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(B) When a particle of charge $e$ is accelerated through a potential difference $V$,its kinetic energy is given by $K = eV = \frac{p^2}{2m}$.Thus,the

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$\lambda \sqrt{\frac{m}{M}}$

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(B) When a particle of charge $e$ is accelerated through a potential difference $V$,its kinetic energy is given by $K = eV = \frac{p^2}{2m}$.Thus,the momentum is $p = \sqrt{2meV}$.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.Since $h$,$e$,and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.For an electron of mass $m$,$\lambda_e = \lambda = \frac{k}{\sqrt{m}}$ (where $k = \frac{h}{\sqrt{2eV}}$).For a proton of mass $M$,$\lambda_p = \frac{k}{\sqrt{M}}$.Dividing the two equations: $\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m}}{\sqrt{M}} = \sqrt{\frac{m}{M}}$.Therefore,$\lambda_p = \lambda \sqrt{\frac{m}{M}}$.

An electron of mass $m$ is accelerated through a potential difference of $V$. Its de Broglie wavelength is $\lambda$. If a proton of mass $M$ is accelerated through the same potential difference,its de Broglie wavelength will be ..............

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