Last line of brackett series for H atom has `lambda_(1) A wavelength 2^(nd)` line of lyman series has wavelength `lambda_(2) A` then

Options

A. `(128)/(lambda_(1))=(6)/(lambda_(2))`
B. `(16)/(lambda_(1))=(9)/(8lambda_(2))`
C. `(4)/(lambda_(1))=(1)/(lambda_(2))`
D. `(128)/(lambda_(1))=(8)/(lambda_(2))`

Answer

To solve the problem, we need to find the relationship between the wavelengths of the last line of the Brackett series and the second line of the Lyman series for the hydrogen atom.

### Step-by-Step Solution:

1. **Identify the Brackett Series:**
The Brackett series corresponds to transitions where the final energy level \( n_1 = 4 \). The last line of this series occurs when the electron transitions from \( n_2 = \infty \) to \( n_1 = 4 \).

2. **Calculate Wavelength for Brackett Series:**
Using the Rydberg formula for hydrogen:
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 4 \) and \( n_2 = \infty \) (where \( \frac{1}{\infty^2} = 0 \)):
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4^2} - 0 \right) = R_H \left( \frac{1}{16} \right) \]
Thus,
\[ \lambda_1 = \frac{16}{R_H} \]
This is given as \( \lambda_1 = \lambda_{1a} \).

3. **Identify the Lyman Series:**
The Lyman series corresponds to transitions where the final energy level \( n_1 = 1 \). The second line of this series corresponds to the transition from \( n_2 = 3 \) to \( n_1 = 1 \).

4. **Calculate Wavelength for Lyman Series:**
Using the Rydberg formula again:
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting \( n_1 = 1 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \]
Thus,
\[ \lambda_2 = \frac{9}{8R_H} \]
This is given as \( \lambda_2 = \lambda_{2a} \).

5. **Find the Ratio of Wavelengths:**
Now we can find the ratio of \( \lambda_1 \) to \( \lambda_2 \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{R_H}}{\frac{9}{8R_H}} = \frac{16 \cdot 8}{9} = \frac{128}{9} \]

6. **Final Relationship:**
Rearranging gives:
\[ \frac{16}{\lambda_1} = \frac{9}{8\lambda_2} \]

### Conclusion:
The relationship between the wavelengths is:
\[ \frac{16}{\lambda_1} = \frac{9}{8\lambda_2} \]