NEET · Physics · STD 12 - 12. atoms
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 \(n(h / e)\) 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 \(\frac{h e}{\pi m}\)
- B \(\frac{h e}{2 \pi m}\)
- C \(\frac{h e B}{\pi m}\)
- D \(\frac{h e B}{2 \pi m}\)
Answer & Solution
Correct Answer
(B) \(\frac{h e}{2 \pi m}\)
Step-by-step Solution
Detailed explanation
Magnetic moment
\(M=I A=I\left(\pi r^2\right)\)
\(M=\left(\frac{e u }{2 \pi r}\right)\left(\pi r^2\right)\)
Given \(B\left(\pi r^2\right)=n\left(\frac{h}{e}\right)\)
\(\Rightarrow r ^2=\frac{ h }{ B \pi e } \quad \ldots(2) \quad(\because n =1)\)
And when charge moving in external magnetic field then \(r=\frac{m u}{q B}\)
\(\frac{ v }{ r }=\frac{e B}{ m } \ldots(3) \quad(\because q =e)\)
Put value from equation (2) and (3) in equation (1)
\(M =\left(\frac{e v}{2 \pi r}\right)\left(\pi r^2\right)\)
\(M =\frac{e}{2 \pi}\left(\frac{e B}{m}\right) \pi\left(\frac{h}{B \pi e}\right)\)
\(M =\frac{e h}{2 \pi m}\)
\(M=I A=I\left(\pi r^2\right)\)
\(M=\left(\frac{e u }{2 \pi r}\right)\left(\pi r^2\right)\)
Given \(B\left(\pi r^2\right)=n\left(\frac{h}{e}\right)\)
\(\Rightarrow r ^2=\frac{ h }{ B \pi e } \quad \ldots(2) \quad(\because n =1)\)
And when charge moving in external magnetic field then \(r=\frac{m u}{q B}\)
\(\frac{ v }{ r }=\frac{e B}{ m } \ldots(3) \quad(\because q =e)\)
Put value from equation (2) and (3) in equation (1)
\(M =\left(\frac{e v}{2 \pi r}\right)\left(\pi r^2\right)\)
\(M =\frac{e}{2 \pi}\left(\frac{e B}{m}\right) \pi\left(\frac{h}{B \pi e}\right)\)
\(M =\frac{e h}{2 \pi m}\)
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