MCQ
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
    $\sqrt{2}: 1$
  • $2 \sqrt{2}: 1$
  • C
    $4 \sqrt{2}: 1$
  • D
    $8: 1$

Answer

Correct option: B.
$2 \sqrt{2}: 1$
b
$p =\sqrt{2 mE }=\sqrt{2 mqV }$

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

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

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