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 Formula**:
The formula for the wavelength in the Pfund series is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{5^2} - \frac{1}{n^2} \right) \]
where \( R \) is the Rydberg constant, \( n \) is the principal quantum number, and for the Pfund series, the lower level is \( n = 5 \).

2. **Set Up for Shortest Wavelength**:
The shortest wavelength corresponds to the transition from \( n = \infty \) to \( n = 5 \). Therefore, we set \( n \) to infinity:
\[ \frac{1}{\lambda} = R \left( \frac{1}{25} - 0 \right) = \frac{R}{25} \]

3. **Substitute the Rydberg Constant**:
The Rydberg constant \( R \) is given as \( 1.97 \times 10^7 \, \text{m}^{-1} \). Plugging this value into the equation:
\[ \frac{1}{\lambda} = \frac{1.97 \times 10^7}{25} \]

4. **Calculate \( \frac{1}{\lambda} \)**:
Now, calculate \( \frac{1}{\lambda} \):
\[ \frac{1}{\lambda} = \frac{1.97 \times 10^7}{25} = 7.88 \times 10^5 \, \text{m}^{-1} \]

5. **Find \( \lambda \)**:
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{7.88 \times 10^5} \approx 1.27 \times 10^{-6} \, \text{m} \]

6. **Convert to Nanometers**:
Since \( 1 \, \text{m} = 10^9 \, \text{nm} \):
\[ \lambda \approx 1.27 \times 10^{-6} \, \text{m} \times 10^9 \, \text{nm/m} = 1270 \, \text{nm} \]

### Final Answer:
The wavelength limit present in the Pfund series is approximately \( 1270 \, \text{nm} \).

---