The potential difference that must be applied to stop the fastest photo electrons emitted by a nickel surface, having work function $5.01\text{eV}$, when ultraviolet light of $200\text{nm}$ falls on it, must be

The correct option is C $\text{1.2 V}$
Given:
Work function $=5.01\text{eV}$,

$\lambda =200\text{nm}$

Energy of incident photon,

$E=h\nu =\frac{12400}{2000}=6.2\text{eV}$

We know that,
$E={\varphi }_0+e{V}_0\Rightarrow V_0=\frac{E-\varphi }{e}$

$\Rightarrow V_0=6.2-5.01=1.19\approx 1.20\text{V}$

Hence, option $\left(D\right)$ is correct.