Question

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 V$. After this, their de Broglie wavelengths are $\lambda_{\alpha}$ and $\lambda_{ p }$ respectively. The ratio $\frac{\lambda_{ p }}{\lambda_{a}}$ is

• A. 3.8
• B. 8
• C. 7.8
• D. 2.8

Answer 1

Answer: (D)

$\lambda=\frac{ h }{ p }=\frac{ h }{\sqrt{2 m q V }}$
$\frac{\lambda_{ p }}{\lambda_{\alpha}}=\sqrt{\frac{ m _{\alpha} q _{\alpha}}{ m _{ p } q _{ p }}}=\sqrt{\frac{4 m _{ p } \times 2 e }{ m _{ p } \times e }}=\sqrt{8}=2 \sqrt{2}$
$=2 \sqrt{2}$
$\frac{\lambda_{ p }}{\lambda_{\alpha}}=2 \times 1.4=2.8$