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Look up the wavelength intervals for each region of the spectrum to see where 1094 nm falls. The electromagnetic spectrum can be divided into these regions: 1. Radio waves: greater than 1 mm 2. Microwaves: 1 mm - 100 µm 3. Infrared: 100 µm - 750 nm 4. Visible light: 750 nm - 400 nm 5. Ultraviolet: 400 nm - 10 nm 6. X-rays: 10 nm - 0.1 nm 7. Gamma rays: less than 0.1 nm 1094 nm falls within the range of infrared light, so this absorption occurs in the infrared region of the spectrum.

The Rydberg formula 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 (\(1.097373 \times 10^7 \mathrm{m^{-1}}\)), and \(n_1\) and \(n_2\) are the initial and final principal quantum numbers, respectively. Since the hydrogen atom absorbs light, we know that an electron transitions to a higher energy level, so \(n_2 > n_1\). First, let's convert the wavelength to meters and calculate the reciprocal: \[\lambda = 1094 \mathrm{~nm} = 1.094 \times 10^{-6} \mathrm{m}\] \[\frac{1}{\lambda} = \frac{1}{1.094 \times 10^{-6}} = 9.137 \times 10^6 \mathrm{m^{-1}}\] Now, plug in the known values into the Rydberg formula: \[(9.137 \times 10^6) = (1.097373 \times 10^7) \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] Divide by the Rydberg constant so that we can solve for the difference in \(n_1^2\) and \(n_2^2\): \[\frac{9.137 \times 10^6}{1.097373 \times 10^7} = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] \[0.833 = \frac{1}{n_1^2} - \frac{1}{n_2^2}\] Now we need to test different integer values of \(n_1\) and \(n_2\) to find a pair of numbers that satisfies this equation. After testing, we find that: \[0.833 \approx \frac{1}{1^2} - \frac{1}{3^2} = \frac{8}{9}\] So, the initial value, \(n_1\), is 1, and the final value, \(n_2\), is 3. This electron transition corresponds to the absorption of light with a wavelength of 1094 nm.