Question

An electron of mass ?m? is accelerated by a potential difference V and the corresponding de-Broglie wavelength is $\lambda$. The de Broglie wavelength of a proton of mass M if it is accelerated by the same potential difference is ,

• A. $\lambda\left(\frac{m}{M}\right)$
• B. $\lambda\left(\frac{M}{m}\right)$
• C. $\lambda\sqrt{\frac{m}{M}}$
• D. $\lambda\sqrt{\frac{M}{m}}$

Answer 1

Answer: (C)

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

Answer 2

For electron $\lambda=\frac{h}{\sqrt{2meV}}$
For Proton $\lambda_{p} =\frac{h}{\sqrt{2MeV}}$
$\frac{\lambda_{p}}{\lambda} =\sqrt{\frac{m}{M}}$