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)

$\lambda =\frac{h}{\sqrt{2Em}}$
$\lambda \propto \frac{1}{\sqrt{m}}$
$\therefore {\lambda }_e>{\lambda }_p>{\lambda }_n>{\lambda }_{\alpha }$