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 wavelength of the first line to that of the second line of the Paschen series of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Paschen Series
The Paschen series corresponds to transitions where the final energy level (nf) is 3. The initial energy levels (ni) for the first and second lines of the Paschen series are 4 and 5, respectively.

### Step 2: Use the Rydberg Equation
The Rydberg equation for the wavelength (λ) of emitted light 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,
- \( n_i \) is the initial energy level.

### Step 3: Calculate the Wavelength for the First Line (λ1)
For the first line (ni = 4, nf = 3):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_1} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_1 = \frac{144}{7R} \]

### Step 4: Calculate the Wavelength for the Second Line (λ2)
For the second line (ni = 5, nf = 3):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \]
Calculating this gives:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{9} - \frac{1}{25} \right) \]
Finding a common denominator (225):
\[ \frac{1}{\lambda_2} = R \left( \frac{25 - 9}{225} \right) = R \left( \frac{16}{225} \right) \]
Thus,
\[ \lambda_2 = \frac{225}{16R} \]

### Step 5: Find the Ratio of Wavelengths (λ1/λ2)
Now, we need to find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{144}{7R}}{\frac{225}{16R}} = \frac{144 \cdot 16}{225 \cdot 7} \]
The \( R \) cancels out:
\[ \frac{\lambda_1}{\lambda_2} = \frac{2304}{1575} \]

### Step 6: Simplify the Ratio
Now we simplify \( \frac{2304}{1575} \):
- The greatest common divisor (GCD) of 2304 and 1575 is 9.
- Dividing both by 9 gives:
\[ \frac{256}{175} \]

### Final Answer
Thus, the ratio of the wavelength of the first line to that of the second line of the Paschen series of the hydrogen atom is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{256}{175} \]