Question
An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be.
✓
Answer
$p =\sqrt{2 mE }=\sqrt{2 mqV }$
$\frac{ p _{\alpha}}{ p _{ p }}=\sqrt{\frac{ m _{\alpha} q _{\alpha}}{ m _{ p } q _{ p }}}=\sqrt{\frac{4}{1} \times \frac{2}{1}}$
$=\frac{2 \sqrt{2}}{1}$
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