The wave number of the spectral line in the emission spectrum of hydrogen will be equal to `(8)/(9)` times the Rydberg's constant if the electron jumps from …………..:-

Options

A. n=3 to n=1
B. n=10 to n=1
C. n=9 to n=1
D. n=2 to n=1

Answer

To solve the question regarding the wave number of the spectral line in the emission spectrum of hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula**:
The wave number (denoted as \( \bar{\nu} \)) of a spectral line in the emission spectrum of hydrogen can be expressed using the Rydberg formula:
\[ \bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Setting Up the Equation**:
According to the problem, the wave number is given as:
\[ \bar{\nu} = \frac{8}{9} R_H \]
We can set this equal to the Rydberg formula:
\[ \frac{8}{9} R_H = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

3. **Cancelling Out the Rydberg Constant**:
Since \( R_H \) appears on both sides of the equation, we can cancel it out:
\[ \frac{8}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{8}{9} \]

5. **Finding Suitable Values for \( n_1 \) and \( n_2 \)**:
To satisfy this equation, we can try different integer values for \( n_1 \) and \( n_2 \). We know that \( n_1 \) should be less than \( n_2 \).

Let's try \( n_1 = 1 \) and \( n_2 = 3 \):
\[ \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \]
This satisfies the equation.

6. **Conclusion**:
Therefore, the electron jumps from \( n_2 = 3 \) to \( n_1 = 1 \).

### Final Answer:
The electron jumps from \( n_2 = 3 \) to \( n_1 = 1 \).

---