Question

A proton and an $\alpha -$particle are accelerated through a potential difference of $200\text{V}$. The ratio of the de-Broglie wavelengths associated with the proton to that with an $\alpha -$particle is,

• A. $\sqrt{2}:1$
• B. $2:1$
• C. $2\sqrt{2}:1$
• D. $\frac{1}{2\sqrt{2}}:1$

Answer 1

The correct option is C $2\sqrt{2}:1$

A point charge $q$ of mass $m$, when accelerated through a potential difference $V$ has kinetic energy $E=qV$ and momentum, $p=\sqrt{2m(K.E)}$
$\therefore p=\sqrt{2mqV}$

So, de-Broglie wavelength is given as
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqV}}$

Now,
$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}=\sqrt{\frac{4m_p\left(2e\right)}{m_p\left(e\right)}}=2\sqrt{2}$

Hence, $\left(C\right)$ is the correct answer.