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(B) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.The linear momentu

Vedclass · Accepted Answer

$2\sqrt{2} : 1$

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(B) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is given by $K = qV$.The linear momentum $p$ is related to kinetic energy by the formula $p = \sqrt{2mK} = \sqrt{2mqV}$.For an $\alpha$ particle,the mass $m_{\alpha} = 4m_p$ and the charge $q_{\alpha} = 2e$,where $m_p$ is the mass of a proton and $e$ is the elementary charge.For a proton,the mass is $m_p$ and the charge is $q_p = e$.The ratio of their momenta is $\frac{p_{\alpha}}{p_p} = \sqrt{\frac{m_{\alpha} q_{\alpha}}{m_p q_p}} = \sqrt{\frac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{4 \cdot 2} = \sqrt{8} = 2\sqrt{2}$.Thus,the ratio is $2\sqrt{2} : 1$.

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by the above two particles will be:

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