Question

An electron is accelerated through a potential difference of $V$ volt. It has a wavelength $\lambda$ associated with it. Through what potential difference an electron must be accelerated so that its de Broglie wavelengths is the same as that of a proton? Take mass of proton to be $1837$ times larger than the mass of electron

• A. $Vvolt$
• B. $1837Vvolt$
• C. $V/1837volt$
• D. $\sqrt{1837}Vvolt$

Answer 1

The correct option is C $V/1837volt$

In first case, wavelength $\lambda =\frac{h}{\sqrt{2m_{e}eV}}...\left(1\right)$

In second case, ${\lambda }^{\prime }=\frac{h}{\sqrt{2m_{e}e{V}^{\prime }}}...\left(2\right)$

Given, ${\lambda }^{\prime }={\lambda }_p$ and ${\lambda }_p=\frac{h}{\sqrt{2m_{p}e{V}^{\prime }}}=\frac{h}{\sqrt{2\left(1837m_e\right)V^{\prime }}}$

Thus, ${\lambda }^{\prime }=\frac{h}{\sqrt{2\left(1837m_e\right)V^{\prime }}}...\left(3\right)$

Comparing (1) and (3), $1837V^{\prime }=V$ or $V^{\prime }=V/1837$ V