The magnetic moment (mu) of an electron revol | vedclass

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(A) The magnetic moment $(\mu)$ of a revolving electron is given by the relation $\mu = \frac{e}{2m} L$,where $L$ is the orbital angular momentum.Acco

Vedclass · Accepted Answer

$\mu \propto n$

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(A) The magnetic moment $(\mu)$ of a revolving electron is given by the relation $\mu = \frac{e}{2m} L$,where $L$ is the orbital angular momentum.According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.Substituting this value into the magnetic moment formula,we get:$\mu = \frac{e}{2m} \left( \frac{nh}{2\pi} \right)$$\mu = n \left( \frac{eh}{4\pi m} \right)$Since $e$,$h$,and $m$ are constants,the term $\frac{eh}{4\pi m}$ is a constant (specifically,the Bohr magneton $\mu_B$ is $\frac{eh}{4\pi m}$).Therefore,$\mu \propto n$.

The magnetic moment $(\mu)$ of an electron revolving around the nucleus varies with the principal quantum number $n$ as

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