In a hydrogen atom, the electron revolves round the nucleus in a circular orbit of radius `0.5 Å` at a frequency of `5xx10^(15)` rev/second. What is the effective magnetic dipole moment?

Options

A. `4.28xx10^(-24) Am^(2)`
B. `5.28xx10^(-24) Am^(2)`
C. `6.28xx10^(-24) Am^(2)`
D. `8.28xx10^(-24) Am^(2)`

Explanation

To find the effective magnetic dipole moment of the hydrogen atom, we can follow these steps:

### Step 1: Convert the radius from angstroms to meters
Given:
- Radius \( r = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \)

### Step 2: Calculate the current due to the revolving electron
The current \( I \) can be calculated using the formula:
\[ I = \frac{q}{T} \]
where \( q \) is the charge of the electron and \( T \) is the period of revolution.

The charge of the electron \( q = 1.6 \times 10^{-19} \, \text{C} \).

The frequency \( f \) is given as \( 5 \times 10^{15} \, \text{rev/s} \). The period \( T \) can be calculated as:
\[ T = \frac{1}{f} = \frac{1}{5 \times 10^{15}} \, \text{s} \]

Substituting \( T \) into the current formula:
\[ I = \frac{1.6 \times 10^{-19}}{\frac{1}{5 \times 10^{15}}} = 1.6 \times 10^{-19} \times 5 \times 10^{15} = 8 \times 10^{-4} \, \text{A} \]

### Step 3: Calculate the area of the circular orbit
The area \( A \) of the circular orbit is given by:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (0.5 \times 10^{-10})^2 = \pi \times 0.25 \times 10^{-20} \approx 7.85 \times 10^{-21} \, \text{m}^2 \]

### Step 4: Calculate the magnetic dipole moment
The magnetic dipole moment \( M \) is given by:
\[ M = NIA \]
where \( N \) is the number of turns (which is 1 for a single electron). Thus:
\[ M = 1 \times (8 \times 10^{-4}) \times (7.85 \times 10^{-21}) \approx 6.28 \times 10^{-24} \, \text{A m}^2 \]

### Final Answer
The effective magnetic dipole moment of the hydrogen atom is approximately:
\[ M \approx 6.28 \times 10^{-24} \, \text{A m}^2 \]

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