Question

Find the ratio of de Broglie wavelength of a proton and an $\alpha -particle$ which have been accelerated through same potential difference

• A. $2\sqrt{2}:1$
• B. 3:2
• C. $3\sqrt{2}:1$
• D. 2:1

Answer 1

The correct option is A $2\sqrt{2}:1$

The de Broglie wavelength is given by, $\lambda =\frac{h}{\sqrt{2mqV}}$

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

Thus, ${\lambda }_p=\frac{h}{\sqrt{2m_p{q}_{p}V}}$ and

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }V}}=\frac{h}{\sqrt{2\left(4m_p\right)\left(2q_p\right)V}}$

Thus, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{2\sqrt{2}}{1}$