The work function of a surface of a photosensitive material is $$6.2\ eV$$. The wavelength of the incident radiation for which the stopping potential is $$5\ V$$ lies in the :-
Correct option is C. Ultraviolet region
$$\begin{array}{l}\text { work function }(\phi)=6.2\mathrm{ev} \\\text { Stopping poleutial }\left(v_{s}\right)=5 v \\\lambda \text { incident }=?\end{array}$$
$$\begin{array}{l}E=\phi+e V_{s} \\E=b \cdot 2 e v+s e v \\f=11 \cdot 2 e v\end{array}$$
$$\text { and } \begin{aligned}\underline{f}=\frac{h c}{\lambda} & \Rightarrow 11.2\mathrm{ev}=\frac{6.62\times 10^{-34} \times 3\times 10^{8}}{\lambda} \\\Rightarrow \quad \lambda &=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{11.2 \times 1.6 \times 10^{-19}} . \\\lambda &=1.108\times 10^{-7} \mathrm{~m} \\\lambda &=110.8\mathrm{nm}\end{aligned}$$
$$\text { This wavelength lies in the UV region. }$$