Question

A model for a quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is \( \phi = n h / e \) where \(n\) is an integer, \(h\) is Planck's constant and \(e\) is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (\(m\) is the mass of the electron):

• A. \( \frac{he}{4 \pi m} \)
• B. \( \frac{he}{2 \pi m} \)
• C. \( \frac{he}{2 m} \)
• D. \( \frac{he}{4 m} \)

Hint:

Remember the quantization of magnetic flux and the expressions for magnetic moment and angular momentum of an orbiting charge. The relationship between magnetic moment and angular momentum is \( \mu = \frac{e}{2m} L \).

Answer 1

The Correct Option is A

To determine the magnetic moment of an electron in its lowest energy state according to the model described, we start with the quantization of magnetic flux. The quantization condition given is:

\(\phi = n \frac{h}{e}\)

where \( \phi \) is the magnetic flux through the electron's orbit, \( n \) is an integer, \( h \) is Planck's constant, and \( e \) is the charge of the electron. For the lowest energy state, we consider \( n = 1 \). Thus:

\(\phi = \frac{h}{e}\)

The magnetic moment \( \mu \) of an electron moving in a circular orbit due to a magnetic field is given by:

\(\mu = \frac{IA}{c}\)

where \( I \) is the current, \( A \) is the area of the orbit, and \( c \) is the speed of light. The current \( I \) is given by the electron's charge divided by the period of its orbit.

The area \( A \) of the orbit is related to the magnetic flux by:

\(A = \phi/B\)

Substituting \(\phi = \frac{h}{e}\) from earlier:

\(A = \frac{h}{eB}\)

In quantum mechanics, for the lowest energy state, the angular momentum is quantized and equals:

\(mvr = \frac{nh}{2\pi}\) with \( n = 1 \), giving \( mvr = \frac{h}{2\pi}\). The speed \( v \) can be expressed as:

\(v = \frac{rB}{m}\)

From \( mvr = \frac{h}{2\pi} \), we solve for \( r \):

\(r = \frac{h}{2\pi mv}\)

Combining all expressions, the magnetic moment \( \mu \) in terms of known quantities is:

\(\mu = \frac{he}{4\pi m}\)

Thus, the magnetic moment of an electron in its lowest energy state is:

\(\frac{he}{4\pi m}\)

This matches the given option \( \frac{he}{4 \pi m} \), confirming it as the correct answer.