What will be the ratio of wavelength of the first line to that of the second line of paschen series of H - atom ?

Options

A. `256:175`
B. `175:256`
C. `15:16`
D. `24:27`

Answer

To find the ratio of the wavelengths of the first and second lines of the Paschen series for the hydrogen atom, we can follow these steps:

### Step 1: Identify the transitions
The Paschen series corresponds to electronic transitions where the electron falls to the n=3 energy level. The first line corresponds to the transition from n=4 to n=3, and the second line corresponds to the transition from n=5 to n=3.

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength of light emitted during a transition in a hydrogen-like atom is given by:

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

Where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate the wavelength for the first line (n=4 to n=3)
For the first line (n=4 to n=3):
- \( n_1 = 3 \)
- \( n_2 = 4 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_1} = R \cdot \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \cdot \left( \frac{1}{9} - \frac{1}{16} \right) \]

Calculating the difference:

\[ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \]

Thus,

\[ \frac{1}{\lambda_1} = R \cdot \frac{7}{144} \]

### Step 4: Calculate the wavelength for the second line (n=5 to n=3)
For the second line (n=5 to n=3):
- \( n_1 = 3 \)
- \( n_2 = 5 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_2} = R \cdot \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \cdot \left( \frac{1}{9} - \frac{1}{25} \right) \]

Calculating the difference:

\[ \frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225} \]

Thus,

\[ \frac{1}{\lambda_2} = R \cdot \frac{16}{225} \]

### Step 5: Find the ratio of wavelengths
Now, we need to find the ratio \( \frac{\lambda_1}{\lambda_2} \). Using the relationship \( \frac{1}{\lambda} \):

\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\lambda_2}}{\frac{1}{\lambda_1}} = \frac{R \cdot \frac{16}{225}}{R \cdot \frac{7}{144}} = \frac{16}{225} \cdot \frac{144}{7} \]

Simplifying this:

\[ \frac{\lambda_1}{\lambda_2} = \frac{16 \cdot 144}{225 \cdot 7} = \frac{2304}{1575} \]

### Step 6: Simplify the ratio
Now we can simplify the ratio \( \frac{2304}{1575} \).

After simplification, we find:

\[ \frac{\lambda_1}{\lambda_2} = \frac{256}{175} \]

### Final Answer
Thus, the ratio of the wavelength of the first line to that of the second line of the Paschen series for the hydrogen atom is:

\[ \frac{\lambda_1}{\lambda_2} = \frac{256}{175} \]