Question

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

Answer 1

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}$

$\therefore \frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}$

=$\sqrt{\frac{4m_p\times 2c}{m_p\times c}}$

$=2\sqrt{2}$ ~ 3