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 the problem of finding the ratio of the magnitude of initial to final acceleration of the electron in a hydrogen atom as it transitions from the M shell to the L shell, we can follow these steps:

### Step 1: Understand the relationship between acceleration and quantum numbers
The acceleration \( a \) of an electron in a hydrogen atom can be expressed in terms of its angular frequency \( \omega \) and radius \( r \):
\[ a = \omega^2 r \]
We know that:
- The angular frequency \( \omega \) is inversely proportional to the cube of the principal quantum number \( n \):
\[ \omega \propto \frac{1}{n^3} \]
- The radius \( r \) of the electron's orbit is directly proportional to the square of the principal quantum number \( n \):
\[ r \propto n^2 \]

### Step 2: Combine the relationships to find acceleration
From the above relationships, we can express acceleration in terms of \( n \):
\[ a \propto \frac{1}{n^3} \cdot n^2 = \frac{1}{n} \]
Thus, we can conclude that:
\[ a \propto \frac{1}{n^4} \]

### Step 3: Set up the ratio of accelerations
Let \( n_1 \) be the principal quantum number for the M shell and \( n_2 \) be the principal quantum number for the L shell. For hydrogen:
- M shell corresponds to \( n_1 = 3 \)
- L shell corresponds to \( n_2 = 2 \)

The ratio of the initial acceleration \( a_1 \) (for M shell) to the final acceleration \( a_2 \) (for L shell) is given by:
\[ \frac{a_1}{a_2} = \frac{n_2^4}{n_1^4} \]

### Step 4: Substitute the values of \( n_1 \) and \( n_2 \)
Substituting the values:
\[ \frac{a_1}{a_2} = \frac{2^4}{3^4} = \frac{16}{81} \]

### Step 5: Conclusion
The ratio of the magnitude of initial to final acceleration of the electron as it transitions from the M shell to the L shell is:
\[ \frac{a_1}{a_2} = \frac{16}{81} \]

### Final Answer
Thus, the answer to the question is \( \frac{16}{81} \).