Q195 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
A photon of energy 8 eV is incident on a metal surface of work function 3.75 eV. The stopping potential of the metal is
- A$3.25 \text{ V}$
- B$3.5 \text{ V}$
- C$4.15 \text{ V}$
- Correct answerD$4.25 \text{ V}$
Solution
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by:
$K_{\max} = E - \Phi$
where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
Substituting the given values:
$K_{\max} = 8 \text{ eV} - 3.75 \text{ eV} = 4.25 \text{ eV}$
The stopping potential $V_0$ is related to the maximum kinetic energy by $K_{\max} = e V_0$.
Therefore, $e V_0 = 4.25 \text{ eV}$
$V_0 = 4.25 \text{ V}$
Answer: $4.25 \text{ V}$
$K_{\max} = E - \Phi$
where $E$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
Substituting the given values:
$K_{\max} = 8 \text{ eV} - 3.75 \text{ eV} = 4.25 \text{ eV}$
The stopping potential $V_0$ is related to the maximum kinetic energy by $K_{\max} = e V_0$.
Therefore, $e V_0 = 4.25 \text{ eV}$
$V_0 = 4.25 \text{ V}$
Answer: $4.25 \text{ V}$
Hence, the correct answer is option: 'D'.