Question

Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of helium and the hydrogen ions are:

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

Answer 1

Answer: (B)

Mass of helium is four times of hydrogen atom.
When a voltage (potential drop) $V$ is applied then energy of ion with charge $q$ is
$E=q V \,\,\,...(1)$
Also a particle of mass $m$ moving with velocity $v$,
has kinetic energy is $K=\frac{1}{2} m v^{2} \,\,\,\,....(2)$
From Eqs. (1) and (2), we have
$\frac{\frac{1}{2} M_{H e}\left(v_{H e}\right)^{2}}{\frac{1}{2} M_{H}\left(v_{H}\right)^{2}}=\frac{q V}{q'V}$
$\Rightarrow \left(\frac{v_{H e}}{v_{H}}\right)^{2}=\frac{q}{q'}\left(\frac{M_{H}}{M_{H e}}\right)$
$=\frac{2}{1} \times \frac{1}{4}=\frac{1}{2}$
$\Rightarrow \frac{v_{H} e}{v_{H}}=\frac{1}{\sqrt{2}}$