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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.Since the particles are accelerated by the same potential

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$\lambda_{p} = \frac{\lambda_{e}}{20 \sqrt{5}}$

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m(KE)}}$.Since the particles are accelerated by the same potential difference $V$, the kinetic energy $KE = qV$.Thus, $\lambda = \frac{h}{\sqrt{2mqV}}$.Since $h$, $q$, and $V$ are constants for both particles, we have $\lambda \propto \frac{1}{\sqrt{m}}$.Therefore, $\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{m_{p}}}$.Given $m_{p} = 2000 m_{e}$, we substitute this into the ratio:$\frac{\lambda_{p}}{\lambda_{e}} = \sqrt{\frac{m_{e}}{2000 m_{e}}} = \sqrt{\frac{1}{2000}} = \frac{1}{\sqrt{400 	imes 5}} = \frac{1}{20 \sqrt{5}}$.Hence, $\lambda_{p} = \frac{\lambda_{e}}{20 \sqrt{5}}$.

$A$ proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is $2000$ times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(\lambda_{p})$ and that of the electron $(\lambda_{e})$?

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