$\begin{array}{1 1}(a)\;0.216\times 10^{-19}J&(b)\;0.312\times 10^{-10}J\\(c)\;1.345\times 10^{-19}J&(d)\;0.114\times 10^{-14}J\end{array}$

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Energy given to $I_2$ molecule $=\large\frac{hc}{\lambda}$

$\Rightarrow \large\frac{6.626\times 10^{-34}\times 3\times 10^8}{4500\times 10^{-10}}$

$\Rightarrow 4.417\times 10^{-19}J$

Energy used for breaking up of $I_2$ molecule=$\large\frac{2.4\times 10^3}{6.023\times 10^{23}}$

$\Rightarrow 3.984\times 10^{-19}J$

Energy used in importing kinetic energy to two I atoms

$\Rightarrow [4.417-3.984]\times 10^{-19}J$

Kinetic energy/iodine atom =$[(4.417-3.984)/2]\times 10^{-19}$

$\Rightarrow 0.216\times 10^{-19}J$

Hence (a) is the correct answer.

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