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 solve the problem of how many spectral lines are produced when an electron in a hydrogen atom transitions from the fourth excited state to the ground state in the Lyman series, we can follow these steps:

### Step 1: Identify the Energy Levels
In the hydrogen atom, the energy levels are denoted by principal quantum numbers (n). The ground state corresponds to n = 1, the first excited state to n = 2, the second excited state to n = 3, the third excited state to n = 4, and the fourth excited state to n = 5.

### Step 2: Determine the Transition
The electron is moving from the fourth excited state (n = 5) to the ground state (n = 1). However, during this transition, the electron can also pass through intermediate energy levels (n = 4, n = 3, n = 2) before reaching the ground state.

### Step 3: Calculate the Number of Spectral Lines
To find the total number of spectral lines produced during this transition, we can use the formula for the number of spectral lines produced when an electron transitions between energy levels:

\[ \text{Number of spectral lines} = \frac{n(n-1)}{2} \]

Where \( n \) is the number of energy levels involved in the transition. In this case, the electron can transition from n = 5 down to n = 1, which involves the levels n = 5, n = 4, n = 3, n = 2, and n = 1.

### Step 4: Determine the Value of n
Here, the electron is transitioning from n = 5 to n = 1, which means we have 5 levels (n = 5, 4, 3, 2, 1). Thus, we can substitute \( n = 5 \) into the formula:

\[ \text{Number of spectral lines} = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \]

### Conclusion
The total number of spectral lines produced when the electron transitions from the fourth excited state (n = 5) to the ground state (n = 1) in the Lyman series is **10**.

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