An electrons is revolving in n=3 orbit. What will be the magnetic field at the centre of hydrogen atom.

Options

A. a) 0.1 T
B. b) 5T
C. c) 0.5T
D. d) 0.05T

Explanation

To find the magnetic field at the center of a hydrogen atom when an electron is revolving in the n=3 orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Magnetic Field Formula**:
The magnetic field \( B \) at the center of a circular loop of current can be calculated using the formula:
\[ B = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the current, and \( R \) is the radius of the orbit.

2. **Finding the Current \( I \)**:
The current \( I \) due to the revolving electron can be expressed as:
\[ I = \frac{e}{T} \]
where \( e \) is the charge of the electron and \( T \) is the time period of revolution. The time period \( T \) can be calculated using:
\[ T = \frac{2\pi R}{v} \]
where \( v \) is the velocity of the electron.

3. **Calculating the Velocity \( v \)**:
For the hydrogen atom, the velocity of the electron in the nth orbit is given by:
\[ v = \frac{e^2}{2\epsilon_0 h} \cdot \frac{1}{n^2} \]
For \( n=3 \):
\[ v = \frac{2.18 \times 10^6}{3^2} = 0.73 \times 10^6 \, \text{m/s} \]

4. **Finding the Radius \( R \)**:
The radius of the nth orbit in a hydrogen atom is given by:
\[ R_n = n^2 \cdot R_1 \]
where \( R_1 \) (the radius of the first orbit) is approximately \( 0.529 \, \text{Å} = 0.529 \times 10^{-10} \, \text{m} \).
For \( n=3 \):
\[ R = 3^2 \cdot 0.529 \times 10^{-10} \, \text{m} = 7.461 \times 10^{-10} \, \text{m} \]

5. **Calculating the Current \( I \)**:
Now substituting \( R \) and \( v \) into the expression for \( I \):
\[ T = \frac{2\pi R}{v} = \frac{2\pi (7.461 \times 10^{-10})}{0.73 \times 10^6} \approx 6.428 \times 10^{-16} \, \text{s} \]
Thus, the current \( I \) becomes:
\[ I = \frac{e}{T} = \frac{1.6 \times 10^{-19}}{6.428 \times 10^{-16}} \approx 2.49 \times 10^{-4} \, \text{A} \]

6. **Calculating the Magnetic Field \( B \)**:
Finally, substituting \( I \) and \( R \) into the magnetic field formula:
\[ B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \cdot 2.49 \times 10^{-4}}{2 \cdot 7.461 \times 10^{-10}} \approx 0.0515 \, \text{T} \]

### Final Answer:
The magnetic field at the center of the hydrogen atom when the electron is in the n=3 orbit is approximately \( 0.0515 \, \text{T} \).

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