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 the electron to that with the proton is :

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

Answer 1

The correct option is A $\sqrt{\frac{m_p}{m_e}}$

The relation between de Broglie wavelength $\lambda$ and potential difference $\left(V\right)$ is given by,

${\lambda }_e=\frac{h}{\sqrt{2m_{e}eV}}$

${\lambda }_p=\frac{h}{\sqrt{2m_{p}eV}}$

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

Hence, $\left(C\right)$ is the correct answer.