Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron from the \(\mathrm{H}\) atom is \(1460 \mathrm{~nm}\). What is the initial excited state for the electron \((n=?)\) ?

In this case, the electron is being ionized, meaning it is completely removed from the atom. This corresponds to a final quantum number of \(n_\mathrm{f} = \infty\). Plugging this into the Balmer-Rydberg formula, we get: \(\dfrac{1}{\lambda} = R_\mathrm{H} \left( \dfrac{1}{n_\mathrm{f}^2} - \dfrac{1}{n_\mathrm{i}^2} \right) = R_\mathrm{H} \left( 0 - \dfrac{1}{n_\mathrm{i}^2} \right) = - \dfrac{R_\mathrm{H}}{n_\mathrm{i}^2}\) Now, we have a modified formula: \(\dfrac{1}{\lambda} = - \dfrac{R_\mathrm{H}}{n_\mathrm{i}^2}\)

We are given the maximum wavelength of electromagnetic radiation that can ionize the electron (\(\lambda = 1460 \mathrm{ ~nm}\)) and the Rydberg constant for hydrogen \(R_\mathrm{H} = 1.097 \times 10^7 \mathrm{ ~m}^{-1}\). Convert the wavelength from nm to m: \(1460 \mathrm{ ~nm} = 1.460 \times 10^{-6} \mathrm{ ~m}\) Use the modified formula to solve for \(n_\mathrm{i}\): \(\dfrac{1}{1.460 \times 10^{-6} \mathrm{ ~m}} = - \dfrac{1.097 \times 10^7 \mathrm{ ~m}^{-1}}{n_\mathrm{i}^2}\) \(n_\mathrm{i}^2 = - \dfrac{1.097 \times 10^7 \mathrm{ ~m}^{-1}}{1.460 \times 10^{-6} \mathrm{ ~m}}\) \(n_\mathrm{i}^2 \approx 7.52\) \(n_\mathrm{i} \approx \sqrt{7.52} \approx 2.74\) Since the principal quantum number must be an integer, we round up to the nearest whole number because the maximum wavelength corresponds to the closest energy level to ionization: \(n_\mathrm{i} = 3\)

The initial excited state for the electron in the hydrogen atom is \(n = 3\).