An electron of mass ' m ' accelerated through a potential...

The de Broglie wavelength of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2mqV}}$.

For an electron of mass $m$ and charge $e$, the wavelength is $\lambda = \dfrac{h}{\sqrt{2meV}}$.

For a proton of mass $M$ and charge $e$ accelerated through the same potential difference $V$, the wavelength is $\lambda_p = \dfrac{h}{\sqrt{2MeV}}$.

Taking the ratio of the two wavelengths:

$\dfrac{\lambda_p}{\lambda} = \dfrac{\sqrt{2meV}}{\sqrt{2MeV}} = \sqrt{\dfrac{m}{M}}$

$\lambda_p = \left(\sqrt{\dfrac{m}{M}}\right)\lambda$

Answer: \left(\sqrt{\dfrac{m}{M}}\right)\lambda