Question

A proton when accelerated through a potential difference of $V$ volts has a wave length $\lambda$ associated with it. An $\alpha$ -particle in order to have the same wavelength $\lambda$ , must be accelerated through a potential difference (in volts)

• A. $2V$
• B. $V$
• C. $\frac{V}{4}$
• D. $\frac{V}{8}$

Answer 1

Answer: (D)

For a charged particle
$\lambda =\frac{h}{\sqrt{2mqV}}$
Since,${\lambda }_p={\lambda }_{\alpha }$ we have
$\therefore \frac{h}{\sqrt{2m_p{q}_{p}V}}=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }{V}_{\alpha }}}$
$V_{\alpha }=\frac{m_p{q}_{p}V}{m_{\alpha }{q}_{\alpha }}$
$m_p{q}_{p}V=m_{\alpha }{q}_{\alpha }{V}_{\alpha }$
$=\frac{m_p{q}_{p}V}{4m_p\cdot 2q_p}=\frac{V}{8}$