The electron in a hydrogen atom makes a transition from M shell to L-shell. The ratio of magnitude of initial to final acceleration of the electron is

Options

A. `9:4`
B. `81:16`
C. `4:9`
D. `16:81`

Explanation

To solve this problem, we need to find the ratio of the magnitudes of the initial and final accelerations of the electron in a hydrogen atom as it transitions from the M shell to the L shell.

### Step-by-Step Solution:

1. **Identify the Shells and Quantum Numbers**:
- The M shell corresponds to the principal quantum number \( n_1 = 3 \).
- The L shell corresponds to the principal quantum number \( n_2 = 2 \).

2. **Understand the Relationship of Acceleration**:
- The centripetal acceleration \( a \) of an electron in a hydrogen atom is given by the formula:
\[ a \propto \frac{n^2 h^2}{4 \pi^2 m r^3} \]
- The radius \( r \) of the electron's orbit is given by:
\[ r \propto n^2 \]
- Therefore, we can express acceleration in terms of \( n \):
\[ a \propto \frac{n^2}{(n^2)^3} = \frac{n^2}{n^6} = \frac{1}{n^4} \]
- This means that the centripetal acceleration is inversely proportional to the fourth power of the principal quantum number \( n \).

3. **Calculate the Ratio of Accelerations**:
- The ratio of the initial acceleration \( a_1 \) (when \( n_1 = 3 \)) to the final acceleration \( a_2 \) (when \( n_2 = 2 \)) can be expressed as:
\[ \frac{a_1}{a_2} = \frac{n_2^4}{n_1^4} \]
- Substituting the values:
\[ \frac{a_1}{a_2} = \frac{2^4}{3^4} = \frac{16}{81} \]

4. **Conclusion**:
- The ratio of the magnitude of initial to final acceleration of the electron is:
\[ \frac{a_1}{a_2} = \frac{16}{81} \]
- Therefore, the correct answer is option D: \( 16:81 \).