The ratio of the wavelengths of `H_(alpha)` and `H_(beta)` lines of Paschen series is of the order of

Options

A. 10
B. `1//10`
C. 1.5
D. 100

Explanation

To find the ratio of the wavelengths of the H-alpha and H-beta lines of the Paschen series, we can use the Rydberg formula. Let's go through the solution step by step.

### Step 1: Understand the Rydberg Formula
The Rydberg formula for the wavelengths of spectral lines in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 2: Identify the Quantum Numbers for H-alpha and H-beta
For the H-alpha line (first line of the Paschen series):
- \( n_1 = 3 \) (lower level)
- \( n_2 = 4 \) (higher level)

For the H-beta line (second line of the Paschen series):
- \( n_1 = 3 \) (lower level)
- \( n_2 = 5 \) (higher level)

### Step 3: Calculate the Wavelengths
Using the Rydberg formula, we can calculate the wavelengths for both lines.

**For H-alpha:**
\[ \frac{1}{\lambda_{\alpha}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_{\alpha}} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_{\alpha} = \frac{144}{7R} \]

**For H-beta:**
\[ \frac{1}{\lambda_{\beta}} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{9} - \frac{1}{25} \right) \]
Finding a common denominator (225):
\[ \frac{1}{\lambda_{\beta}} = R \left( \frac{25 - 9}{225} \right) = R \left( \frac{16}{225} \right) \]
Thus,
\[ \lambda_{\beta} = \frac{225}{16R} \]

### Step 4: Calculate the Ratio of Wavelengths
Now, we can find the ratio \( \frac{\lambda_{\alpha}}{\lambda_{\beta}} \):
\[ \frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{\frac{144}{7R}}{\frac{225}{16R}} = \frac{144 \cdot 16}{225 \cdot 7} \]
Simplifying this gives:
\[ \frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{2304}{1575} \]

### Step 5: Approximate the Ratio
Calculating the approximate value:
\[ \frac{2304}{1575} \approx 1.46 \approx 1.5 \]

### Final Answer
Thus, the ratio of the wavelengths of H-alpha to H-beta is approximately \( 1.5 \).