Question

A proton, a neutron, an electron and an $\alpha$-particle have same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e}$ and $\lambda_{\alpha}$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following :

• A. $\lambda_{ p }=\lambda_{ n } >\lambda_{ e } >\lambda_{\alpha}$
• B. $\lambda_{\alpha} < \lambda_{ n }< \lambda_{ p }< \lambda_{ e }$
• C. $\lambda_{ e }< \lambda_{ p }=\lambda_{ n } >\lambda_{\alpha}$
• D. $\lambda_{ e }=\lambda_{ p }=\lambda_{ n }=\lambda_{\alpha}$

Answer 1

Answer: (B)

$\lambda=\frac{ h }{\sqrt{2 Em }}$
$\lambda \propto \frac{1}{\sqrt{ m }}$
$\therefore \lambda_{ e }>\lambda_{ p }>\lambda_{ n }>\lambda_{\alpha}$