Question

A proton and $\alpha -$ particle are accelerated through the same potential. The ratio of de Broglie wavelength of the proton to $\alpha -$ particle will be is $X.$ Then, the value of $X^2$ is

Answer 1

de-Broglie wavelength of proton $\left({\lambda }_p\right)$ and $\alpha -$ particle $\left({\lambda }_{\alpha }\right)$ are given as

${\lambda }_p=\frac{h}{\sqrt{2m_p\times eV}};{\lambda }_{\alpha }=\frac{h}{\sqrt{2\left(4m_p\right)\times 2eV}}$

$X=\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{8};X^2=8$