The energy of a photon emitted from the n = 3 to the n = 1 state from a singly ionized helium atom is
A) 37.4 eV.
B) 48.8 eV.
C) 54.4 eV.
D) 60.5 eV.

Question

Quizplus · Accepted Answer

The energy of a photon emitted during a transition between two energy levels in an atom can be calculated using the formula derived from the Rydberg formula for hydrogen-like atoms: $E = R_H Z^2 \left( \frac{1}{n_{\text{lower}}^2} - \frac{1}{n_{\text{upper}}^2} \right)$, where $R_H$ is the Rydberg constant (approximately 13.6 eV), $Z$ is the atomic number (for helium, $Z = 2$), $n_{\text{lower}}$ and $n_{\text{upper}}$ are the principal quantum numbers of the lower and upper energy levels, respectively. For a transition from $n = 3$ to $n = 1$, the formula becomes $E = 13.6 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$, which calculates to approximately 48.8 eV.

The Energy of a Photon Emitted from the N =