Question

In the Bohr model, an electron moves in a circular orbit around the nucleus. Considering an 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 ($e$ = electronic charge, $m_e$ = mass of the electron, $h$ = Planck's constant)

• A. $\left(\frac{e}{m_e}\right)\frac{nh}{2\pi}$
• B. $\left(\frac{e}{m_e}\right)\frac{n^2h}{2\pi}$
• C. $\left(\frac{e^2}{m_e}\right)\frac{n^2h}{2\pi}$
• D. $\left(\frac{e}{2m_e}\right)\frac{nh}{2\pi}$

Hint:

The ratio of magnetic moment to angular momentum ($\frac{\mu}{L}$) for any revolving particle of charge $q$ and mass $m$ is a constant known as the gyromagnetic ratio, equal to $\frac{q}{2m}$. For an electron, $\mu = \frac{e}{2m_e}L$. Simply multiply this ratio by Bohr's quantized angular momentum value $\frac{nh}{2\pi}$ to get the answer.

Answer 1

The Correct Option is D

Step 1: Model the orbit as a current loop.
An electron of charge $e$ circling the nucleus is an electric current, and a current loop has a magnetic moment $\mu = I A$.
Step 2: Write the current.
One electron passes a point once per period $T = \dfrac{2\pi r}{v}$, so $I = \dfrac{e}{T} = \dfrac{ev}{2\pi r}$.
Step 3: Multiply by the loop area.
$A = \pi r^2$, so $\mu = \dfrac{ev}{2\pi r}\cdot\pi r^2 = \dfrac{evr}{2}$.
Step 4: Bring in angular momentum.
Multiply and divide by $m_e$: $\mu = \dfrac{e}{2m_e}(m_e v r) = \dfrac{e}{2m_e}L$, where $L = m_e v r$.
Step 5: Apply Bohr's quantisation.
Bohr requires $L = \dfrac{nh}{2\pi}$.
Step 6: Combine.
$\mu = \left(\dfrac{e}{2m_e}\right)\dfrac{nh}{2\pi}$, which is option (4).
\[ \boxed{\mu = \left(\frac{e}{2m_e}\right)\frac{nh}{2\pi}} \]