Question

An electron of mass $m_e$, and a proton of mass $m_p$ are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is

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

Answer 1

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

Given,
$m_e$ = mass of the electron
$m_p$ = mass of proton
By de- Broglie wavelength of matter waves
$\therefore \lambda =\frac{h}{p}(\because p=mv)$
$\therefore \lambda =\frac{h}{mv}$
If $K$ be the kinetic energy of the electron, then
$K=\frac{1}{2}m{v}^2$
or $v=\sqrt{2K/m}$
Now, $\lambda =\frac{h}{m\sqrt{2K/m}}$
$\lambda =\frac{h}{\sqrt{2Km}}$
Since kinetic energy gained by each particle $K=eV=$ same
So, $\lambda \propto \frac{1}{\sqrt{m}}$
Thus, ${\lambda }_e\propto \frac{1}{\sqrt{m_e}}$ ...(i)
${\lambda }_p\propto \frac{1}{\sqrt{m_p}}$ ...(ii)
From eq. (i) and (ii)
$\frac{{\lambda }_e}{{\lambda }_p}=\frac{1/\sqrt{m_e}}{1/\sqrt{m_p}}$
$⟹\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$