Question

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?

Short answer

Answer: The wavelength of the emitted photon for the transition from the \(n_i = 4\) state to the ground state (\(n_f = 1\)) is approximately \(97.24\: \mathrm{nm}\).

Step-by-step solution

Rydberg Formula for Hydrogen Atom

We'll start by writing down the Rydberg formula for the hydrogen atom, which relates the wavelengths of emitted photons to the initial (\(n_i\)) and final (\(n_f\)) energy levels of the electron in the atom: \(\displaystyle \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\) where \(\lambda\) is the wavelength of the emitted photon, \(R_H\) is the Rydberg constant for hydrogen (\(1.097373 \times 10^7\: \mathrm{m}^{-1}\)), and \(n_i\) and \(n_f\) are the initial and final energy levels, respectively. In this case, we have \(\displaystyle n_i = 4\) and \(\displaystyle n_f =1\), since the electron is transitioning back to the ground state.

Determine Transitions and Apply the Rydberg Formula

The electron will move from the \(\displaystyle n_i = 4\) state to the ground state \(\displaystyle (n_f = 1)\). Therefore, we have one transition occurring: \(\displaystyle 4 \rightarrow 1\). Now we can use the Rydberg formula to find the wavelength of the emitted photon for this transition: \(\displaystyle \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{4^2} \right)\)

Solve for the Wavelength

Next, we'll solve for the wavelength of the emitted photon, after plugging in the Rydberg constant value: \(\displaystyle \frac{1}{\lambda} = \left(1.097373\times 10^7\: \mathrm{m}^{-1}\right) \left( \frac{1}{1} - \frac{1}{16} \right)\) \(\displaystyle \frac{1}{\lambda} = \left(1.097373\times 10^7\: \mathrm{m}^{-1}\right) \left( \frac{15}{16} \right)\) Now, find the wavelength by inverting this expression: \(\displaystyle \lambda = \frac{1}{\left(1.097373\times 10^7\: \mathrm{m}^{-1}\right) \left( \frac{15}{16} \right)}\) \(\displaystyle \lambda \approx 9.724\times 10^{-8}\: \mathrm{m} = 97.24\: \mathrm{nm}\) The wavelength of the emitted photon for the transition from the \(\displaystyle n_i = 4\) state to the ground state (\(\displaystyle n_f = 1\)) is approximately \(\displaystyle 97.24\: \mathrm{nm}\).