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(D) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv

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$r_e < r_p = r_{He}$

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(D) The radius of a charged particle moving 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 $r = \frac{\sqrt{2mK}}{qB}$.Given that $K$ and $B$ are the same for all particles,$r \propto \frac{\sqrt{m}}{q}$.For a proton $(p)$: $m_p = m, q_p = e$,so $r_p \propto \frac{\sqrt{m}}{e}$.For an electron $(e)$: $m_e \approx \frac{m}{1836}, q_e = e$,so $r_e \propto \frac{\sqrt{m/1836}}{e} = \frac{r_p}{\sqrt{1836}}$. Thus,$r_e For a Helium nucleus $(He^{2+})$: $m_{He} \approx 4m, q_{He} = 2e$,so $r_{He} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e} = r_p$.Therefore,$r_e < r_p = r_{He}$.

$A$ proton,an electron,and a Helium nucleus have the same kinetic energy. They are moving in circular orbits in a plane due to a magnetic field perpendicular to the plane. Let $r_p, r_e$,and $r_{He}$ be their respective radii. Then:

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