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

We have, $\frac{1}{2}m{v}^2=eV$

De broglie wavelength= $\lambda =\frac{h}{mv}$

$\lambda =\frac{h}{\sqrt{2meV}}$

${\lambda }_p=\frac{h}{\sqrt{2\left(m\right)eV}}$....(1)

${\lambda }_{\alpha }=\frac{h}{2\left(4m\right)2eV}$....(2)

Now, $\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{8}\simeq 3$