A proton, a deuteron and an $α -$ particle are accelerated through the same potential difference of V volt. The velocities acquired by them are in the ratio

1 : 1 : \sqrt{2}

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1 : \sqrt{2} : 1

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1 : 1 : 1

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\sqrt{2} : 1 : 1

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Solution

The correct option is C $\sqrt{2} : 1 : 1$
Charge on proton $= α$
Charge on deuteron $= α$
Charge on $α$ -particle $= 2 α$
So, K.E. on proton, deuteron and $α$ -particle are in the ration $α V, α V$ and $2 α V$ respectively.
So, let the velocities be in the ratio of $V_{1}, V_{2}$ and $V_{3}$
So, for Proton ,
mass $= β$
$\Rightarrow \frac{1}{2} β V_{1}^{2} = α V \Rightarrow V_{1} = \sqrt{\frac{2 α V}{β}}$

For deuteron,
mass $= 2 β$
$\Rightarrow β V_{2}^{2} = α V \Rightarrow V_{2} = \sqrt{\frac{α V}{β}}$
For $α$ -particle,
mass $= 4 β$
$\Rightarrow 2 α V = 2 α V_{3}^{2} \Rightarrow V_{3} = \sqrt{\frac{α V}{β}}$
So, Ratio of $V_{1} : V_{2} : V_{3} = \sqrt{2} : 1 : 1$
So, the answer is option (D).

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A proton, a deuteron and an α−particle are accelerated through the same potential difference of V volt. The velocities acquired by them are in the ratio

A proton, a deuteron and an $α -$ particle are accelerated through the same potential difference of V volt. The velocities acquired by them are in the ratio