The wavelength of a spectral line emmited by hydrogen atom in the lyman series is `16/15R` cm. what is the value of `n_(2)`

Options

A. 2
B. 3
C. 4
D. 1

Explanation

To solve the problem, we need to find the value of \( n_2 \) for the given wavelength of a spectral line emitted by a hydrogen atom in the Lyman series, which is given as \( \frac{16}{15} R \) cm.

### Step-by-Step Solution:

1. **Identify the Formula**:
The formula for the wavelength in a hydrogen atom's spectral series is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Set the Values**:
For the Lyman series, the lower energy level \( n_1 \) is always 1. Therefore, we can substitute \( n_1 = 1 \) into the formula:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_2^2} \right) \]

3. **Convert Wavelength**:
The given wavelength \( \lambda \) is \( \frac{16}{15} R \) cm. We need to express \( \frac{1}{\lambda} \):
\[ \frac{1}{\lambda} = \frac{15R}{16} \]

4. **Substitute into the Formula**:
Now, we can substitute \( \frac{1}{\lambda} \) into the Rydberg formula:
\[ \frac{15R}{16} = R \left( 1 - \frac{1}{n_2^2} \right) \]

5. **Cancel R**:
We can cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ \frac{15}{16} = 1 - \frac{1}{n_2^2} \]

6. **Rearrange the Equation**:
Rearranging gives:
\[ \frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16} \]

7. **Solve for \( n_2^2 \)**:
Taking the reciprocal gives:
\[ n_2^2 = 16 \]

8. **Find \( n_2 \)**:
Taking the square root:
\[ n_2 = 4 \]

### Final Answer:
The value of \( n_2 \) is \( 4 \).

---