximum kinetic energy (in eV) of the ejected electrons.

The work function of a metal is 2 eV. What is the threshold frequency for the photoelectric emission? (Take Planck's constant $ h = 6.63 \times 10^{-34} \, \text{Js} $, $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $)

The threshold frequency for a given metal is \( 3.6 \times 10^{14} \) Hz. If monochromatic radiations of frequency \( 6.8 \times 10^{14} \) Hz are incident on this metal, find the cut-off potential for the photoelectrons.
Given: - Threshold frequency, \( \nu_0 = 3.6 \times 10^{14} \) Hz
- Frequency of incident radiation, \( \nu = 6.8 \times 10^{14} \) Hz
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

Draw a plot of frequency \( \nu \) of incident radiations as a function of stopping potential \( V_0 \) for a given photoemissive material. What information can be obtained from the value of the intercept on the stopping potential axis?

(a) Briefly explain Einstein’s photoelectric equation.
(b) Four metals with their work functions are listed below:
K = 2.3 eV, Na = 2.75 eV, Mo = 4.17 eV, Ni = 5.15 eV.
The radiation of wavelength 330 nm from a laser source placed 1 m away, falls on these metals.
Which of these metals will not show photoelectric emission?
What will happen if the laser source is brought closer to a distance of 50 cm?

(a) Define 'work function' of a metal. How can its value be determined from a graph between stopping potential and frequency of the incident radiation?
(b) The work function of a metal is 2.4 eV. A stopping potential of 0.6 V is required to reduce the photocurrent to zero, in a photoelectric experiment. Calculate the wavelength of light used.

Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C), and (D) as given below.
Assertion (A): For monochromatic incident radiation, the emitted photoelectrons from a given metal have speed ranging from zero to a certain maximum value.
Reason (R): Each metal has a definite work function.

Answer the following giving reasons:
(a) The maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
(b) Photoelectric current increases with the increase in the intensity of the incident radiation.
(c) The stopping potential \( V_0 \) varies linearly with the frequency \( \nu \) of the incident radiation for a given photosensitive surface.

Assertion (A): The minimum negative potential applied to the anode in a photoelectric experiment at which photoelectric current becomes zero, is called cut-off voltage.
Reason (R): The threshold frequency for a metal is the minimum frequency of incident radiation below which emission of photoelectrons does not take place.

Explain the following observations using Einstein’s photoelectric equation:
(a) Photoelectric emission does not occur from a surface when the frequency of the light incident on it is less than a certain minimum value.
(b) It is the frequency, and not the intensity of the incident light which affects the maximum kinetic energy of the photoelectrons.
(c) The cut-off voltage \( V_0 \) versus frequency \( \nu \) of the incident light curve is a straight line with a slope \( \frac{h}{e} \).

A photo-emissive substance is illuminated with a radiation of wavelength $ \lambda_i $ so that it releases electrons with de-Broglie wavelength $ \lambda_e $. The longest wavelength of radiation that can emit photoelectron is $ \lambda_0 $. Expression for de-Broglie wavelength is given by : (m : mass of the electron, h : Planck's constant and c : speed of light)

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is