Question

A helium ion and a hydrogen ion are accelerated from rest through a potential difference of $V$ to velocities $v_{He}$ and $v_H$ respectively. If helium has lost one electron, the ratio of $v_{He}/v_H$ is:

• A. 1/4
• B. 1/2
• C. 1
• D. $\sqrt{2}$

Answer 1

Answer: (B)

1/2
Since the charge on both He ion and ion is same, they will gain same kinetic energy after passing through the field equivalent to $qV$

Therefore,
$\frac{1}{2}{m}_{He}{v}_{He}^2=\frac{1}{2}{m}_H{v}_H^2$
$⟹\frac{v_{He}^2}{v_H^2}=\frac{m_H}{m_{He}}$
now,

$\frac{m_H}{m_{He}}=\frac{1}{4}$

$⟹\frac{v_{He}^2}{v_H^2}=\frac{1}{4}$
$⟹\frac{v_{He}}{v_H}=\frac{1}{2}$