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 (μ) produced due to the revolution of an electron is given by:

μ = I · A

where:

• I is the current produced by the moving electron
• A is the area of the electron’s circular orbit

Step 1: Electron-induced current

For an electron moving in a circular orbit, the current is defined as the charge passing a point per unit time. If T is the time period of revolution, then:

f = 1 / T = v / (2πr)

Hence, the current I is:

I = e / T = e · (v / 2πr) = ev / (2πr)

Step 2: Orbital area

The area enclosed by the circular orbit is:

A = πr²

Step 3: Magnetic dipole moment

μ = I · A = (ev / 2πr) · (πr²)

μ = evr / 2

Thus, the magnetic dipole moment of an atom is:

μ = evr / 2

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(b) To show: μ⃗ = −(e / 2m) L⃗

Solution:

The angular momentum of the electron is:

L⃗ = mvr

Substituting vr from this expression into μ:

μ = (e / 2) · (vr) = (e / 2m) · (mvr)

Since the electron carries negative charge, the direction of μ⃗ is opposite to L⃗:

μ⃗ = −(e / 2m) L⃗