In a hydrogen atom, the electron is in `n^(th)` excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of n ?

Options

A. `6`
B. `7`
C. `8`
D. `5`

Explanation

To solve the problem, we start by understanding the context of the hydrogen atom and the concept of energy levels.

### Step-by-Step Solution:

1. **Identify the States**:
- The second excited state corresponds to \( n_1 = 3 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), and the second excited state is \( n = 3 \)).
- The electron is initially in the \( n^{th} \) excited state, which corresponds to \( n_2 = n + 1 \).

2. **Use the Formula for Spectral Lines**:
- The number of different wavelengths (spectral lines) emitted when an electron transitions from a higher energy level to a lower energy level can be calculated using the formula:
\[ \text{Number of spectral lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
- Here, \( n_1 = 3 \) and \( n_2 = n + 1 \).

3. **Set Up the Equation**:
- We know that the number of emitted wavelengths is 10, so we can set up the equation:
\[ \frac{(n + 1 - 3)(n + 1 - 3 + 1)}{2} = 10 \]
- Simplifying this gives:
\[ \frac{(n - 2)(n - 1)}{2} = 10 \]

4. **Clear the Fraction**:
- Multiply both sides by 2:
\[ (n - 2)(n - 1) = 20 \]

5. **Expand the Equation**:
- Expanding the left side gives:
\[ n^2 - 3n + 2 = 20 \]
- Rearranging this leads to:
\[ n^2 - 3n - 18 = 0 \]

6. **Factor the Quadratic Equation**:
- We can factor this quadratic equation:
\[ (n - 6)(n + 3) = 0 \]

7. **Solve for n**:
- Setting each factor to zero gives us two potential solutions:
\[ n - 6 = 0 \quad \Rightarrow \quad n = 6 \]
\[ n + 3 = 0 \quad \Rightarrow \quad n = -3 \]
- Since \( n \) must be a positive integer, we discard \( n = -3 \).

8. **Conclusion**:
- Therefore, the value of \( n \) is \( 6 \).

### Final Answer:
The value of \( n \) is \( 6 \).

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