A collection of hydrogen atoms have all been placed into the \(n=4\) excited state. What wavelengths of photons will be emitted by the hydrogen atoms as they transition back to the ground state?

The Rydberg formula for hydrogen is given by: \( \dfrac{1}{\lambda} = R_H (\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}) \) where \(\lambda\) is the wavelength of the emitted photon, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097 \times 10^7 \, \text{m}^{-1}\)), \(n_1\) is the principal quantum number of the lower energy level, and \(n_2\) is the principal quantum number of the higher energy level. Since the hydrogen atoms are initially in the \(n=4\) excited state, we know that \(n_2 = 4\). The ground state of an atom corresponds to \(n_1 = 1\).

Now we can use the Rydberg formula to calculate the wavelength of the emitted photon when the hydrogen atom transitions from the \(n=4\) state to the \(n=1\) ground state: \( \dfrac{1}{\lambda_{4 \rightarrow 1}} = R_H (\dfrac{1}{1^2} - \dfrac{1}{4^2}) \) \( \dfrac{1}{\lambda_{4 \rightarrow 1}} = R_H (\dfrac{1}{1} - \dfrac{1}{16}) \) \( \dfrac{1}{\lambda_{4 \rightarrow 1}} = R_H (\dfrac{15}{16}) \) Now, we'll find the value of \(\lambda_{4 \rightarrow 1}\): \( \lambda_{4 \rightarrow 1} = \dfrac{1}{R_H (\dfrac{15}{16})} \) \( \lambda_{4 \rightarrow 1} = \dfrac{1}{(1.097 \times 10^7 \, \text{m}^{-1}) (\dfrac{15}{16})} \) \( \lambda_{4 \rightarrow 1} \approx 9.723 \times 10^{-8} \, \text{m} \) So, the wavelength of the photon emitted when the hydrogen atom transitions from the \(n=4\) state to the \(n=1\) ground state is approximately \(9.723 \times 10^{-8} \, \text{m}\).