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  • An particle and a proton are accelerated from rest through the same potential difference.

An \mathrm{\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:
 

Option: 1

\sqrt{2}: 1


Option: 2

2 \sqrt{2}: 1


Option: 3

4 \sqrt{2}: 1


Option: 4

8: 1


Answers (1)

\mathrm{q_\alpha=2 e, m_\alpha=4 m_p }

\mathrm{q_{p}=e}

\mathrm{P=\sqrt{2m\left ( kE \right )}=\sqrt{2mqv}}

\mathrm{\frac{P_{\alpha }}{P_{p}}=\sqrt{\frac{2 m_\alpha q_\alpha V}{2 m_p q_p V}} }

\mathrm{\frac{p_\alpha}{p_p}=\sqrt{4\times 2} =2\sqrt{2}}

Hence 2 is correct option



 

Posted by

Ramraj Saini

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