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(B) The kinetic energy $KE$ of a charged particle accelerated through a potential $V$ is $KE = qV$.The radius $r$ of a charged particle moving in a un

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$\sqrt{2} : 1 : \sqrt{2}$

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(B) The kinetic energy $KE$ of a charged particle accelerated through a potential $V$ is $KE = qV$.The radius $r$ of a charged particle moving in a uniform magnetic field $B$ is given by $r = \frac{mv}{qB}$.Since $KE = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the radius formula: $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.Since $B$ and $V$ are constant for all particles,$r \propto \sqrt{\frac{m}{q}}$.For an $\alpha$-particle: $m_{\alpha} = 4m_p, q_{\alpha} = 2e$. Thus,$r_{\alpha} \propto \sqrt{\frac{4m_p}{2e}} = \sqrt{2} \sqrt{\frac{m_p}{e}}$.For a proton: $m_p = m_p, q_p = e$. Thus,$r_p \propto \sqrt{\frac{m_p}{e}}$.For a deuteron: $m_d = 2m_p, q_d = e$. Thus,$r_d \propto \sqrt{\frac{2m_p}{e}} = \sqrt{2} \sqrt{\frac{m_p}{e}}$.The ratio $r_{\alpha} : r_p : r_d = \sqrt{2} : 1 : \sqrt{2}$.

An $\alpha$-particle,a proton,and a deuteron enter a uniform transverse magnetic field $B$ with the same accelerating potential $V$. Find the ratio of the radii of the paths followed by these particles.

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