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 need to determine the value of \( n \) in a hydrogen atom where the electron is in the \( n^{th} \) excited state and can transition to the second excited state (which corresponds to \( n = 3 \)) by emitting ten different wavelengths.

### Step-by-Step Solution:

1. **Understanding the Excited States**:
- The ground state of a hydrogen atom is \( n = 1 \).
- The first excited state is \( n = 2 \).
- The second excited state is \( n = 3 \).
- The \( n^{th} \) excited state corresponds to \( n = n + 1 \).

2. **Transitioning Between States**:
- The electron can transition from the \( n^{th} \) excited state to the second excited state (\( n = 3 \)).
- The number of different wavelengths (or spectral lines) emitted during these transitions can be calculated using the formula for combinations, which is given by:
\[ \text{Number of transitions} = \binom{n}{2} = \frac{n(n-1)}{2} \]

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

4. **Solving the Equation**:
- Multiply both sides by 2 to eliminate the fraction:
\[ n(n-1) = 20 \]
- Rearranging gives us a quadratic equation:
\[ n^2 - n - 20 = 0 \]
- We can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -1, c = -20 \).

5. **Calculating the Discriminant**:
- Calculate the discriminant:
\[ b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-20) = 1 + 80 = 81 \]
- Now substitute back into the quadratic formula:
\[ n = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2} \]
- This gives us two possible solutions:
\[ n = \frac{10}{2} = 5 \quad \text{and} \quad n = \frac{-8}{2} = -4 \]
- Since \( n \) must be a positive integer, we have:
\[ n = 5 \]

6. **Conclusion**:
- The value of \( n \) is 5, which means the electron is in the 5th excited state.

### Final Answer:
The value of \( n \) is **5**.