Question

An electron of mass $m_e$ and a proton of mass $m_p$ are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

• A. 1
• B. $m_p/m_e$
• C. $m_e/m_p$
• D. $\sqrt{m_p/m_e}$

Answer 1

Answer: (D)

$\sqrt{m_p/m_e}$

The relation between de Broglie wavelength $\left(\lambda \right)$ and potential difference $\left(V\right)$ is given by, $\lambda =\frac{h}{\sqrt{2meV}}$

As $V$ is constant so $\lambda \propto \frac{1}{\sqrt{m}}$

Thus, $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$