In a hydrogen atom , the electron atom makes a transition from `n = 2 to n = 1`. The magnetic field produced by the circulating electron at the nucleus

Options

A. decrease 16 time
B. increase 4 time
C. decrease 4 time
D. increase 32 time

Explanation

To solve the problem of the magnetic field produced by the circulating electron in a hydrogen atom during a transition from \( n = 2 \) to \( n = 1 \), we will follow these steps:

### Step 1: Understanding the Magnetic Field Formula
The magnetic field \( B \) produced by a circulating charge can be expressed 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 orbit.

### Step 2: Finding the Current \( I \)
The current \( I \) due to the electron can be calculated as:
\[ I = \frac{q}{T} \]
where \( q \) is the charge of the electron (denoted as \( e \)), and \( T \) is the time period of rotation. The time period \( T \) can be expressed as:
\[ T = \frac{2\pi R}{v} \]
where \( v \) is the velocity of the electron.

### Step 3: Relating Velocity and Radius to Principal Quantum Number \( n \)
For a hydrogen atom, the velocity \( v \) and radius \( R \) of the electron are related to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \quad \text{and} \quad R \propto n^2 \]
Thus, substituting for \( T \):
\[ T \propto \frac{R}{v} \propto \frac{n^2}{\frac{1}{n}} = n^3 \]
This implies:
\[ I \propto \frac{e}{T} \propto \frac{e}{n^3} \]

### Step 4: Finding the Magnetic Field \( B \)
Now substituting \( I \) and \( R \) into the magnetic field formula:
\[ B \propto \frac{I}{R} \propto \frac{\frac{e}{n^3}}{n^2} = \frac{e}{n^5} \]
This shows that the magnetic field \( B \) is inversely proportional to \( n^5 \).

### Step 5: Calculating the Magnetic Fields for \( n = 1 \) and \( n = 2 \)
Let’s denote the magnetic field when \( n = 1 \) as \( B_1 \) and when \( n = 2 \) as \( B_2 \):
\[ B_1 \propto \frac{1}{1^5} = 1 \]
\[ B_2 \propto \frac{1}{2^5} = \frac{1}{32} \]

### Step 6: Finding the Ratio of Magnetic Fields
Now we can find the ratio of the magnetic fields:
\[ \frac{B_1}{B_2} = \frac{1}{\frac{1}{32}} = 32 \]
This indicates that the magnetic field at \( n = 1 \) is 32 times greater than that at \( n = 2 \).

### Conclusion
Thus, when the electron transitions from \( n = 2 \) to \( n = 1 \), the magnetic field at the nucleus increases by a factor of 32.