One of the emission lines of the hydrogen atom has a wavelength of $94.974 \mathrm{nm}$. (a) In what region of the electromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

The given wavelength of the emission line is \( 94.974 \mathrm{nm} \). To find out the region of the electromagnetic spectrum this emission falls into, we can refer to the range of values associated with each region: - Ultraviolet (UV): 10 - 400 nm - Visible light: 400 - 700 nm - Infrared (IR): 700 nm - 1 mm Since the given wavelength falls within the range of 10 - 400 nm, this emission is found in the ultraviolet region.

We can use the Rydberg formula to relate the wavelength of the emitted light to the initial and final values of the principal quantum number n: \[ \frac{1}{\lambda} = R_\mathrm{H} \left( \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} \right) \] where: - \( \lambda \) is the wavelength of the emitted light - \( R_\mathrm{H} \) is the Rydberg constant for hydrogen (\( 1.0973 \times 10^7 \mathrm{m}^{-1} \)) - \( n_\mathrm{f} \) is the final value of the principal quantum number - \( n_\mathrm{i} \) is the initial value of the principal quantum number We can insert the given wavelength into the formula: \[ \frac{1}{94.974 \times 10^{-9}\mathrm{m}} = 1.0973 \times 10^7 \mathrm{m}^{-1} \left( \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} \right) \] Solving for the term in the parentheses: \[ \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} = \frac{1}{94.974 \times 10^{-9}\mathrm{m} \times 1.0973 \times 10^7 \mathrm{m}^{-1}} \approx 1.013 \times 10^{-2} \] By analyzing the energy level transitions in a hydrogen atom, we note that the maximum possible value for \( n_\mathrm{f} \) is 1 since ultraviolet emissions come from the highest energy transitions. So we should try different initial n values to find the one that satisfies the above equation: For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 2 \), the equation gives: \[ 1 - \frac{1}{4} = 0.75 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 3 \), the equation gives: \[ 1 - \frac{1}{9} \approx 0.889 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 4 \), the equation gives: \[ 1 - \frac{1}{16} \approx 0.938 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 5 \), the equation gives: \[ 1 - \frac{1}{25} \approx 0.96 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 6 \), the equation gives: \[ 1 - \frac{1}{36} \approx 0.972 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 7 \), the equation gives: \[ 1 - \frac{1}{49} \approx 0.980 \] which is not equal to \( 1.013 \times 10^{-2} \). Finally, for \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 8 \), the equation gives: \[ 1 - \frac{1}{64} \approx 0.984 \] which is very close to \( 1.013 \times 10^{-2} \). This means that the initial and final values of the principal quantum number associated with this emission are \( n_\mathrm{i} = 8 \) and \( n_\mathrm{f} = 1 \), respectively.