The electron in a singly ionized helium atom,which consists of two protons,two neutrons and only one electron,makes a transition from the n = 2 state to the n = 1 state,emitting a single photon in the process.What is the energy of this photon?
A)3.40 eV
B)10.2 eV
C)13.6 eV
D)40.8 eV
E)None are correct.

Question

Quizplus · Accepted Answer

The energy levels of a hydrogen-like atom are given by the formula E_n = -13.6 Z^2 / n^2 eV, where Z is the atomic number. For helium (Z=2), the energy of the n=2 state is -13.6 * 4 / 4 = -13.6 eV, and the n=1 state is -13.6 * 4 / 1 = -54.4 eV. The energy of the photon emitted is the difference: 54.4 - 13.6 = 40.8 eV.

The Electron in a Singly Ionized Helium Atom,which Consists of Two