$\begin{array}{1 1}(a)\;2.63\times 10^6m^{-1}&(b)\;4.62\times 10^6m^{-1}\\(c)\;3.63\times 10^6m^{-1}&(d)\;3.63\times 10^2m^{-1}\end{array}$

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Energy of an accelerated electron =$Q.V=1.602\times 10^{-19}\times 4.5$

$\Rightarrow 7.209\times 10^{-19}J$

This energy is completely converted into light.

$\large\frac{hc}{\lambda}$$=7.209\times 10^{-19}$

$\large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{\lambda}$$=7.209\times 10^{-19}$

$\large\frac{1}{\lambda}$=wave no =$3.63\times 10^6m^{-1}$

Hence (c) is the correct answer.

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