Physics - Modern Physics Question with Solution | TestHub

Here,
$K_{\max }=3 \text{eV}$

Maximum kinetic energy is given by:

$\begin{array}{c}{K}_{\max }=e{V}_0\\ 3 \mathrm{eV}=e\times V_0\\ 3 \mathrm{V}=V_0\end{array}$                        

$\therefore$  Stopping potential  $V_0=3 \mathrm{V}$