Question

If the angular momentum of an electron of mass m is J, then the magnitude of the magnetic moment will be
A. \[\dfrac{{{\text{eJ}}}}{{\text{m}}}\]
B. \[\dfrac{{{\text{eJ}}}}{{{\text{2m}}}}\]
C. \[{\text{2eJm}}\]
D. \[\dfrac{{{\text{2m}}}}{{{\text{eJ}}}}\]

Answer 1

Hint:Express the period of revolution of the electron in the circular loop in terms of its orbital velocity. Then express the current associated with the electron which is the ratio of charge per unit time. The magnetic moment is the product of the current and the area of the circular loop. Use the expression for the angular momentum of the electron in the equation that you obtained in the previous step.

Formula used:
1. Orbital velocity, \[v = \dfrac{{2\pi r}}{T}\], where, r is the radius and T is the periodic time.
2. \[{m_{orb}} = I\,A\], where I is the current and A is the area.
3. Angular momentum, \[J = mvr\], where, m is the mass of the electron.

Complete step by step answer:
Let’s consider an electron moving with constant speed v in a circular orbit of radius r about the nucleus. We assume that the electron has travelled a circumference of the circle in time T. Therefore, let’s express the orbital velocity of the electron as follows,
\[v = \dfrac{{2\pi r}}{T}\]
\[ \Rightarrow T = \dfrac{{2\pi r}}{v}\] …… (1)

Now, let’s express the current associated with the motion of electron as follows,
\[I = \dfrac{e}{T}\]
Here, e is the charge of an electron.
Substituting the value of T from equation (1) in the above equation, we get,
\[I = \dfrac{e}{{\dfrac{{2\pi r}}{v}}}\]
\[ \Rightarrow I = \dfrac{{ev}}{{2\pi r}}\] …… (2)

We know that the orbital magnetic moment associated with orbital current loop is expressed as,
\[{m_{orb}} = I\,A\] ……. (3)
The area of circumference of the orbital loop is given as,
\[A = \pi {r^2}\] ……. (4)
Substituting equation (2) and (4) in equation (3), we get,
\[{m_{orb}} = \left( {\dfrac{{ev}}{{2\pi r}}} \right)\left( {\pi {r^2}} \right)\]
\[ \Rightarrow {m_{orb}} = \dfrac{{evr}}{2}\] ……. (5)

We have the angular momentum J of the electron is,
\[J = mvr\]
\[ \Rightarrow \dfrac{J}{m} = vr\]
Here, m is the mass of the electron.
Substituting the above equation in equation (5), we get,
\[\therefore{m_{orb}} = \dfrac{{eJ}}{{2m}}\]
This is the required expression for the magnitude of the magnetic moment of an electron.

So, the correct answer is option B.

Note:The magnetic moment and angular momentum are scalar quantities. Since the electron has negative charge, the vector form of magnetic moment is expressed as, \[{\vec m_{orb}} = - \dfrac{{e\vec J}}{{2m}}\]. This implies that the magnetic moment and angular momentum are in opposite directions to each other. Also, the magnetic moment and angular momentum are perpendicular to the plane of the loop.