The rartio `("in" S 1 units)` of magnetic dipole moment to that of the angular momentum of an electron of mass `m kg` and charge `e` coulomb in Bohr's orbit of hydrogen atom is

Options

A. `(e)/(2 m)`
B. `(e)/(m)`
C. `(2 e)/(m)`
D. none of these

Explanation

To find the ratio of the magnetic dipole moment (μ) to the angular momentum (L) of an electron in a Bohr orbit of a hydrogen atom, we can follow these steps:

### Step 1: Define the magnetic dipole moment
The magnetic dipole moment (μ) for an electron in a circular orbit can be expressed as:
\[ \mu = \frac{1}{2} e V R \]
where:
- \( e \) is the charge of the electron,
- \( V \) is the velocity of the electron,
- \( R \) is the radius of the orbit.

### Step 2: Define the angular momentum
The angular momentum (L) of the electron in a circular orbit is given by:
\[ L = m V R \]
where:
- \( m \) is the mass of the electron,
- \( V \) is the velocity of the electron,
- \( R \) is the radius of the orbit.

### Step 3: Express velocity in terms of angular momentum
From the equation for angular momentum, we can express \( V R \) as:
\[ V R = \frac{L}{m} \]

### Step 4: Substitute \( V R \) in the magnetic dipole moment equation
Substituting \( V R \) into the equation for magnetic dipole moment, we get:
\[ \mu = \frac{1}{2} e \left(\frac{L}{m}\right) \]
This simplifies to:
\[ \mu = \frac{e L}{2m} \]

### Step 5: Find the ratio of magnetic dipole moment to angular momentum
Now, we can find the ratio of the magnetic dipole moment to the angular momentum:
\[ \frac{\mu}{L} = \frac{e L / 2m}{L} \]
This simplifies to:
\[ \frac{\mu}{L} = \frac{e}{2m} \]

### Final Answer
Thus, the ratio of the magnetic dipole moment to the angular momentum of an electron in Bohr's orbit of a hydrogen atom is:
\[ \frac{e}{2m} \]