Question

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. $\lambda \frac{m}{M}$
• C. $\lambda \sqrt{\frac{M}{m}}$
• D. $\lambda \sqrt{\frac{m}{M}}.$

Answer 1

Answer: (D)

Momentum of electrons $\left(p_e\right)=\sqrt{2meV}$
and momentum for proton $\left(p_p\right)=\sqrt{2MeV}$
Therefore, $\frac{{\lambda }_p}{{\lambda }_e}=\frac{h/p_p}{h/p_e}=\frac{p_e}{p_p}=\frac{\sqrt{2meV}}{\sqrt{2MeV}}=\sqrt{\left(\frac{m}{M}\right)}$.
Therefore ${\lambda }_p=\lambda \sqrt{\left(\frac{m}{M}\right)}$.