One of the emission lines of the hydrogen atom has a wavelength of 93.07 nm. (a) In what region of the elertromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

The given wavelength is 93.07 nm. We can use this information to identify the region of the electromagnetic spectrum to which this emission belongs. The electromagnetic spectrum is divided into various regions, some of which are: - Ultraviolet (UV) region: 10 nm to 400 nm - Visible region: 400 nm to 700 nm - Infrared (IR) region: 700 nm to 1 mm The given wavelength of 93.07 nm falls in the ultraviolet (UV) region. Answer to (a): The emission line with a wavelength of 93.07 nm is in the ultraviolet region of the electromagnetic spectrum.

The Rydberg formula for the hydrogen atom's emission spectra is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (approximately 1.097373 x \(10^7\) m\(^{-1}\)), \(n_1\) is the final principal quantum number, and \(n_2\) is the initial principal quantum number. It is important to note that \(n_2 > n_1\) for emission. Let's plug in the given value of the wavelength and solve for the combination of \(n_1\) and \(n_2\).

First, we need to convert the wavelength from nm to meters: \(\lambda = 93.07 \times 10^{-9} \text{m}\) Now, plug in the value of the wavelength and the Rydberg constant into the Rydberg formula and solve for the difference of inverse squares of quantum numbers: \[\frac{1}{93.07 \times 10^{-9} \text{m}} = (1.097373 \times 10^7 \text{m}^{-1}) \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] \[\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{93.07 \times 10^{-9} \text{m} \times 1.097373 \times 10^7 \text{m}^{-1}} \approx 0.122\] We will now create a table for values of \(n_1\) and \(n_2\) and their corresponding differences in inverse squares of quantum numbers to observe the possible combinations of quantum numbers. | \(n_1\) | \(n_2\) | \(\Delta(\frac{1}{n^2})\) |------|------|----------------| | 1 | 2 | 0.750 | | 1 | 3 | 0.889 | | 1 | 4 | 0.937 | | 1 | 5 | 0.960 | | 1 | 6 | 0.972 | | 2 | 3 | 0.139 | | 2 | 4 | 0.187 | | 2 | 5 | 0.210 | | 2 | 6 | 0.222 | The closest value to 0.122 in our table is 0.139, which corresponds to \(n_1 = 2\) and \(n_2 = 3\). Answer to (b): The initial and final values of n associated with this emission are \(n_1 = 2\) (final) and \(n_2 = 3\) (initial).