Question

A proton when accelerated through a potential difference of $V$ volt has a wavelength $\lambda$ associated with it. An $\alpha -particle$ in order to have the same $\lambda$ must be accelerated through a potential difference of:

• A. $V$ volt
• B. $4V$ volt
• C. $2V$ volt
• D. $\left(V/8\right)$ volt

Answer 1

The correct option is D $\left(V/8\right)$ volt

We know that, wavelength $\lambda =\frac{h}{\sqrt{mqV}}$ or $V=\frac{h^2}{2mq{\lambda }^2}$

Mass, $m_{\alpha }=4m_p$ and charge , $q_{\alpha }=2q_p$

So, $V_p=V=\frac{h^2}{2m_p{q}_p{\lambda }^2}$ and

$V_{\alpha }=\frac{h^2}{2m_{\alpha }{q}_{\alpha }{\lambda }^2}=\frac{h^2}{2\left(4m_p\right)\left(2q_p\right){\lambda }^2}=V_p/8=V/8$ volt