To solve this problem, we need to find the ratio of the wavelength of the last line of the Balmer series to the last line of the Lyman series.$\\$The wavelength of the spectral lines in the hydrogen atom can be calculated using the Rydberg formula:$\\ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \\$where $R$ is the Rydberg constant, $n_1$ and $n_2$ are integers with $n_2 > n_1. \\$For the Balmer series, $n_1 = 2$ and the last line corresponds to $n_2$ to $\infty. \\$For the Lyman series, $n_1 = 1$ and the last line corresponds to $n_2 \to \infty. \\$Let's calculate the wavelength for the last line of each series:$\\$• For the Balmer series:$\\ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} \right) \\ \lambda_B = \frac{4}{R} \\$• For the Lyman series:$\\ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 \right) \\ \lambda_L = \frac{1}{R} \\$Now, calculate the ratio of the wavelengths:$\\ \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{1}{R}} = 4 \\$Therefore, the ratio of the wavelength of the last line of the Balmer series to the last line of the Lyman series is $4. \\$This corresponds to Option 2.