Question:18 A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(h / \mathrm{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)
(1) \frac{h e}{\pi m}
(2) \frac{h e}{2 \pi m}
(3) \frac{h e B}{\pi m}
(4) \frac{h e B}{2 \pi m}
Answer: Option (2)
Explanation:
The model states that the magnetic flux through the electron’s circular orbit is quantized as
\Phi = n(h/e). For the lowest energy state,
n = 1, so flux is \Phi = h/e.
The flux through a circular orbit of radius r in magnetic field
B is given by \Phi = B \pi r^2.
Equating the two expressions gives B \pi r^2 = h/e.
Thus, the radius of the orbit is r^2 = \frac{h}{e \pi B}.
The magnetic moment of an electron in circular motion is \mu = I A,
where current I = e/T and area A = \pi r^2.
The time period of revolution is T = \frac{2\pi r}{v}.
For an electron in a magnetic field, v = r(eB/m),
so T = \frac{2\pi m}{eB}.
Thus, current is I = \frac{e}{T} = \frac{e^2 B}{2\pi m}.
Magnetic moment becomes \mu = I A = \frac{e^2 B}{2\pi m} \cdot \pi r^2.
Substituting r^2 = \frac{h}{e \pi B},
we get:
\mu = \frac{e^2 B}{2\pi m} \cdot \pi \cdot \frac{h}{e \pi B} = \frac{he}{2\pi m}.
Thus, the magnetic moment in the lowest state is
\frac{he}{2\pi m}.