In hydrogen atom, an electron is revolving in the orbit of radius `0.53 Å` with `6.6xx10^(15) rotations//second`. Magnetic field produced at the centre of the orbit is

Options

A. `0.125 Wb//m^(2)`
B. `1.25 Wb//m^(2)`
C. `12.5 Wb//m^(2)`
D. `125 Wb//m^(2)`

Explanation

To find the magnetic field produced at the center of the orbit of an electron in a hydrogen atom, we can follow these steps:

### Step 1: Identify the Given Values
- Radius of the orbit (r) = 0.53 Å = \(0.53 \times 10^{-10}\) m
- Frequency of rotation (f) = \(6.6 \times 10^{15}\) rotations/second

### Step 2: Calculate the Current (I)
The current produced by the revolving electron can be calculated using the formula:
\[ I = n \cdot e \]
where:
- \( n \) = number of electrons passing a point per second (which is equal to the frequency in this case)
- \( e \) = charge of an electron = \(1.6 \times 10^{-19}\) C

Substituting the values:
\[ I = 6.6 \times 10^{15} \times 1.6 \times 10^{-19} \]
\[ I = 1.056 \times 10^{-3} \text{ A} \]

### Step 3: Apply the Biot-Savart Law
The magnetic field (B) at the center of a circular loop carrying current can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2r} \]
where:
- \( \mu_0 \) (permeability of free space) = \(4\pi \times 10^{-7} \text{ T m/A}\)
- \( r \) = radius of the orbit = \(0.53 \times 10^{-10}\) m

Substituting the values:
\[ B = \frac{(4\pi \times 10^{-7}) \cdot (1.056 \times 10^{-3})}{2 \cdot (0.53 \times 10^{-10})} \]

### Step 4: Calculate the Magnetic Field
First, calculate the numerator:
\[ 4\pi \times 10^{-7} \times 1.056 \times 10^{-3} \approx 1.33 \times 10^{-9} \text{ T m} \]

Now, calculate the denominator:
\[ 2 \cdot (0.53 \times 10^{-10}) = 1.06 \times 10^{-10} \text{ m} \]

Now, calculate B:
\[ B = \frac{1.33 \times 10^{-9}}{1.06 \times 10^{-10}} \approx 12.5 \text{ T} \]

### Final Answer
The magnetic field produced at the center of the orbit is approximately \(12.5 \text{ T}\).

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