Question

Figure 39-26 indicates the lowest energy levels (in electronvolts) for five situations in which an electron is trapped in a one-dimensional infinite potential well. In wells B, C, D, and E, the electron is in the ground state. We shall excite the electron in well A to the fourth excited state (at 25 eV). The electron can then de-excite to the ground state by emitting one or more photons, corresponding to one long jump or several short jumps. Which photon emission energies of this de-excitation match a photon absorption energy (from the ground state) of the other four electrons? Give then values.

Short answer

The photon energy of the given de-excitation matches the absorption energy of the electron in n = 1 of D which jumps to n = 3 of D.

Step-by-step solution

Given data

The energy carried away by a photon when the electron in A jumps down from the fourth excited state to the first excited state is,

$E_{A14}=25\text{eV}$.

Energy levels in B are as follow.

$\begin{array}{rcl}{E}_{B1}& =& 2\text{eV}\\ E_{B2}& =& 8\text{eV}\\ E_{B3}& =& 19\text{eV}\end{array}$

Energy levels in C are given below.

$\begin{array}{rcl}{E}_{C1}& =& 4\text{eV}\\ E_{C2}& =& 16\text{eV}\end{array}$

Energy levels in D are as follow.

$\begin{array}{rcl}{E}_{D1}& =& 3\text{eV}\\ E_{D2}& =& 12\text{eV}\\ E_{D3}& =& 27\text{eV}\end{array}$

Energy levels in E are,

$\begin{array}{rcl}{E}_{E1}& =& 5\text{eV}\\ E_{E2}& =& 20\text{eV}\end{array}$

Photon absorption condition

A photon can be absorbed by an electron in a particular energy state if the energy carried by the photon is exactly equal to the difference in energy of a higher energy state to the one the electron is in. The electron will then excite to the higher level.

Determining the energy level of the electron which absorbed the emitted photon

By inspection it can be checked that the energy of the emitted photon $E_{A14}$ is exactly equal to the difference in energy between the third excited state and the ground state of D.

$\begin{array}{rcl}{E}_{D3}-E_{D1}& =& 27\text{eV -}3\text{eV}\\ & =& 24\text{eV}\\ & =& E_{A14}\end{array}$

Hence, the electron in the ground state (n = 1) of D absorbed the photon and jumped to n = 3 of D.