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Q.
A photon of energy $4\, eV$ is incident on a metal surface whose work function is $2\, eV$. The minimum reverse potential to be applied for stopping the emission of electrons is
AIIMSAIIMS 2004Dual Nature of Radiation and Matter
Solution:
Kinetic energy of photoelectron is $e V_{0}$ where $V_{0}$ is stopping potential.
From Einsteins photoelectric equation
$E_{k}=\frac{1}{2} m v_{\max }^{2}=h v-W$
where $E_{k}$ is maximum kinetic energy of electron, $v$ is frequency and $W$ is work function.
$\therefore \frac{1}{2} m v_{\max }^{2}=4 \,eV -2 \,eV =2\, eV$
but $\frac{1}{2} m v_{\max }^{2}=e V_{0}$
where $V_{0}$ is stopping potential.
Thus, $eV_0 = 2\,eV$
$\Rightarrow V_0 = 2\,V$