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):
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- \( \frac{he}{4 \pi m} \)
- \( \frac{he}{2 \pi m} \)
- \( \frac{he}{2 m} \)
- \( \frac{he}{4 m} \)
The Correct Option is A
Solution and Explanation
The magnetic moment of an electron in its lowest energy state is determined starting with the quantization of magnetic flux, given by \( \phi = n \frac{h}{e} \). Here, \( \phi \) is the magnetic flux through the electron's orbit, \( n \) is an integer, \( h \) is Planck's constant, and \( e \) is the electron's charge. For the lowest energy state, \( n = 1 \), resulting in \( \phi = \frac{h}{e} \).
The magnetic moment \( \mu \) of an electron in a circular orbit is \( \mu = \frac{IA}{c} \), where \( I \) is the current, \( A \) is the orbit's area, and \( c \) is the speed of light. The current \( I \) is the electron's charge divided by its orbital period.
The orbit's area \( A \) relates to the magnetic flux by \( A = \phi/B \). Substituting \( \phi = \frac{h}{e} \), we get \( A = \frac{h}{eB} \).
In quantum mechanics, the lowest energy state's angular momentum is quantized as \( mvr = \frac{nh}{2\pi} \). For \( n = 1 \), this becomes \( mvr = \frac{h}{2\pi} \). The speed \( v \) is \( v = \frac{rB}{m} \).
Solving \( mvr = \frac{h}{2\pi} \) for \( r \) gives \( r = \frac{h}{2\pi mv} \).
Combining these expressions, the magnetic moment \( \mu \) is found to be \( \mu = \frac{he}{4\pi m} \).
Therefore, the magnetic moment of an electron in its lowest energy state is \( \frac{he}{4\pi m} \). This result matches the provided option \( \frac{he}{4 \pi m} \).