• # question_answer An electron of mass m when accelerated through a potential difference V has de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be A) $\lambda \frac{m}{M}$ B) $\lambda \sqrt{\frac{m}{M}}$ C) $\lambda \frac{M}{m}$ D) $\lambda \sqrt{\frac{M}{m}}$

[b] $\lambda =\frac{h}{\sqrt{2mE}}$ $\Rightarrow$ $\lambda \propto \frac{1}{\sqrt{m}}$ (E = same)