An α particle and a proton are accelerated from rest through same potential difference. ratio of li

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2. Electric Potential and Capacitance

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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.

A

$\sqrt{2}: 1$

B

$2 \sqrt{2}: 1$

C

$4 \sqrt{2}: 1$

D

$8: 1$

(JEE MAIN-2022)

Solution

$p =\sqrt{2 mE }=\sqrt{2 mqV }$

$\frac{ p _{\alpha}}{ p _{ p }}=\sqrt{\frac{ m _{\alpha} q _{\alpha}}{ m _{ p } q _{ p }}}=\sqrt{\frac{4}{1} \times \frac{2}{1}}$

$=\frac{2 \sqrt{2}}{1}$

Standard 12

Physics

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