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Step-by-Step Solution
Step 1: Write down the energy gained by each ion
When an ion is accelerated from rest through a potential difference $V$, the work done on the ion becomes its kinetic energy. Mathematically,
$
q\,V = \tfrac{1}{2} m v^2
$
where
• $q$ is the charge of the ion,
• $m$ is the mass of the ion,
• $v$ is the final speed of the ion.
Step 2: Express the speed
Rearrange to solve for $v$:
$
v = \sqrt{\tfrac{2\,q\,V}{m}}
$
Step 3: Identify $q$ and $m$ for the ions
1. Hydrogen ion ($\text{H}^+$):
• Charge $q_H = e$
• Mass $m_H \approx m_p$ (mass of a proton)
2. Singly ionized helium atom ($\text{He}^+$):
• Charge $q_{\text{He}} = e$ (because it has lost one electron, leaving net charge +$e$)
• Mass $m_{\text{He}} \approx 4\,m_p$ (about four times a proton’s mass)
Step 4: Form the ratio of final speeds
Since both ions are accelerated through the same potential difference $V$,
$
\frac{v_H}{v_{\text{He}}}
=
\frac{\sqrt{\tfrac{2\,q_H\,V}{m_H}}}{\sqrt{\tfrac{2\,q_{\text{He}}\,V}{m_{\text{He}}}}}
=
\sqrt{
\frac{q_H/m_H}{q_{\text{He}}/m_{\text{He}}}
}.
$
Given that $q_H = q_{\text{He}} = e$, and $m_{\text{He}} = 4\,m_p$, $m_H = m_p$, we get:
$
\frac{v_H}{v_{\text{He}}}
=
\sqrt{\frac{e/m_p}{\,e/(4\,m_p)}}
=
\sqrt{4}
=
2.
$
Step 5: Conclude the ratio
Thus, the final speeds of hydrogen ion and singly ionized helium ion are in the ratio
$
2 : 1.
$
