An electron in a hydrogen atom makes a transition from the n = 4 to the n = 3 energy state. Determine the energy (in eV) of the emitted photon.
A) 0.54
B) 0.66
C) 0.85
D) 1.51
E) 10.2

Question

Quizplus · Accepted Answer

The energy difference between two energy levels in a hydrogen atom can be calculated using the formula: $$E = -13.6\,eV \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$ where $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively. Plugging in the values for the transition from $n = 4$ to $n = 3$, we get: $$E = -13.6\,eV \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = -13.6\,eV \left( \frac{1}{9} - \frac{1}{16} \right)$$ $$E = -13.6\,eV \left( \frac{16 - 9}{144} \right) = -13.6\,eV \left( \frac{7}{144} \right) \approx -0.66\,eV$$ Since the energy is emitted, we take the absolute value, giving 0.66 eV.

An Electron in a Hydrogen Atom Makes a Transition from the N