An electron in the ground state of hydrogen atom is revolving in anticlockwise direction in circular orbit of radius R. The orbital magnetic dipole moment of the electron will be

Options

A. `(eh)/(4pim)`
B. `(eh)/(2pim)`
C. `(eh^2)/(4pim)`
D. `(e^2h)/(4pim)`

Explanation

To find the orbital magnetic dipole moment of an electron in the ground state of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Formula for Magnetic Dipole Moment
The orbital magnetic dipole moment (\( \mu \)) is given by the formula:
\[ \mu = i \cdot A \]
where \( i \) is the current and \( A \) is the area of the circular orbit.

### Step 2: Calculate the Area of the Orbit
The area \( A \) of a circular orbit with radius \( R \) is:
\[ A = \pi R^2 \]

### Step 3: Determine the Current \( i \)
The current \( i \) due to the revolving electron can be calculated using the charge of the electron (\( e \)) and the time period (\( T \)) of the revolution:
\[ i = \frac{e}{T} \]
The time period \( T \) is given by:
\[ T = \frac{2\pi R}{v} \]
where \( v \) is the velocity of the electron.

### Step 4: Find the Velocity of the Electron
For an electron in a circular orbit, we can use the quantization of angular momentum:
\[ m v R = n \frac{h}{2\pi} \]
For the ground state of hydrogen, \( n = 1 \):
\[ m v R = \frac{h}{2\pi} \]
From this, we can express the velocity \( v \):
\[ v = \frac{h}{2\pi m R} \]

### Step 5: Substitute Velocity into Time Period
Now substituting \( v \) into the expression for \( T \):
\[ T = \frac{2\pi R}{\frac{h}{2\pi m R}} = \frac{4\pi^2 m R^2}{h} \]

### Step 6: Substitute Time Period into Current
Now substituting \( T \) back into the expression for current \( i \):
\[ i = \frac{e}{T} = \frac{e h}{4\pi^2 m R^2} \]

### Step 7: Substitute Current and Area into Magnetic Dipole Moment
Now substituting \( i \) and \( A \) into the magnetic dipole moment formula:
\[ \mu = i \cdot A = \left(\frac{e h}{4\pi^2 m R^2}\right) \cdot (\pi R^2) \]
This simplifies to:
\[ \mu = \frac{e h}{4\pi m} \]

### Final Answer
Thus, the orbital magnetic dipole moment of the electron in the ground state of the hydrogen atom is:
\[ \mu = \frac{e h}{4\pi m} \]