In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron…
In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in $n^{\text {th }}$ excited state, is :
$\left(\frac{e}{2 m} \frac{n^2 h}{2 \pi}\right)$
$\left(\frac{e}{m}\right) \frac{n h}{2 \pi}$
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$
$\left(\frac{e}{m}\right) \frac{n^2 h}{2 \pi}$
Solution
Magnetic moment of the hydrogen atom, when the electron is in $\mathrm{n}^{\text {th }}$ excited state, i.e., $n^{\prime}=(n+1)$
As magnetic moment $\mathrm{M}_{\mathrm{n}}=\mathrm{I}_{\mathrm{n}} \mathrm{A}=\mathrm{i}_{\mathrm{n}}\left(\pi \mathrm{r}_{\mathrm{n}}^2\right)$
$\begin{aligned}
& i_n=e V_n=\frac{m z^2 e^5}{4 \varepsilon_0^2 n^3 h^3} \\
& r_n=\frac{n^2 h^2}{4 \pi^2 k z m e^2}\left(k=\frac{1}{4 \pi \epsilon_0}\right)
\end{aligned}$
Solving we get magnetic moment of the hydrogen atom for $\mathrm{n}^{\text {th }}$ excited state
$M_{n^{\prime}}=\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$