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

To solve this problem, we need to determine the maximum kinetic energy of photoelectrons ejected during a photoelectric experiment.

The photoelectric effect equation is given by:

KE_max = hf - φ

Where:

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

We are also given that the stopping potential (V₀) is related to the maximum kinetic energy by:

KE_max = eV₀

Where e is the charge of an electron (approximately 1.6 × 10⁻¹⁹ C).

In the problem, the stopping potential is 2.5 V, thus:

KE_max = 2.5 eV

Hence, the maximum kinetic energy of the ejected photoelectrons is 2.5 eV.