If the series limit frequency of the Lyman series is `v_(L)`, then the series limit frequency of the Pfund seriesis :

Options

A. `v_(L)//25`
B. `25 v_(L)`
C. `16 v_(L)`
D. `v_(L)//16`

Answer

To find the series limit frequency of the Pfund series given the series limit frequency of the Lyman series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Series Limit Frequencies**:
The series limit frequency for a hydrogen atom transition can be calculated using the formula:
\[ \nu \propto \frac{1}{n_f^2 - n_i^2} \]
where \( n_f \) is the final energy level and \( n_i \) is the initial energy level.

2. **Identify the Values for Lyman Series**:
For the Lyman series, the transitions end at \( n_f = 1 \) (ground state). The initial levels can be any \( n_i \geq 2 \). The series limit corresponds to the transition from \( n_i \to \infty \):
\[ \nu_L \propto \frac{1}{1^2 - \infty^2} = \frac{1}{1 - \infty} \rightarrow \text{(approaches zero)} \]

3. **Identify the Values for Pfund Series**:
For the Pfund series, the transitions end at \( n_f = 5 \). The initial levels can be any \( n_i \geq 6 \). The series limit corresponds to the transition from \( n_i \to \infty \):
\[ \nu_P \propto \frac{1}{5^2 - \infty^2} = \frac{1}{25 - \infty} \rightarrow \text{(approaches zero)} \]

4. **Setting Up the Ratio**:
To find the relationship between the frequencies of the two series:
\[ \frac{\nu_P}{\nu_L} = \frac{1/(5^2 - \infty^2)}{1/(1^2 - \infty^2)} \]
Simplifying this gives:
\[ \frac{\nu_P}{\nu_L} = \frac{1/(25 - \infty)}{1/(1 - \infty)} = \frac{1/25}{1} = \frac{1}{25} \]

5. **Finding the Series Limit Frequency of Pfund**:
If we denote the series limit frequency of the Lyman series as \( \nu_L \), then:
\[ \nu_P = \nu_L \cdot \frac{1}{25} \]
This means:
\[ \nu_P = \frac{\nu_L}{25} \]

### Final Answer:
Thus, the series limit frequency of the Pfund series is:
\[ \nu_P = \frac{\nu_L}{25} \]