The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

Options

A. `e//m`
B. `e//2m`
C. `m//e`
D. `2m//e`

Answer

To solve the problem of finding the ratio of the magnetic dipole moment of an electron in a Bohr orbit to its angular momentum, we can follow these steps:

### Step-by-Step Solution:

1. **Define Magnetic Dipole Moment (μ)**:
The magnetic dipole moment (μ) of a current loop is given by the formula:
\[ \mu = I \cdot A \]
where \(I\) is the current and \(A\) is the area of the loop.

2. **Calculate the Current (I)**:
The current \(I\) due to the moving electron can be defined as the charge passing through a point per unit time. The charge of the electron is \(e\), and the time period \(T\) for one complete revolution around the orbit is:
\[ T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi r}{v} \]
where \(v\) is the velocity of the electron. The current can then be expressed as:
\[ I = \frac{e}{T} = \frac{e \cdot v}{2\pi r} \]

3. **Determine the Velocity (v)**:
From Bohr's model, the velocity of the electron in the nth orbit is given by:
\[ v = \frac{n h}{2 \pi m r} \]
where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(r\) is the radius of the orbit.

4. **Substitute Velocity into Current**:
Substituting \(v\) into the expression for current \(I\):
\[ I = \frac{e \cdot \frac{n h}{2 \pi m r}}{2 \pi r} = \frac{n h e}{4 \pi m r^2} \]

5. **Calculate the Area (A)**:
The area \(A\) of the circular orbit is:
\[ A = \pi r^2 \]

6. **Calculate the Magnetic Dipole Moment (μ)**:
Now substituting \(I\) and \(A\) into the formula for magnetic dipole moment:
\[ \mu = I \cdot A = \left(\frac{n h e}{4 \pi m r^2}\right) \cdot (\pi r^2) = \frac{n h e}{4 m} \]

7. **Define Angular Momentum (L)**:
The angular momentum \(L\) of the electron in the nth orbit is given by:
\[ L = m v r = m \cdot \left(\frac{n h}{2 \pi m r}\right) \cdot r = \frac{n h}{2 \pi} \]

8. **Calculate the Ratio (μ/L)**:
Now, we can find the ratio of the magnetic dipole moment to the angular momentum:
\[ \frac{\mu}{L} = \frac{\frac{n h e}{4 m}}{\frac{n h}{2 \pi}} = \frac{e}{2 m} \cdot \frac{2 \pi}{4} = \frac{e}{2 m} \]

### Final Result:
The ratio of the magnetic dipole moment of an electron in the Bohr orbit to its angular momentum is:
\[ \frac{\mu}{L} = \frac{e}{2m} \]