the wavelength limit present in the pfund series is `(R=1. 097xx10^7m^-1)`

Options

A. 1572 nm
B. 1898 nm
C. 2278 nm
D. 2535 nm

Explanation

To find the wavelength limit present in the Pfund series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Series and Formula**:
The Pfund series corresponds to transitions where the final energy level \( n_f = 5 \). The formula for the wavelength \( \lambda \) is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R = 1.097 \times 10^7 \, \text{m}^{-1} \).

2. **Set the Final State**:
For the Pfund series, we set \( n_f = 5 \).

3. **Determine the Initial State for Wavelength Limit**:
To find the wavelength limit, we consider the case where the electron transitions from an infinitely high energy level to \( n_f = 5 \). Thus, we set \( n_i \) to approach infinity:
\[ \frac{1}{n_i^2} \to 0 \quad \text{as} \quad n_i \to \infty \]

4. **Substitute Values into the Formula**:
Substitute \( n_f = 5 \) and \( n_i \to \infty \) into the Rydberg formula:
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{5^2} - 0 \right) = R \left( \frac{1}{25} \right) \]

5. **Calculate the Wavelength Limit**:
Plugging in the value of \( R \):
\[ \frac{1}{\lambda_{\text{min}}} = 1.097 \times 10^7 \times \frac{1}{25} \]
\[ \lambda_{\text{min}} = \frac{25}{1.097 \times 10^7} \]

6. **Perform the Calculation**:
Calculate \( \lambda_{\text{min}} \):
\[ \lambda_{\text{min}} = \frac{25}{1.097 \times 10^7} \approx 2.28 \times 10^{-6} \, \text{m} \]

7. **Convert to Nanometers**:
To convert meters to nanometers, multiply by \( 10^9 \):
\[ \lambda_{\text{min}} \approx 2.28 \times 10^{-6} \times 10^9 \, \text{nm} = 2280 \, \text{nm} \]

8. **Final Answer**:
Thus, the wavelength limit present in the Pfund series is approximately:
\[ \lambda_{\text{min}} \approx 2278 \, \text{nm} \]