What is the maximum number of emission lines when the excited electron of a H atom in `n = 6` drop to the ground state?

Options

A. 6
B. 15
C. 30
D. 10

Explanation

To find the maximum number of emission lines when an excited electron of a hydrogen atom in the n = 6 state drops to the ground state (n = 1), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial State**: The electron is initially in the n = 6 energy level.

2. **Determine the Ground State**: The ground state of the hydrogen atom corresponds to n = 1.

3. **Use the Formula for Maximum Emission Lines**: The maximum number of emission lines can be calculated using the formula:
\[ \text{Maximum Emission Lines} = \frac{n(n - 1)}{2} \]
where \( n \) is the principal quantum number of the initial state.

4. **Substitute the Value of n**: In this case, \( n = 6 \). Plugging this value into the formula gives:
\[ \text{Maximum Emission Lines} = \frac{6(6 - 1)}{2} \]

5. **Calculate the Result**:
\[ = \frac{6 \times 5}{2} = \frac{30}{2} = 15 \]

6. **Conclusion**: The maximum number of emission lines when the excited electron drops from n = 6 to the ground state is 15.

### Final Answer:
The maximum number of emission lines is **15**.

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