Question

Figure 40-24 shows the energy levels of two types of atoms. Atoms A are in one tube, and atoms B are in another tube. The energies (relative to ground-state energy of zero) are indicated; the average lifetime of atoms in each level is also indicated. All the atoms are initially pumped to levels higher than the levels shown in the figure. The atoms then drop down through the levels, and many become “stuck” on certain levels, leading to population inversion and lasing. The light emitted by A illuminates B and can cause stimulated emission of B. What is the energy per photon of that stimulated emission of B?

Short answer

The energy per photon of that stimulated emission of B is 3eV.

Step-by-step solution

The given data

There are two atoms whose energies and average lifetime at each level are indicated.

Understanding the concept of energy of a stimulated emission:

Stimulated emission is a process by which an incoming photon of a certain frequency can interact with an electron atomic electron (or other exciting molecular structure), causing it to plummet.

For stimulated emission to take place, need a long-lived state above a short-lived state in both the atoms. In addition for the light emitted by A to cause stimulated emission of B, energy matching the transition is required. Now, using the figure, get the transition states in both cases. Now, for the case of energy per photon in only B, the energy difference is taken.

Calculation of the energy per photon of B:

The above conditions given in the concept are fulfilled for the transition from the 6.9 eV state (lifetime 3 ms) to 3.9 eV state $\left(3\mathrm{\mu s}\right)$in A and the transition from 10.8 eV state (lifetime 3 ms) to 7.8 eV state $\left(3\mathrm{\mu s}\right)$in B considering the given figure.

Thus, the energy per photon of the stimulated emission of B is given by:

$$\begin{gathered} \mathrm{E}=\left(10.8-7.8\right)\mathrm{eV} \\ =3\mathrm{eV} \end{gathered}$$

Hence, the value of the energy per photon for emission of B is 3 eV.