An orbital electron in the ground state of hy | vedclass

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(D) The magnetic moment $M$ of an orbital electron is given by $M = \frac{qL}{2m}$,where $L$ is the orbital angular momentum.According to Bohr's quant

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$4\mu_1 = \mu_2$

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(D) The magnetic moment $M$ of an orbital electron is given by $M = \frac{qL}{2m}$,where $L$ is the orbital angular momentum.According to Bohr's quantization condition,$L = \frac{nh}{2\pi}$,where $n$ is the principal quantum number.Thus,$M \propto L \propto n$.For the ground state,$n_1 = 1$,so $\mu_1 \propto 1$.For the $3^{rd}$ excited state,$n_2 = 1 + 3 = 4$,so $\mu_2 \propto 4$.Taking the ratio,$\frac{\mu_2}{\mu_1} = \frac{n_2}{n_1} = \frac{4}{1} = 4$.Therefore,$\mu_2 = 4\mu_1$.

An orbital electron in the ground state of hydrogen has a magnetic moment $\mu_1$. This orbital electron is excited to the $3^{rd}$ excited state by some energy transfer to the hydrogen atom. If the new magnetic moment of the electron is $\mu_2$,then:

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