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

We consider a uniform magnetic field that is perpendicular to the plane of motion of the charged particle. For a charged particle of charge $$q$$ and mass $$m$$ moving with speed $$v$$ in a circle of radius $$r$$, the magnetic Lorentz force provides the required centripetal force. The relation is

$$q\,v\,B = \dfrac{m v^{2}}{r}$$

First we isolate the radius $$r$$. Cancelling one factor of $$v$$ from both sides, we obtain

$$q B = \dfrac{m v}{r} \quad \Longrightarrow \quad r = \dfrac{m v}{q B}$$

The question tells us that the proton, the electron, and the Helium nucleus all possess the same kinetic energy, which we denote by $$E$$. The kinetic energy of a particle is given by

$$E = \dfrac{1}{2} m v^{2}$$

Solving this expression for the speed $$v$$, we have

$$v = \sqrt{\dfrac{2E}{m}}$$

We now substitute this expression for $$v$$ into the formula for the radius:

$$r = \dfrac{m}{q B} \,\sqrt{\dfrac{2E}{m}} = \dfrac{\sqrt{2 E}\, \sqrt{m}}{q B}$$

The factors $$\sqrt{2E}$$ and $$B$$ are common to all three particles because the energy and the magnetic field are the same for each. Hence, for comparison of radii, we may write the proportionality

$$r \;\propto\; \dfrac{\sqrt{m}}{q}$$

Therefore, the ordering of the radii depends solely on the ratio $$\sqrt{m}/q$$ for the proton, the electron, and the Helium nucleus. We list their masses and charges (magnitude of charge only, because radius is a positive quantity):

• Electron: $$m = m_e,\; q = e$$
• Proton: $$m = m_p,\; q = e$$
• Helium nucleus (alpha particle): $$m = 4 m_p,\; q = 2e$$

Now we compute the proportional factors:

$$ \begin{aligned} r_e &\propto \dfrac{\sqrt{m_e}}{e},\\[4pt] r_p &\propto \dfrac{\sqrt{m_p}}{e},\\[4pt] r_{He} &\propto \dfrac{\sqrt{4 m_p}}{2e} = \dfrac{2\sqrt{m_p}}{2e} = \dfrac{\sqrt{m_p}}{e}. \end{aligned} $$

We immediately observe that

$$r_{He} \propto \dfrac{\sqrt{m_p}}{e} = r_p.$$

Thus $$r_{He} = r_p.$$ To compare with the electron, note that the electron’s mass is enormously smaller than the proton’s mass, i.e.

$$m_e \ll m_p \quad\Longrightarrow\quad \sqrt{m_e} \ll \sqrt{m_p}.$$

Consequently

$$r_e \propto \dfrac{\sqrt{m_e}}{e} \ll \dfrac{\sqrt{m_p}}{e} \propto r_p = r_{He}.$$

Collecting these results, we obtain the order

$$r_e \lt r_p = r_{He}.$$

This ordering corresponds to Option B.

Hence, the correct answer is Option B.