If an electron of hydrogen atom moves from fourth excited state to ground state in Lyman series find the total number of spectral lines?

Answer

To find the total number of spectral lines when an electron in a hydrogen atom moves from the fourth excited state to the ground state in the Lyman series, we can follow these steps:

### Step 1: Identify the Energy Levels
The fourth excited state corresponds to the principal quantum number \( n = 5 \). The ground state corresponds to \( n = 1 \).

### Step 2: Determine the Transitions
When an electron transitions from \( n = 5 \) to \( n = 1 \), it can pass through intermediate energy levels. The possible transitions are:
- From \( n = 5 \) to \( n = 4 \)
- From \( n = 5 \) to \( n = 3 \)
- From \( n = 5 \) to \( n = 2 \)
- From \( n = 5 \) to \( n = 1 \)
- From \( n = 4 \) to \( n = 3 \)
- From \( n = 4 \) to \( n = 2 \)
- From \( n = 4 \) to \( n = 1 \)
- From \( n = 3 \) to \( n = 2 \)
- From \( n = 3 \) to \( n = 1 \)
- From \( n = 2 \) to \( n = 1 \)

### Step 3: Calculate the Number of Spectral Lines
The formula to calculate the total number of spectral lines formed when an electron transitions from a higher energy level \( n_2 \) to a lower energy level \( n_1 \) is given by:
\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]
where \( n \) is the total number of energy levels involved in the transition.

In this case, the electron transitions from \( n = 5 \) (fourth excited state) to \( n = 1 \) (ground state). The total number of energy levels involved is:
- \( n = 1, 2, 3, 4, 5 \) (which gives us 5 levels)

Substituting \( n = 5 \) into the formula:
\[ \text{Number of spectral lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \]

### Conclusion
Thus, the total number of spectral lines when an electron moves from the fourth excited state to the ground state in the Lyman series is **10**.