Question

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 \[{r_p}\], \[{r_e}\], \[{r_{He}}\] be their respective radii, then
(A) \[{r_e} > {r_p} > {r_{He}}\]
(B) \[{r_e} < {r_p} < {r_{He}}\]
(C) \[{r_e} < {r_p} = {r_{He}}\]
(D) \[{r_e} > {r_p} = {r_{He}}\]

Answer 1

Hint: The radius of the circular radius is directly proportional to the momentum but inversely proportional to the charge of the particle. The helium nucleus it’s about 4 times as heavy as a proton and twice as much charge. The mass of the electron is far less than the mass of both.
Formula used: In this solution we will be using the following formulae;
\[r = \dfrac{p}{{qB}}\] where \[r\] is the radius of the circular orbit of a particle in a perpendicular magnetic field, \[p\] is the momentum of the particle, \[q\] is the charge of the particle, and \[B\] is the magnetic field.
\[p = \sqrt {2mE} \] where \[m\] is the mass of a particle, \[E\] is the kinetic energy of the particle.

Complete Step-by-Step Solution:
To find the relationship of their radii, we shall recall the formula
\[r = \dfrac{p}{{qB}}\] where \[r\] is the radius of the circular orbit of a particle in a perpendicular magnetic field, \[p\] is the momentum of the particle, \[q\] is the charge of the particle, and \[B\] is the magnetic field.
The momentum
\[p = \sqrt {2mE} \]where \[m\] is the mass of a particle, \[E\] is the kinetic energy of the particle.
Magnetic fields do not possess potential energy, hence all its energy are kinetic.
\[r = \dfrac{{\sqrt {2mE} }}{{qB}}\]
So if they have the same energy, it is obvious that since the electron has a mass far less than the proton or helium then the radius of its path is smaller.
Helium however is only about 4 times as heavy as the proton, and twice as much charge, hence,
\[{r_{He}} = \dfrac{{\sqrt {2\left( {4{m_p}} \right)E} }}{{2eB}} = 2\dfrac{{\sqrt {2{m_p}E} }}{{2eB}}\]
Hence, by simplification, we have
\[{r_{He}} = \dfrac{{\sqrt {2{m_p}E} }}{{eB}} = {r_p}\]
Hence, \[{r_e} < {r_p} = {r_{He}}\]

The correct option is C

Note: For clarity, the mass of the helium is about 4 times as much as the proton because it comprises two neutrons and two protons. The neutron and proton have very similar mass, and only slightly difference, hence, the mass of helium can be approximated as 4 times the mass of proton.