Question

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

• A. 1.8V
• B. 1.3V
• C. 0.5 V
• D. 2.3 V

Answer 1

The Correct Option is C

The question pertains to the photoelectric effect, where electrons are emitted from a metal surface when it is illuminated by light. The energy of the incident photons is used to overcome the work function of the metal and provide kinetic energy to the emitted electrons.

Let's break down the problem:

1. The work function of the metal (\phi) is given as 1.8\, eV. This is the minimum energy required to remove an electron from the surface of the metal.
2. The kinetic energy (K.E.) of the most energetic electrons emitted is 0.5\, eV.

According to the equation for the photoelectric effect, we have:

h\nu = \phi + K.E.

Here, h\nu represents the energy of the incident photons.

The stopping potential (V_s) is the potential difference required to stop the most energetic electrons. The stopping potential is given by the kinetic energy of these electrons:

eV_s = K.E.

where e is the charge of an electron (1 elementary charge). Thus, the stopping potential is simply:

V_s = \frac{K.E.}{e} = 0.5\, V

Therefore, the stopping potential is 0.5\, V. Hence, the correct answer is:

• Option 3: 0.5 V