Find the ratio of wavelength of 3rd member of lyman to 1st member of paschen series

Answer

To find the ratio of the wavelength of the 3rd member of the Lyman series to the 1st member of the Paschen series, we will follow these steps:

### Step 1: Identify the transitions for the Lyman and Paschen series
- **Lyman series**: Transitions end at \( n = 1 \).
- The 3rd member corresponds to the transition from \( n = 4 \) to \( n = 1 \).

- **Paschen series**: Transitions end at \( n = 3 \).
- The 1st member corresponds to the transition from \( n = 4 \) to \( n = 3 \).

### Step 2: Use the Rydberg formula for wavelengths
The Rydberg formula for the wavelength of light emitted during transitions in a hydrogen atom is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate the wavelength for the 3rd member of the Lyman series
For the 3rd member of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 4 \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{16 - 1}{16} \right) = R \left( \frac{15}{16} \right) \]
Thus,
\[ \lambda_1 = \frac{16}{15R} \]

### Step 4: Calculate the wavelength for the 1st member of the Paschen series
For the 1st member of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = 4 \)

Using the Rydberg formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_2} = R \left( \frac{16}{144} - \frac{9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_2 = \frac{144}{7R} \]

### Step 5: Find the ratio of the wavelengths
Now, we can find the ratio of \( \lambda_1 \) to \( \lambda_2 \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{15R}}{\frac{144}{7R}} = \frac{16 \cdot 7}{15 \cdot 144} = \frac{112}{2160} \]
Simplifying this fraction:
\[ \frac{112}{2160} = \frac{7}{135} \]

### Final Answer
The ratio of the wavelength of the 3rd member of the Lyman series to the 1st member of the Paschen series is:
\[ \frac{7}{135} \]