The longest ` lambda` for the Lyamn series is …………… (Given ` R_H = 109678 cm^(-1))` :

Options

A. ` 1215`
B. ` 1315`
C. `1415`
D. `1515`

Explanation

To find the longest wavelength (λ) for the Lyman series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Lyman Series**: The Lyman series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 2) to the lowest energy level (n = 1).

2. **Identify the Formula**: The formula to calculate the wavelength (λ) in a hydrogen atom for any transition is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant (given as \( 109678 \, \text{cm}^{-1} \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Determine the Longest Wavelength**: The longest wavelength corresponds to the smallest energy difference, which occurs when the transition is between the closest energy levels. For the Lyman series, this means:
- \( n_1 = 1 \) (the lowest energy level)
- \( n_2 = 2 \) (the next higher energy level)

4. **Substitute Values into the Formula**:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) \]
\[ \frac{1}{\lambda} = R_H \left( \frac{3}{4} \right) \]

5. **Calculate λ**:
\[ \frac{1}{\lambda} = \frac{3}{4} R_H \]
\[ \lambda = \frac{4}{3 R_H} \]

6. **Substitute the Value of \( R_H \)**:
\[ \lambda = \frac{4}{3 \times 109678 \, \text{cm}^{-1}} \]

7. **Final Calculation**:
- Calculate \( 3 \times 109678 = 329034 \).
- Therefore, \( \lambda = \frac{4}{329034} \).

### Final Answer:
\[ \lambda = \frac{4}{329034} \, \text{cm} \]