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(C) The radius $r$ of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}m

Vedclass · Accepted Answer

$4$

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(C) The radius $r$ of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Substituting this into the radius formula,we get $r = \frac{\sqrt{2mK}}{qB} = \frac{1}{B} \sqrt{\frac{2K}{q^2/m}}$.Since specific charge $\alpha = \frac{q}{m}$,we can write $r = \frac{1}{B} \sqrt{\frac{2K}{m \alpha^2}}$.Given the ratio of specific charges $\frac{\alpha_1}{\alpha_2} = \frac{2}{1}$ and masses $\frac{m_1}{m_2} = \frac{1}{4}$.Since kinetic energy $K$ and magnetic field $B$ are the same for both,the ratio of radii is $\frac{r_1}{r_2} = \sqrt{\frac{m_1}{m_2}} \cdot \frac{\alpha_2}{\alpha_1}$.Substituting the values: $\frac{r_1}{r_2} = \sqrt{\frac{1}{4}} \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.Thus,the ratio of the radii is $1:4$.

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