According to Bohr model, magnetic field at the centre (at the nucleus) of a hydrogen atom due to the motion of the electron in nth orbit is proportional to `1//n^(x)`, find the value of x

Answer

To find the value of \( x \) in the expression for the magnetic field at the center of a hydrogen atom according to the Bohr model, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Current Due to Electron Motion**:
- The electron in the nth orbit can be treated as a circular loop of current. The current \( I \) generated by the electron moving in a circular path is given by:
\[ I = \frac{q}{T} \]
where \( q \) is the charge of the electron and \( T \) is the time period of one complete revolution.

2. **Finding the Time Period \( T \)**:
- The time period \( T \) can be expressed in terms of the frequency \( f \):
\[ T = \frac{1}{f} \]
- The frequency \( f \) can be expressed as:
\[ f = \frac{v}{2\pi R_n} \]
where \( v \) is the speed of the electron and \( R_n \) is the radius of the nth orbit.

3. **Using Bohr's Model for Radius**:
- According to Bohr's model, the radius of the nth orbit is given by:
\[ R_n \propto n^2 \]
- This implies that \( R_n = k \cdot n^2 \) for some constant \( k \).

4. **Finding the Speed of the Electron**:
- The speed \( v \) of the electron can be derived from the centripetal force and the electrostatic force acting on it:
\[ v = \frac{e^2}{n \cdot h} \]
- Here, \( e \) is the charge of the electron and \( h \) is Planck's constant.

5. **Substituting Values**:
- Now substituting \( v \) and \( R_n \) back into the expression for frequency:
\[ f = \frac{\frac{e^2}{n \cdot h}}{2\pi (k \cdot n^2)} = \frac{e^2}{2\pi k h n^3} \]

6. **Finding the Current \( I \)**:
- Substituting \( f \) into the current equation:
\[ I = q \cdot f = e \cdot \frac{e^2}{2\pi k h n^3} = \frac{e^3}{2\pi k h n^3} \]

7. **Magnetic Field at the Center**:
- The magnetic field \( B \) at the center of the loop due to the current \( I \) is given by:
\[ B \propto \frac{\mu_0 I}{2R_n} \]
- Substituting \( I \) and \( R_n \):
\[ B \propto \frac{\mu_0 \cdot \frac{e^3}{2\pi k h n^3}}{2(k \cdot n^2)} = \frac{\mu_0 e^3}{4\pi k h n^5} \]

8. **Final Expression**:
- Thus, the magnetic field \( B \) at the center is proportional to:
\[ B \propto \frac{1}{n^5} \]
- Therefore, we find that \( x = 5 \).

### Conclusion:
The value of \( x \) is \( 5 \).