A proton and an alpha-particle (with their ma | vedclass

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(A) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is $K = qV$.For a proton: $m_p = m$,$q_

Vedclass · Accepted Answer

$1 : \sqrt{2}$

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(A) The kinetic energy $K$ acquired by a particle of charge $q$ accelerated through a potential difference $V$ is $K = qV$.For a proton: $m_p = m$,$q_p = q$,$K_p = qV$.For an $\alpha$-particle: $m_{\alpha} = 4m$,$q_{\alpha} = 2q$,$K_{\alpha} = (2q)V = 2qV$.The radius $r$ of the circular path of a charged particle in a magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Taking the ratio of the radii:$\frac{r_p}{r_{\alpha}} = \frac{\sqrt{2m_p K_p} / q_p B}{\sqrt{2m_{\alpha} K_{\alpha}} / q_{\alpha} B} = \frac{\sqrt{m_p K_p}}{q_p} \cdot \frac{q_{\alpha}}{\sqrt{m_{\alpha} K_{\alpha}}}$Substituting the values:$\frac{r_p}{r_{\alpha}} = \frac{\sqrt{m \cdot qV}}{q} \cdot \frac{2q}{\sqrt{4m \cdot 2qV}} = \frac{\sqrt{mqV}}{q} \cdot \frac{2q}{\sqrt{8mqV}} = \frac{2}{\sqrt{8}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

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