$\begin{array}{1 1}(a)\;3.65eV&(b)\;2.65eV\\(c)\;4.65eV&(d)\;1.65eV\end{array}$

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Energy of photon =work function $\large\frac{1}{2}$$mu^2$

Energy of photon =work function $ev_0$------(1)

Where e is electronic charge and $v_0$ is stopping potential and $ev_0$ is equal to energy required to stop the ejection of electron.

$E_{photon}=\large\frac{hc}{\lambda}$

$\Rightarrow \large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{253.7\times 10^{-9}}$

$\Rightarrow 7.834\times 10^{-19}J$

$\Rightarrow \large\frac{7.834\times 10^{-19}}{1.602\times 10^{-19}}$eV=4.89eV

By equation (1)

$4.89$=work function +0.24

Work function =(4.89-0.24)eV

Work function =4.65eV

Hence (c) is the correct answer.

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