$(a)\;62.3v\qquad(b)\;63.3v\qquad(c)\;61.5v\qquad(d)\;60v$

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For an electron $\large\frac{1}{2}$$mu^2=e.V$

and $\lambda=\large\frac{h}{mu}$

Thus $\large\frac{1}{2}m$$\large\frac{h^2}{m^2\lambda^2}$$=e.V$

$V=\large\frac{1\times (6.62\times 10^{-34})^2}{2\times 9.108\times 10^{-31}\times (1.54\times 10^{-10})^2\times 1.602\times 10^{-19}}$

$\Rightarrow 63.3Volt$

Hence (b) is the correct answer.

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