In a hydrogen `(H)` sample on excitation electron jumps into `n^(th)` energy level and comes back to its ground energy level giving 6 different spectrum lines. Determine value of n.

Options

A. `4`
B. `5`
C. `3`
D. `2`

Explanation

To determine the value of \( n \) for the hydrogen atom where an electron jumps to the \( n^{th} \) energy level and produces 6 different spectral lines upon returning to the ground state, we can use the formula for the number of spectral lines produced when an electron transitions between energy levels.

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the value of \( n \) such that the number of spectral lines produced when the electron returns to the ground state is 6.

2. **Use the Formula**: The number of different spectral lines produced when an electron transitions from level \( n \) to the ground state (level 1) is given by the formula:
\[ \text{Number of lines} = \frac{n(n-1)}{2} \]
This formula arises because each transition from a higher energy level to a lower one produces a spectral line, and we need to consider all possible pairs of transitions.

3. **Set Up the Equation**: Since we know the number of spectral lines is 6, we can set up the equation:
\[ \frac{n(n-1)}{2} = 6 \]

4. **Solve for \( n \)**:
- Multiply both sides by 2 to eliminate the fraction:
\[ n(n-1) = 12 \]
- Rearranging gives us the quadratic equation:
\[ n^2 - n - 12 = 0 \]

5. **Factor the Quadratic**: We can factor the quadratic equation:
\[ (n - 4)(n + 3) = 0 \]
This gives us two potential solutions:
\[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \]
\[ n + 3 = 0 \quad \Rightarrow \quad n = -3 \quad (\text{not a valid solution since } n \text{ must be positive}) \]

6. **Conclusion**: The only valid solution is:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is 4.

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