if the wavelength of first member of Lyman series is `lambda` then calculate the wavelength of first member of Pfund series

Options

A. `675/11 lambda`
B. `245/11 lambda`
C. `322/13 lambda`
D. `289/11 lambda`

Answer

To solve the problem, we need to calculate the wavelength of the first member of the Pfund series given that the wavelength of the first member of the Lyman series is λ.

### Step-by-Step Solution:

1. **Understanding the Series**:
- The Lyman series corresponds to transitions where the final energy level (nf) is 1. The first member of the Lyman series corresponds to a transition from n=2 to n=1 (ni = 2).
- The Pfund series corresponds to transitions where the final energy level (nf) is 5. The first member of the Pfund series corresponds to a transition from n=6 to n=5 (ni = 6).

2. **Using the Rydberg Formula**:
- The Rydberg formula for the wavelength of emitted light is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
- Where R is the Rydberg constant, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

3. **Calculate for Lyman Series**:
- For the first member of the Lyman series:
- \( n_f = 1 \)
- \( n_i = 2 \)
- Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore, we have:
\[ \frac{1}{\lambda} = \frac{3R}{4} \quad \text{(1)} \]

4. **Calculate for Pfund Series**:
- For the first member of the Pfund series:
- \( n_f = 5 \)
- \( n_i = 6 \)
- Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) = R \left( \frac{1}{25} - \frac{1}{36} \right) \]
- Finding a common denominator (which is 900):
\[ \frac{1}{\lambda_2} = R \left( \frac{36 - 25}{900} \right) = R \left( \frac{11}{900} \right) \]
- Therefore, we have:
\[ \frac{1}{\lambda_2} = \frac{11R}{900} \quad \text{(2)} \]

5. **Finding the Ratio of Wavelengths**:
- Now, we can find the ratio of \( \lambda_2 \) to \( \lambda \) by dividing equation (2) by equation (1):
\[ \frac{\lambda_2}{\lambda} = \frac{\frac{900}{11R}}{\frac{4}{3R}} = \frac{900 \cdot 3}{11 \cdot 4} = \frac{2700}{44} = \frac{675}{11} \]
- Therefore, we can express \( \lambda_2 \) in terms of \( \lambda \):
\[ \lambda_2 = \frac{675}{11} \lambda \]

### Final Answer:
The wavelength of the first member of the Pfund series is:
\[ \lambda_2 = \frac{675}{11} \lambda \]