Question

An $\alpha -$particle and a proton are accelerated from rest by a potential difference of $100V$. After this, their de-Broglie wavelengths are ${\lambda }_{\alpha }$and ${\lambda }_p$ respectively. The ratio $\frac{{\lambda }_p}{{\lambda }_{\alpha }}$ , to the nearest integer, is

Answer 1

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}$
$\lambda =\frac{h}{\sqrt{2\times m\times eV}}$
${\lambda }_{\alpha }=\frac{h}{\sqrt{2\times \left(4m\right)\times \left(2e\right)V}}=\frac{{\lambda }_p}{2\sqrt{2}}$
$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=2\sqrt{2}\approx 3$