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(B) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Since the particle is accelera

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$4$

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(B) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Since the particle is accelerated through a potential $V$,the kinetic energy $K = qV$.Substituting this,we get $r = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.For the $\alpha$-particle: $m_\alpha = 4 	ext{ amu}$,$q_\alpha = 2e$.For the sulfur ion: $m_s = 32 	ext{ amu}$,$q_s = 1e$.The ratio of the radii is $\frac{r_s}{r_\alpha} = \frac{\sqrt{m_s/q_s}}{\sqrt{m_\alpha/q_\alpha}} = \sqrt{\frac{m_s}{m_\alpha} \cdot \frac{q_\alpha}{q_s}}$.Substituting the values: $\frac{r_s}{r_\alpha} = \sqrt{\frac{32}{4} \cdot \frac{2}{1}} = \sqrt{8 \cdot 2} = \sqrt{16} = 4$.

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