Question

In a photoelectric experiment with a material of work function 2.1 eV, the stopping potential is found to be 2.5 V. The maximum kinetic energy of ejected photoelectrons is:

• A. 0.4 eV
• B. 2.1 eV
• C. 2.5 eV
• D. 4.6 eV

Hint:

The stopping potential is the minimum voltage required to stop the most energetic photoelectrons, and it directly measures their maximum kinetic energy.

Answer 1

The Correct Option is C

The objective is to determine the maximum kinetic energy of photoelectrons emitted in a photoelectric experiment.

The photoelectric equation is:

KE_max = hf − φ

where:

• KE_max is the maximum kinetic energy of the emitted electrons.
• hf is the energy of the incident photons.
• φ is the work function of the material.

The maximum kinetic energy is also related to the stopping potential (V₀) by:

KE_max = eV₀

Here, e is the charge of an electron (\(1.6 \times 10^{-19}\,\text{C}\)).

Given the stopping potential, V₀ = 2.5\,\text{V},

KE_max = 2.5\,\text{eV}

Therefore, the maximum kinetic energy of the emitted photoelectrons is 2.5 eV.