In hydrogen atom, the electron is making `6.6xx10^(15) rev//sec` around the nucleus in an orbit of radius `0.528 A`. The magnetic moment `(A-m^(2)) will be

Options

A. `1xx10^(-15)`
B. `1xx10^(-10)`
C. `1xx10^(-23)`
D. `1xx10^(-27)`

Answer

To find the magnetic moment of the electron in a hydrogen atom, we can follow these steps:

### Step 1: Identify Given Values
- Frequency \( f = 6.6 \times 10^{15} \) revolutions per second
- Radius \( r = 0.528 \) Ångström \( = 0.528 \times 10^{-10} \) meters

### Step 2: Calculate Angular Frequency
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \times 3.14 \times (6.6 \times 10^{15}) \approx 4.14 \times 10^{16} \text{ radians/second} \]

### Step 3: Calculate Current
The current \( I \) due to the moving electron can be calculated using the formula:
\[ I = q \cdot \omega \]
Where \( q \) is the charge of the electron, \( q = 1.6 \times 10^{-19} \) coulombs. Substituting the values:
\[ I = (1.6 \times 10^{-19}) \times (4.14 \times 10^{16}) \approx 5.35 \times 10^{-3} \text{ A} \]

### Step 4: Calculate Area of the Orbit
The area \( A \) of the circular orbit can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi \times (0.528 \times 10^{-10})^2 \approx \pi \times (0.278784 \times 10^{-20}) \approx 8.76 \times 10^{-21} \text{ m}^2 \]

### Step 5: Calculate Magnetic Moment
The magnetic moment \( M \) can be calculated using the formula:
\[ M = n \cdot I \cdot A \]
For a single electron, \( n = 1 \):
\[ M = 1 \cdot (5.35 \times 10^{-3}) \cdot (8.76 \times 10^{-21}) \approx 4.72 \times 10^{-23} \text{ A m}^2 \]

### Final Answer
The magnetic moment of the electron in a hydrogen atom is approximately:
\[ M \approx 4.72 \times 10^{-23} \text{ A m}^2 \]

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