An excited hydrogen atom emits a photon with an energy of 0.661 eV.What was the initial state of the hydrogen atom before emitting the photon?
A)n = 1
B)n = 2
C)n = 3
D)n = 4
E)n = 5

Question

Quizplus · Accepted Answer

The energy difference between the energy levels in a hydrogen atom can be calculated using the formula for the energy of a photon emitted or absorbed when an electron transitions between two energy levels: $$E = -13.6\,eV \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ where $E$ is the energy of the photon, $n_i$ is the initial energy level, $n_f$ is the final energy level, and $13.6\,eV$ is the ionization energy of hydrogen. Given that the energy of the photon is $0.661\,eV$, and assuming the final state is $n = 2$ (since transitions to $n = 1$ would result in energies higher than $0.661\,eV$), we can solve for $n_i$. Rearranging the formula and solving for $n_i$ gives a value that corresponds to $n_i = 4$, indicating the initial state of the hydrogen atom was $n = 4$.

An Excited Hydrogen Atom Emits a Photon with an Energy