The series limit of Balmer series is 6400 Å. The series limit of Paschen series will be

Options

A. a. 3200 Å
B. b. 14400 Å
C. c. 12800 Å
D. d. 64000 Å

Explanation

To find the series limit of the Paschen series given that the series limit of the Balmer series is 6400 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Series Limits**: The series limit of a spectral series corresponds to the transition of an electron from an infinite energy level (n = ∞) to a specific lower energy level (n = n_final). For the Balmer series, the final energy level is n = 2.

2. **Use the Formula for Wavelength**: The formula for the wavelength (λ) in the hydrogen spectrum 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 Balmer Series**:
- For the Balmer series, \( n_f = 2 \) and \( n_i = ∞ \).
- Thus, the equation becomes:
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4} \]
- Therefore, the wavelength for the Balmer series is:
\[ \lambda_B = \frac{4}{R} \]

4. **Calculate for Paschen Series**:
- For the Paschen series, \( n_f = 3 \) and \( n_i = ∞ \).
- The equation becomes:
\[ \frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \]
- Therefore, the wavelength for the Paschen series is:
\[ \lambda_P = \frac{9}{R} \]

5. **Relate the Two Series**:
- We have:
\[ \frac{\lambda_B}{\lambda_P} = \frac{4/R}{9/R} = \frac{4}{9} \]
- Rearranging gives:
\[ \lambda_P = \frac{9}{4} \lambda_B \]

6. **Substitute the Given Value**:
- Given \( \lambda_B = 6400 \, \text{Å} \):
\[ \lambda_P = \frac{9}{4} \times 6400 \, \text{Å} \]
- Calculate:
\[ \lambda_P = 9 \times 1600 \, \text{Å} = 14400 \, \text{Å} \]

### Final Answer:
The series limit of the Paschen series is **14400 Å**.