Photoelectric effect is maximum in
- Cs
- Na
- K
- Li
The Correct Option is A
Solution and Explanation
Top Questions on Photoelectric Effect
- 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, \( e = 1.6 \times 10^{-19} \, \text{C} \)- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
Einstein's Explanation of the Photoelectric Effect:
Einstein explained the photoelectric effect on the basis of Planck’s quantum theory, where light travels in the form of small bundles of energy called photons.
The energy of each photon is hν, where:- ν is the frequency of the incident light
- h is Planck’s constant
The number of photons in a beam of light determines the intensity of the incident light.When a photon strikes a metal surface, it transfers its total energy hν to a free electron in the metal.A part of this energy is used to eject the electron from the metal, and this required energy is called the work function.The remaining energy is carried by the ejected electron as its kinetic energy.
- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- The work function of a material is 2.21 eV. Which of the following cannot produce photoelectrons from it?
- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- 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.- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
- (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?- CBSE CLASS XII - 2025
- Physics
- Photoelectric Effect
Notes on Photoelectric Effect
Concepts Used:
Photoelectric Effect
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Photoelectric Effect Formula:
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,
- h is Planck’s constant.
- ν is the frequency of the incident photon.
- W is a work function.
- E is the maximum kinetic energy of ejected electrons: 1/2 mv².
Laws of Photoelectric Effect:
- The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
- There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
- The maximum kinetic energy of the photoelectrons increases with the increase in the frequency (provided frequency γ > γ Th exceeds the threshold limit) of the incident light. The maximum kinetic energy is free from the intensity of light.
- The process of photo-emission is an instantaneous process.