The wavelength of first members of Lyman series is `1216A^(@)`. Calculate the wavelength of 3rd member of Paschen series.

Explanation

To calculate the wavelength of the 3rd member of the Paschen series, we can follow these steps:

### Step 1: Understand the Lyman Series
The wavelength of the first member of the Lyman series is given as \( \lambda_1 = 1216 \, \text{Å} \). The Lyman series corresponds to transitions where the electron falls to the \( n=1 \) level.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light during electron transitions 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 Rydberg Constant \( R \)
For the first member of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the Rydberg formula, we have:

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

This simplifies to:

\[ \frac{1}{1216} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus, we can express \( R \) as:

\[ R = \frac{4}{3 \times 1216} \]

### Step 4: Calculate the Wavelength for the 3rd Member of the Paschen Series
For the Paschen series, we are interested in the 3rd member, which corresponds to:
- \( n_1 = 3 \)
- \( n_2 = 6 \)

Using the Rydberg formula again:

\[ \frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{6^2} \right) \]

Calculating the terms:

\[ \frac{1}{\lambda'} = R \left( \frac{1}{9} - \frac{1}{36} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda'} = R \left( \frac{4}{36} - \frac{1}{36} \right) = R \left( \frac{3}{36} \right) = R \left( \frac{1}{12} \right) \]

### Step 5: Substitute \( R \) into the Equation
Substituting \( R \) from Step 3 into the equation for \( \lambda' \):

\[ \frac{1}{\lambda'} = \frac{4}{3 \times 1216} \times \frac{1}{12} \]

This simplifies to:

\[ \frac{1}{\lambda'} = \frac{4}{36 \times 1216} = \frac{1}{9 \times 1216} \]

Thus:

\[ \lambda' = 9 \times 1216 \]

### Step 6: Calculate the Final Wavelength
Calculating \( \lambda' \):

\[ \lambda' = 9 \times 1216 = 10944 \, \text{Å} \]

### Final Answer
The wavelength of the 3rd member of the Paschen series is \( \lambda' = 10944 \, \text{Å} \).

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