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: Understanding the Question:
The question asks for the magnetic dipole moment ($\mu$) of a hydrogen atom when its electron is revolving in a quantized Bohr orbit corresponding to the $n^{\text{th}}$ state.

Step 2: Key Formula or Approach:
1. The magnetic moment of a circulating charge loop is:
$$\mu = I \cdot A$$ where $I = \frac{e}{T}$ is the current and $A = \pi r^2$ is the area of the orbit. 2. The orbital angular momentum of the electron is:
$$L = m_e v r$$ 3. According to Bohr's quantization postulate, angular momentum is limited to discrete steps:
$$L = \frac{nh}{2\pi}$$ Combining these equations yields the gyromagnetic ratio: $\frac{\mu}{L} = \frac{e}{2m_e}$.

Step 3: Detailed Explanation:
Let's find the expression for current $I$ using the velocity $v$ and radius $r$ of the electron orbit. The period of revolution is $T = \frac{2\pi r}{v}$:
$$I = \frac{e}{T} = \frac{ev}{2\pi r}$$ Substitute this current into the magnetic moment area equation:
$$\mu = \left(\frac{ev}{2\pi r}\right) \cdot (\pi r^2) = \frac{evr}{2}$$ We can multiply the numerator and denominator by the mass of the electron $m_e$ to introduce the angular momentum term $L$:
$$\mu = \frac{e(m_e v r)}{2m_e} = \frac{e}{2m_e} \cdot L$$ Now, apply Bohr's second postulate by substituting $L = \frac{nh}{2\pi}$:
$$\mu = \left(\frac{e}{2m_e}\right) \cdot \frac{nh}{2\pi}$$ This matches the formulation given in option (D).

Step 4: Final Answer:
The magnetic moment of the hydrogen atom is $\left(\frac{e}{2m_e}\right)\frac{nh}{2\pi}$, which corresponds to option (D).