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The gyromagnetic ratio of an electron in a hydrogen atom, according to Bohr model is
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The correct answer is:
independent of which orbit it is in
$$
\begin{aligned}
& \text { Gyromagnetic ratio }=\frac{\text { Magnetic Moment }}{\text { Angular Momentum }} \\
& =\frac{\mathrm{iA}}{\mathrm{mvr}}=\frac{\left(\frac{\mathrm{e} \omega}{2 \pi}\right)\left(\pi \mathrm{r}^2\right)}{\mathrm{m}(\mathrm{r} \omega) \mathrm{r}}=\frac{\mathrm{e}}{2 \mathrm{~m}}
\end{aligned}
$$
It is independent of the orbit of the electron
\begin{aligned}
& \text { Gyromagnetic ratio }=\frac{\text { Magnetic Moment }}{\text { Angular Momentum }} \\
& =\frac{\mathrm{iA}}{\mathrm{mvr}}=\frac{\left(\frac{\mathrm{e} \omega}{2 \pi}\right)\left(\pi \mathrm{r}^2\right)}{\mathrm{m}(\mathrm{r} \omega) \mathrm{r}}=\frac{\mathrm{e}}{2 \mathrm{~m}}
\end{aligned}
$$
It is independent of the orbit of the electron
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