An α particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be

Correct Option - 2


Detailed Solution:

$k_{\alpha }=2eV\Rightarrow P_{\alpha }=\sqrt[]{2K_{\alpha }{m}_{\alpha }}$

k_P = eV P_P = $\sqrt[]{2K_P{m}_P}$

$\therefore \frac{P_{\alpha }}{P_P}=\sqrt[]{\frac{K_{\alpha }{m}_{\alpha }}{K_P{m}_P}}=\sqrt[]{\frac{2\times \left(4m_P\right)}{m_P}}\begin{array}{cc}\hfill \sqrt[]{8}:1\hfill & \hfill 2\sqrt[]{2}:1\hfill \end{array}$

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