Question

A photon of energy $4eV$ is incident on a metal surface whose work function is $2eV$. The minimum reverse potential to be applied for stopping the emission of electrons is

• A. 2 V
• B. 4 V
• C. 6 V
• D. 8 V

Answer 1

Answer: (A)

Kinetic energy of photoelectron is $e{V}_0$ where $V_0$ is stopping potential.
From Einsteins photoelectric equation
$E_k=\frac{1}{2}m{v}_{\max }^2=hv-W$
where $E_k$ is maximum kinetic energy of electron, $v$ is frequency and $W$ is work function.
$\therefore \frac{1}{2}m{v}_{\max }^2=4eV-2eV=2eV$
but $\frac{1}{2}m{v}_{\max }^2=e{V}_0$
where $V_0$ is stopping potential.
Thus, $e{V}_0=2eV$
$\Rightarrow V_0=2V$