Question

A proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is

• A. $\sqrt{8}$
• B. $\frac{1}{\sqrt{8}}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\sqrt{8}$

de-Broglie $\lambda =\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$ where $K=qV$

$⟹$ $\lambda \propto \frac{1}{\sqrt{mq}}$ as $V$ is same for both

$\therefore$ $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}$

OR $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{\left(4m_p\right)\left(2q_p\right)}{m_p{q}_p}}=\sqrt{8}$