Last line of bracket series for H-atom has wavelength `lambda_(1)"Å"` and 2nd line of Lyman series has wavelength `lambda_(2)"Å"`then:

Options

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

Answer

To solve the problem regarding the wavelengths of the last line of the Brackett series and the second line of the Lyman series for the hydrogen atom, we can follow these steps:

### Step 1: Understand the Brackett Series
The Brackett series corresponds to transitions where the electron falls to the n=4 energy level from higher energy levels (n=5, 6, 7, ...). The formula for the wavelength in the Brackett series is given by:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the last line of the Brackett series, the transition occurs from \(n_2 = \infty\) to \(n_1 = 4\). Thus, we have:

\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{16} - 0 \right) = \frac{R_H}{16} \]

### Step 2: Calculate Wavelength for Brackett Series
From the above, we can express \(R_H\):

\[ R_H = \frac{16}{\lambda_1} \]

### Step 3: Understand the Lyman Series
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level from higher energy levels (n=2, 3, ...). The formula for the wavelength in the Lyman series is:

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

For the second line of the Lyman series, the transition occurs from \(n_2 = 3\) to \(n_1 = 1\):

\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \]

### Step 4: Calculate Wavelength for Lyman Series
From this, we can express \(R_H\) again:

\[ R_H = \frac{9}{8\lambda_2} \]

### Step 5: Equate the Two Expressions for \(R_H\)
Now we have two expressions for \(R_H\):

1. \(R_H = \frac{16}{\lambda_1}\)
2. \(R_H = \frac{9}{8\lambda_2}\)

Setting these equal gives:

\[ \frac{16}{\lambda_1} = \frac{9}{8\lambda_2} \]

### Step 6: Cross Multiply to Solve for Relationship
Cross multiplying yields:

\[ 16 \cdot 8\lambda_2 = 9\lambda_1 \]

This simplifies to:

\[ 128\lambda_2 = 9\lambda_1 \]

### Step 7: Final Expression
Rearranging gives us the final relationship:

\[ \frac{\lambda_1}{\lambda_2} = \frac{128}{9} \]

### Conclusion
Thus, the relationship between the wavelengths of the last line of the Brackett series and the second line of the Lyman series is:

\[ \lambda_1 = \frac{128}{9} \lambda_2 \]