If `He^+` ion undergoes transition n=2 `to` 1 the ratio of final to initial magnetic field due to motion of electron at the nucleus will

Options

A. a.`32:1`
B. b.`1:32`
C. c`16:1`
D. d.`1:16`

Answer

To solve the problem of finding the ratio of the final to the initial magnetic field due to the motion of the electron in the `He^+` ion during the transition from n=2 to n=1, we can follow these steps:

### Step 1: Understand the relationship between magnetic field and quantum numbers
The magnetic field (B) generated by the motion of an electron in an atom is inversely proportional to the fifth power of the principal quantum number (n). This can be expressed as:
\[ B \propto \frac{Z^3}{n^5} \]
where Z is the atomic number (for `He^+`, Z = 2).

### Step 2: Identify the initial and final states
In this case, the initial state corresponds to n=2 and the final state corresponds to n=1. Therefore:
- Initial state (n1) = 2
- Final state (n2) = 1

### Step 3: Write the ratio of the magnetic fields
Using the relationship from Step 1, we can write the ratio of the magnetic fields for the two states:
\[ \frac{B_2}{B_1} = \frac{n_1^5}{n_2^5} \]

### Step 4: Substitute the values of n1 and n2
Substituting the values of n1 and n2 into the equation:
\[ \frac{B_2}{B_1} = \frac{2^5}{1^5} \]

### Step 5: Calculate the ratio
Calculating the powers:
\[ 2^5 = 32 \]
\[ 1^5 = 1 \]
Thus, the ratio becomes:
\[ \frac{B_2}{B_1} = \frac{32}{1} \]

### Conclusion
The ratio of the final to the initial magnetic field due to the motion of the electron at the nucleus during the transition from n=2 to n=1 is:
\[ \frac{B_2}{B_1} = 32:1 \]