The magnetic field induction produced at the centre of orbit due to an electron revolving in `n^(th)` orbit of hydrogen atom is proportional to

Options

A. `n^(-3)`
B. `n^(-5)`
C. `n^(5)`
D. `n^(3)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Concept**:
The magnetic field (B) produced at the center of a circular loop due to a current (I) is given by 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 loop.

2. **Identifying the Current**:
An electron revolving in a circular orbit can be considered as a current loop. The current (I) due to the electron can be calculated using:
\[ I = \frac{q}{T} \]
where \( q \) is the charge of the electron and \( T \) is the time period of one complete revolution.

3. **Calculating the Time Period (T)**:
The time period \( T \) can be expressed as:
\[ T = \frac{\text{Distance}}{\text{Speed}} = \frac{2\pi R_n}{v_n} \]
where \( R_n \) is the radius of the nth orbit and \( v_n \) is the speed of the electron in that orbit.

4. **Substituting for Current (I)**:
Substituting \( T \) into the current formula gives:
\[ I = \frac{q}{T} = \frac{q v_n}{2\pi R_n} \]

5. **Substituting Current into Magnetic Field Formula**:
Now substituting the expression for I into the magnetic field formula:
\[ B = \frac{\mu_0}{2R} \cdot \frac{q v_n}{2\pi R_n} \]

6. **Considering the Radius and Velocity**:
From Bohr's model, we know that the radius of the nth orbit \( R_n \) is proportional to \( n^2 \) and the velocity \( v_n \) is proportional to \( \frac{1}{n} \). Thus:
\[ R_n \propto n^2 \quad \text{and} \quad v_n \propto \frac{1}{n} \]

7. **Substituting Proportional Relationships**:
Substituting these relationships into the magnetic field expression gives:
\[ B \propto \frac{1}{R_n^2} \cdot \frac{1}{n} = \frac{1}{(n^2)^2} \cdot \frac{1}{n} = \frac{1}{n^5} \]

8. **Final Result**:
Therefore, the magnetic field induction produced at the center of the orbit due to an electron revolving in the nth orbit of a hydrogen atom is proportional to:
\[ B \propto \frac{1}{n^5} \]