Question

A proton and an $\alpha -$particle are accelerated through the same potential difference. The ratio of de Broglie wavelength $\lambda_{p}$to that of$\lambda_{\alpha}$is:

• A. $\sqrt{2}:1$
• B. $\sqrt{4}:1$
• C. )$\sqrt{6}:1$
• D. $\sqrt{8}:1$

Answer 1

Answer: (D)

$\sqrt{8}:1$

Answer 2

: As$\lambda = \frac{h}{\sqrt{2mqV}}$

$\therefore\lambda \propto \frac{1}{\sqrt{mq}}\therefore\frac{\lambda_{p}}{\lambda_{\alpha}} = \frac{\sqrt{m_{\alpha}q_{\alpha}}}{m_{p}q_{p}}$

$= \frac{\sqrt{4m_{p} \times 2e}}{\sqrt{m_{p} \times e}} = \sqrt{8}(\therefore m_{\alpha} = 4m_{p},q_{\alpha} = 2q_{p})$