The Lyman series of emission lines of the hydrogen atom are those for which \(n_{1}=1 .\) (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series - those for which \(n_{1}=2,3,\) and \(4 .\)

The formula we will use to determine the wavelength of hydrogen spectrum lines is known as the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] Here, \(\lambda\) is the wavelength of the spectral line, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.09737\times10^7\) m\(^{-1}\)), \(n_1\) and \(n_2\) are positive integers representing the energy levels, with \(n_2 > n_1\). For the Lyman series, \(n_1=1\) so we will be plugging this value into the formula to get the required wavelengths.

To find the region of the electromagnetic spectrum in which the Lyman series lines are observed, we need to find the wavelengths associated with \(n_2 \to \infty\), which gives the shortest wavelength of the series (since shorter wavelength means higher frequency and energy). Plugging \(n_1=1, n_2=\infty\) into the Rydberg formula, we have: \[\frac{1}{\lambda_\text{min}} = R_H \left(\frac{1}{1^2} - \frac{1}{\infty}\right) = R_H\] Thus, the minimum wavelength (\(\lambda_\text{min}\)) of the Lyman series is: \[\lambda_\text{min} = \frac{1}{R_H} \approx \frac{1}{1.09737\times10^7\,\text{m}^{-1}} \approx 91.175\times10^{-9}\,\text{m} \] Since this series lies in the ultraviolet region of the electromagnetic spectrum, we can say that the lines of the Lyman series are observed in the ultraviolet region.

To find the wavelengths of the first three lines in the Lyman series, we need to calculate the wavelengths for \(n_2=2\), \(3\), and \(4\). 1. For \(n_2=2\): \[\frac{1}{\lambda_1} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\] \[\lambda_1 =\frac{1}{R_H\left(\frac{3}{4}\right)}\approx \frac{1}{(1.09737\times10^7\,\text{m}^{-1})\left(\frac{3}{4}\right)} \approx 121.54\times10^{-9}\,\text{m} \] 2. For \(n_2=3\): \[\frac{1}{\lambda_2} = R_H \left(\frac{1}{1^2} - \frac{1}{3^2}\right)\] \[\lambda_2 =\frac{1}{R_H\left(\frac{8}{9}\right)}\approx \frac{1}{(1.09737\times10^7\,\text{m}^{-1})\left(\frac{8}{9}\right)} \approx 102.57\times10^{-9}\,\text{m} \] 3. For \(n_2=4\): \[\frac{1}{\lambda_3} = R_H \left(\frac{1}{1^2} - \frac{1}{4^2}\right)\] \[\lambda_3 =\frac{1}{R_H\left(\frac{15}{16}\right)}\approx \frac{1}{(1.09737\times10^7\,\text{m}^{-1})\left(\frac{15}{16}\right)} \approx 97.243\times10^{-9}\,\text{m} \] Therefore, the wavelengths of the first three lines in the Lyman series are approximately \(121.54\) nm, \(102.57\) nm, and \(97.243\) nm, respectively.