Question

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

Answer 1

Correct option is A. $$1.2\ V$$
$$\begin{array}{l}E=\phi+e V_{s} \\\text { and } E=\frac{n c}{\lambda} \\\therefore \frac{h c}{\lambda}=5.01+e V_{s}\end{array}$$

$$V_{s}=\text { stopping potential }$$

$$\frac{1240}{200}=5.01+\mathrm{eV}_{s}$$

$$\begin{aligned}\therefore e v_{s} &=6 \cdot 21-5.01\\&=1.2\mathrm{ev}\\\therefore\left[v_{s}\right.&=1.2 \mathrm{v}]\end{aligned}$$