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)None are correct.

Question

Quizplus · Accepted Answer

The energy of the photon emitted during the transition can be calculated using the formula for the energy levels of a hydrogen-like atom, $E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2}$, where $Z$ is the atomic number (for helium, $Z=2$) and $n$ is the principal quantum number. The energy difference ($\Delta E$) between two levels is given by $\Delta E = E_2 - E_1$, where $E_2$ and $E_1$ are the energies of the final and initial states, respectively. For a transition from $n=2$ to $n=1$, the energy of the emitted photon for a singly ionized helium atom (He$^+$) is $\Delta E = -\frac{4 \cdot 13.6 \text{ eV}}{1^2} - \left(-\frac{4 \cdot 13.6 \text{ eV}}{2^2}\right) = -54.4 \text{ eV} + 13.6 \text{ eV} = -40.8 \text{ eV}$. The positive value of 40.8 eV (since we are interested in the magnitude of energy released) is not listed among the options, making all provided choices incorrect.

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