In a hydrogen atom, the electron is in nth excited state. It comes down to the first excited state by emitting `10` different wavelength. The value of `n` is

Options

A. 6
B. 7
C. 8
D. 9

Explanation

To solve the problem of determining the value of \( n \) in a hydrogen atom where the electron transitions from the \( n \)-th excited state to the first excited state (n=2) and emits 10 different wavelengths, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the States**:
- The ground state of hydrogen corresponds to \( n = 1 \).
- The first excited state corresponds to \( n = 2 \).
- The \( n \)-th excited state corresponds to \( n = n + 1 \) (where \( n \) is the principal quantum number).

2. **Identifying the Transition**:
- The electron is transitioning from the \( n \)-th excited state (which is \( n \)) to the first excited state (which is \( n = 2 \)).
- This means the electron is falling from \( n \) to \( 2 \).

3. **Using the Formula for Spectral Lines**:
- The number of spectral lines (or different wavelengths) emitted during a transition from \( n_2 \) to \( n_1 \) is given by the formula:
\[ \text{Number of lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} \]
- Here, \( n_2 = n \) and \( n_1 = 2 \).

4. **Setting Up the Equation**:
- According to the problem, the number of emitted wavelengths is 10:
\[ 10 = \frac{(n - 2)(n - 2 + 1)}{2} \]
- Simplifying this gives:
\[ 10 = \frac{(n - 2)(n - 1)}{2} \]
- Multiplying both sides by 2:
\[ 20 = (n - 2)(n - 1) \]

5. **Expanding the Equation**:
- Expanding the right-hand side:
\[ 20 = n^2 - 3n + 2 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ n^2 - 3n - 18 = 0 \]

7. **Solving the Quadratic Equation**:
- We can factor or use the quadratic formula to solve for \( n \):
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1, b = -3, c = -18 \):
\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1} \]
\[ n = \frac{3 \pm \sqrt{9 + 72}}{2} \]
\[ n = \frac{3 \pm \sqrt{81}}{2} \]
\[ n = \frac{3 \pm 9}{2} \]
- This gives us two possible solutions:
\[ n = \frac{12}{2} = 6 \quad \text{and} \quad n = \frac{-6}{2} = -3 \]

8. **Selecting the Valid Solution**:
- Since \( n \) must be a positive integer, we discard \( n = -3 \).
- Thus, the valid solution is \( n = 6 \).

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

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