What is the maximum number of emission lines obtained when the excited electron of a H atom in n = 5 drops to the ground state?

Options

A. 12
B. 15
C. 21
D. 10

Explanation

To find the maximum number of emission lines when an excited electron of a hydrogen atom in the n = 5 state drops to the ground state, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**: When an electron in a hydrogen atom transitions from a higher energy level (n = 5) to a lower energy level (ground state, n = 1), it can emit photons at various wavelengths. The number of distinct emission lines corresponds to the different possible transitions the electron can make.

2. **Use the Formula for Maximum Emission Lines**: The maximum number of emission lines (or spectral 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 excited state.

3. **Substitute the Value of n**: In this case, \( n = 5 \). Plugging this value into the formula gives:
\[ \text{Maximum Emission Lines} = \frac{5(5 - 1)}{2} \]

4. **Calculate the Value**:
- First, calculate \( 5 - 1 = 4 \).
- Then, multiply \( 5 \times 4 = 20 \).
- Finally, divide by 2:
\[ \frac{20}{2} = 10 \]

5. **Conclusion**: The maximum number of emission lines obtained when the excited electron of a hydrogen atom in n = 5 drops to the ground state is **10**.

### Final Answer:
The maximum number of emission lines is **10**.

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