An electron revolving in an orbit of radius `0.5 Å` in a hydrogen atom executes per secon. The magnetic momentum of electron due to its orbital motion will be

Options

A. `1.256 xx 10^(-23) A m^(2)`
B. `653 xx 10^(-26) A m^(2)`
C. `10^(-3) A m^(2)`
D. `256 xx 10^(-26) A m^(2)`

Explanation

To solve the problem of finding the magnetic moment of an electron revolving in an orbit of radius \(0.5 \, \text{Å}\) in a hydrogen atom, we can follow these steps:

### Step 1: Understand the Formula for Magnetic Moment
The magnetic moment \(M\) due to the orbital motion of an electron can be calculated using the formula:
\[ M = I \cdot A \]
where \(I\) is the current and \(A\) is the area of the orbit.

### Step 2: Calculate the Area of the Orbit
The electron is moving in a circular orbit, so the area \(A\) can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Given that the radius \(r = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m}\), we can substitute this value into the area formula:
\[ A = \pi (0.5 \times 10^{-10})^2 \]

### Step 3: Calculate the Current \(I\)
The current \(I\) due to the revolving electron can be expressed in terms of the charge and frequency:
\[ I = e \cdot f \]
where \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)) and \(f\) is the frequency of revolution. For the hydrogen atom in the first orbit, the frequency \(f\) is approximately \(10^{16} \, \text{Hz}\).

### Step 4: Substitute Values into the Magnetic Moment Formula
Now we can substitute the values into the magnetic moment formula:
\[ M = I \cdot A = (e \cdot f) \cdot (\pi r^2) \]
Substituting the known values:
\[ M = (1.6 \times 10^{-19} \, \text{C} \cdot 10^{16} \, \text{Hz}) \cdot \left(\pi (0.5 \times 10^{-10} \, \text{m})^2\right) \]

### Step 5: Calculate the Magnetic Moment
Now we can calculate the area:
\[ A = \pi (0.5 \times 10^{-10})^2 = \pi \cdot (0.25 \times 10^{-20}) = 0.785 \times 10^{-20} \, \text{m}^2 \]
Now substituting this back into the magnetic moment equation:
\[ M = (1.6 \times 10^{-19} \cdot 10^{16}) \cdot (0.785 \times 10^{-20}) \]
Calculating this gives:
\[ M = (1.6 \times 0.785) \times 10^{-23} = 1.256 \times 10^{-23} \, \text{A m}^2 \]

### Conclusion
Thus, the magnetic moment of the electron due to its orbital motion is:
\[ M \approx 1.256 \times 10^{-23} \, \text{A m}^2 \]