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

Explanation

To find the minimum 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. **Understanding the Transition**: The electron can transition from a higher energy level (n = 6) to a lower energy level (n = 1). During this process, it can emit photons corresponding to the energy difference between the levels.

2. **Using the Formula for Emission Lines**: The formula to calculate the number of possible emission lines when an electron drops from a higher energy level (n) to a lower energy level (1) is given by:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
where \( n \) is the principal quantum number of the excited state.

3. **Substituting the Value of n**: For our case, \( n = 6 \):
\[ \text{Number of lines} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = \frac{30}{2} = 15 \]

4. **Identifying the Minimum Emission Lines**: Although there are 15 possible transitions, the minimum number of emission lines corresponds to the largest single transition from n = 6 directly to n = 1. This transition will produce one emission line.

5. **Conclusion**: Therefore, the minimum number of emission lines when the excited electron drops from n = 6 to the ground state (n = 1) is **1**.