A doubly ionised helium ion and a $H_{2}$ ion are accelerated through the same potential. The ratio of the speed of helium and $H_{2}$ ion is:

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Solution

Given that a doubly ionized Helium ion and a $H_{2}$ ion are accelerated through the same potential difference. We have to find the ratio of the speed of Helium and $H_{2}$ ion.
So, a doubly ionized ion is an alpha particle having charge $+ 2 e$ and mass $= 4 \times 1.67 \times 10^{- 27} K g$ .
where, $e =$ electrostatic charge $= 1.6 \times 10^{- 19} C$
$H_{2}$ ion is a proton having charge $1.6 \times 10^{- 19} C$ and mass $1.67 \times 10^{- 27} K g$ .

So, K.E of $α$ -particle $= \frac{1}{2} m_{α} v_{α}^{2} = q_{α} V_{α}$
where, $m_{α}$ , $v_{α}$ , $q_{α}$ and $V_{α}$ are mass, velocity, charge and potential of $a l p h a$ -particle respectively.
So, $v_{α} = \sqrt{\frac{2 q_{α} v_{α}}{m_{α}}}$
For $H_{2}$ ion or a proton, K.E $= \frac{1}{2} m_{p} v_{p}^{2} = q_{p} V_{p}$
where, $m_{p}$ , $v_{p}$ , $q_{p}$ and $V_{p}$ are mass, velocity, charge and potential of $H_{2}$ ion respectively.

So, $v_{p} = \sqrt{\frac{2 q_{p} v_{p}}{m_{p}}}$
$\frac{v_{α}}{v_{p}} = \sqrt{\frac{q_{α} v_{α} \times m_{p}}{m_{α} \times q_{p} v_{p}}}$
Given, $V_{α} = V_{p}$ (Potential is same)
So, $\frac{v_{α}}{v_{p}} = \sqrt{\frac{q_{α} m_{p}}{q_{p} m_{α}}}$

As, $q_{α} = 2 q_{p}$
and, $m_{α} = 4 m_{p}$
So, ratio of velocity is $\frac{v_{α}}{v_{p}} = \sqrt{\frac{2 q_{p} \times m_{p}}{4 m_{p} \times q_{p}}}$
$\frac{v_{α}}{v_{p}} = \sqrt{\frac{1}{2}}$

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