Question

An electron of mass \(m\) and charge \(-e\) is revolving anticlockwise around the nucleus of an atom. (a) Obtain the expression for the magnetic dipole moment \(\mu\) of the atom.

(b) If \( \vec{L} \) is the angular momentum of the electron, show that:
\[ \vec{\mu} = -\frac{e}{2m} \vec{L} \]

Hint:

In the case of an electron moving in a circular orbit, the magnetic dipole moment is proportional to the angular momentum of the electron. The negative sign indicates that the direction of the magnetic dipole moment is opposite to the direction of the angular momentum due to the negative charge of the electron.

Answer 1

The magnetic dipole moment \(\mu\) due to the revolving electron is given by: \[ \mu = I \cdot A \] Where:
- \(I\) is the current associated with the moving electron,
- \(A\) is the area enclosed by the orbit of the electron.
Step 1: Current due to the electron The electron is moving in a circular orbit, so the current is defined as the charge passing through a point per unit time. The time period \(T\) of the electron's revolution is the time it takes to complete one full revolution around the nucleus. The frequency of the electron is: \[ f = \frac{v}{2 \pi r} \] where:
- \(v\) is the velocity of the electron,
- \(r\) is the radius of the orbit. The current \(I\) is given by: \[ I = \frac{e}{T} = \frac{e}{\frac{2 \pi r}{v}} = \frac{e v}{2 \pi r} \]
Step 2: Area enclosed by the orbit The area \(A\) enclosed by the electron's orbit is the area of a circle: \[ A = \pi r^2 \]
Step 3: Magnetic dipole moment Now, the magnetic dipole moment is: \[ \mu = I \cdot A = \left( \frac{e v}{2 \pi r} \right) \cdot (\pi r^2) = \frac{e v r}{2} \] Thus, the expression for the magnetic dipole moment of the atom is: \[ \mu = \frac{e v r}{2} \] (b) If \(\vec{L}\) is the angular momentum of the electron, show that \(\vec{\mu} =
-\left( \frac{e}{2m} \right) \vec{L\).} Solution: The angular momentum of the electron \( \vec{L} \) is given by: \[ \vec{L} = m v r \] Now, substitute \( v r \) from this equation into the expression for the magnetic dipole moment: \[ \mu = \frac{e v r}{2} = \frac{e}{2m} \cdot (m v r) =
- \frac{e}{2m} \cdot \vec{L} \] Thus, we have shown that: \[ \vec{\mu} =
- \frac{e}{2m} \vec{L} \]