What is the energy of the light emitted when an electron falls from the n = 4 to the n = 2 orbit in the hydrogen atom?

Question

Quizplus · Accepted Answer

The energy of the light emitted when an electron falls from a higher energy level to a lower one in a hydrogen atom can be calculated using the formula derived from the Rydberg equation: $\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $ joules, where $n_f$ is the final energy level and $n_i$ is the initial energy level. Substituting $n_f = 2$ and $n_i = 4$ gives $\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4.09 \times 10^{-19}$ joules.

What Is the Energy of the Light Emitted When an Electron