To calculate the wavelengths for the first three lines in the Paschen series, we will use the formula derived from the Rydberg formula for hydrogen spectral lines:
\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
Where:
- \( \lambda \) is the wavelength,
- \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \) (Rydberg constant),
- \( n_1 \) is the lower energy level (for the Paschen series, \( n_1 = 3 \)),
- \( n_2 \) is the higher energy level (for the first three lines, \( n_2 = 4, 5, 6 \)).
### Step 1: Calculate the wavelength for the transition from \( n_2 = 4 \) to \( n_1 = 3 \)
\[
\frac{1}{\lambda_4} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right)
\]
Calculating the right-hand side:
\[
\frac{1}{\lambda_4} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right)
\]
Calculating \( \frac{1}{9} - \frac{1}{16} \):
\[
\frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144} \quad \Rightarrow \quad \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144}
\]
Now substituting back:
\[
\frac{1}{\lambda_4} = 1.097 \times 10^7 \times \frac{7}{144}
\]
Calculating:
\[
\frac{1}{\lambda_4} \approx 5.36 \times 10^5 \, \text{m}^{-1}
\]
Now, taking the reciprocal to find \( \lambda_4 \):
\[
\lambda_4 \approx \frac{1}{5.36 \times 10^5} \approx 1.87 \times 10^{-6} \, \text{m}
\]
### Step 2: Calculate the wavelength for the transition from \( n_2 = 5 \) to \( n_1 = 3 \)
\[
\frac{1}{\lambda_5} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right)
\]
Calculating the right-hand side:
\[
\frac{1}{\lambda_5} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{25} \right)
\]
Calculating \( \frac{1}{9} - \frac{1}{25} \):
\[
\frac{1}{9} = \frac{25}{225}, \quad \frac{1}{25} = \frac{9}{225} \quad \Rightarrow \quad \frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225}
\]
Now substituting back:
\[
\frac{1}{\lambda_5} = 1.097 \times 10^7 \times \frac{16}{225}
\]
Calculating:
\[
\frac{1}{\lambda_5} \approx 7.78 \times 10^5 \, \text{m}^{-1}
\]
Taking the reciprocal to find \( \lambda_5 \):
\[
\lambda_5 \approx \frac{1}{7.78 \times 10^5} \approx 1.28 \times 10^{-6} \, \text{m}
\]
### Step 3: Calculate the wavelength for the transition from \( n_2 = 6 \) to \( n_1 = 3 \)
\[
\frac{1}{\lambda_6} = R_H \left( \frac{1}{3^2} - \frac{1}{6^2} \right)
\]
Calculating the right-hand side:
\[
\frac{1}{\lambda_6} = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{36} \right)
\]
Calculating \( \frac{1}{9} - \frac{1}{36} \):
\[
\frac{1}{9} = \frac{4}{36}, \quad \frac{1}{36} = \frac{1}{36} \quad \Rightarrow \quad \frac{1}{9} - \frac{1}{36} = \frac{4 - 1}{36} = \frac{3}{36} = \frac{1}{12}
\]
Now substituting back:
\[
\frac{1}{\lambda_6} = 1.097 \times 10^7 \times \frac{1}{12}
\]
Calculating:
\[
\frac{1}{\lambda_6} \approx 9.14 \times 10^5 \, \text{m}^{-1}
\]
Taking the reciprocal to find \( \lambda_6 \):
\[
\lambda_6 \approx \frac{1}{9.14 \times 10^5} \approx 1.09 \times 10^{-6} \, \text{m}
\]
### Final Results
1. \( \lambda_4 \approx 1.87 \times 10^{-6} \, \text{m} \)
2. \( \lambda_5 \approx 1.28 \times 10^{-6} \, \text{m} \)
3. \( \lambda_6 \approx 1.09 \times 10^{-6} \, \text{m} \)
