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 ${\lambda }_p/{\lambda }_{\alpha }$, to the nearest integer, is:

Answer 1

Kinetic energy = K.E = $qV$
De Broglie wavelength = $\frac{h}{\sqrt{2mK.E}}$ = $\frac{h}{\sqrt{2mqV}}$
Therefore ratio $\frac{{\lambda }_p}{{\lambda }_{\alpha }}$ = $\frac{\sqrt{m_{\alpha }{q}_{\alpha }}}{\sqrt{m_p{q}_p}}$
Alpha particle has 4 times the mass and twice the charge of a proton .
$\frac{{\lambda }_p}{{\lambda }_{\alpha }}$ = $\sqrt{8}$ $\approx$ 3