Question

The magnitude of the De-Broglie wavelength ($\lambda$) of an electron (e),proton(p),neutron (n) and $\alpha$ - particle ($\alpha$ ) all having the same energy of MeV, in the increasing order will follow the sequence:

• A. ${\lambda }_e,{\lambda }_p,{\lambda }_n,{\lambda }_{\alpha }$
• B. ${\lambda }_{\alpha },{\lambda }_n,{\lambda }_p,{\lambda }_e$
• C. ${\lambda }_e,{\lambda }_n,{\lambda }_p,{\lambda }_{\alpha }$
• D. ${\lambda }_p,{\lambda }_e,{\lambda }_{\alpha },{\lambda }_n$

Answer 1

The correct option is B ${\lambda }_{\alpha },{\lambda }_n,{\lambda }_p,{\lambda }_e$

We know
$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2Em}$
So $\lambda =\frac{h}{\sqrt{2Em}}$
${\lambda }_e=\frac{h}{\sqrt{2E{m}_e}}$
${\lambda }_p=\frac{h}{\sqrt{2E{m}_p}}$
${\lambda }_n=\frac{h}{\sqrt{2E{m}_n}}$
${\lambda }_{\alpha }=\frac{h}{\sqrt{2E{m}_{\alpha }}}$
we know $m_{\alpha }>m_n>m_p>me$
So ${\lambda }_{\alpha }<{\lambda }_n<{\lambda }_p<\lambda e$