$\begin{array}{1 1}(a)\;20.5V&(b)\;32.85V\\(c)\;33.36V&(d)\;31.4V\end{array}$

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$u=\large\frac{h}{m\lambda}$

Mass of proton=$1.67\times 10^{-27}$

$u=\large\frac{6.625\times 10^{-34}}{1.67\times 10^{-27}\times 0.005\times 10^{-9}}$

$\Rightarrow 7.94\times 10^4m/s$

Accelerating potential is V,then velocity (u) acquired by the charge particle having charge Q and mass m.

$QV=\large\frac{1}{2}$$mu^2$

$u=\sqrt{\large\frac{2QV}{m}}$

$\;\;\;=\sqrt{\large\frac{2\times 1.602\times 10^{-19}\times V}{1.67\times 10^{-27}}}$

$7.94\times 10^4=\sqrt{\large\frac{2\times 1.602\times 10^{-19}\times V}{1.67\times 10^{-27}}}$

$V=32.85V$

Hence (b) is the correct answer.

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