An electron in H-atom in M-shell on de-excitation to ground state gives maximum ........... spectrum lines.

Options

A. 3
B. 6
C. 2
D. 1

Answer

To solve the problem of how many spectral lines are produced when an electron in a hydrogen atom transitions from the M shell (n=3) to the ground state (n=1), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Energy Levels:**
- The electron is in the M shell, which corresponds to the principal quantum number \( n = 3 \).
- The ground state corresponds to \( n = 1 \).

2. **Determine the Formula for Spectral Lines:**
- The number of spectral lines produced during a transition from a higher energy level \( n_2 \) to a lower energy level \( n_1 \) can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]

3. **Substitute the Values:**
- Here, \( n_2 = 3 \) (M shell) and \( n_1 = 1 \) (ground state).
- Substitute these values into the formula:
\[ \text{Number of spectral lines} = \frac{(3 - 1)(3 - 1 + 1)}{2} \]

4. **Calculate the Result:**
- Simplifying the expression:
\[ = \frac{(2)(3)}{2} = \frac{6}{2} = 3 \]

5. **Conclusion:**
- Therefore, the maximum number of spectral lines produced when the electron de-excites from the M shell to the ground state is **3**.

### Final Answer:
The maximum number of spectral lines produced is **3**.