According to Bohr's theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the `n^(th)` orbit is proportional to :
(n = principal quantum number)

Options

A. `n^(-3)`
B. `n^(-2)`
C. `n^(-4)`
D. `n^(-5)`

Answer

To solve the problem of finding the time-averaged magnetic field at the center of a hydrogen atom due to the motion of electrons in the \( n^{th} \) orbit according to Bohr's theory, we can follow these steps:

### Step 1: Understand the motion of the electron
The electron in the \( n^{th} \) orbit revolves in a circular path around the nucleus. This motion generates a current due to the charge of the electron moving in a circle.

### Step 2: Calculate the current
The current \( I \) generated by the moving electron can be expressed as:
\[ I = \frac{Q}{T} \]
where \( Q \) is the charge of the electron and \( T \) is the time period of revolution. The time period \( T \) can be calculated as:
\[ T = \frac{2\pi r}{v} \]
where \( r \) is the radius of the orbit and \( v \) is the velocity of the electron. Therefore, substituting for \( T \):
\[ I = \frac{Qv}{2\pi r} \]

### Step 3: Magnetic field at the center of the circular loop
The magnetic field \( B \) at the center of a circular current-carrying loop is given by:
\[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space. Substituting the expression for \( I \) into this equation gives:
\[ B = \frac{\mu_0}{2r} \cdot \frac{Qv}{2\pi r} = \frac{\mu_0 Qv}{4\pi r^2} \]

### Step 4: Relate velocity and radius to the principal quantum number \( n \)
From Bohr's model, we know that:
- The velocity \( v \) of the electron is inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \]
- The radius \( r \) of the orbit is proportional to \( n^2 \):
\[ r \propto n^2 \]

### Step 5: Substitute \( v \) and \( r \) into the magnetic field equation
Now, substituting these relationships into the expression for \( B \):
\[ B \propto \frac{Q}{r^2} \cdot v \]
Substituting \( r \) and \( v \):
\[ B \propto \frac{Q}{(n^2)^2} \cdot \frac{1}{n} = \frac{Q}{n^4} \cdot \frac{1}{n} = \frac{Q}{n^5} \]

### Conclusion
Thus, the time-averaged magnetic field \( B \) at the center of a hydrogen atom due to the motion of electrons in the \( n^{th} \) orbit is proportional to:
\[ B \propto \frac{1}{n^5} \]