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.
(b) If \( \vec{L} \) is the angular momentum of the electron, show that:
\[ \vec{\mu} = -\frac{e}{2m} \vec{L} \]
The sum of the spin-only magnetic moment values (in B.M.) of $[\text{Mn}(\text{Br})_6]^{3-}$ and $[\text{Mn}(\text{CN})_6]^{3-}$ is ____.
Consider a water tank shown in the figure. It has one wall at \(x = L\) and can be taken to be very wide in the z direction. When filled with a liquid of surface tension \(S\) and density \( \rho \), the liquid surface makes angle \( \theta_0 \) (\( \theta_0 < < 1 \)) with the x-axis at \(x = L\). If \(y(x)\) is the height of the surface then the equation for \(y(x)\) is: (take \(g\) as the acceleration due to gravity)
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :