Video Transcript
The diagram represents the relative densities of the electron occupation of relaxed, excited, and metastable states in the atoms of the active medium of a laser that is supplied with external energy to establish a population inversion. A very short time interval Δ𝑡 passes, corresponding to the shortest time within which any of the electrons could change in energy. Which of the following diagrams most correctly represents the energy states of the electrons after the time interval? No external energy is supplied to the active medium during the time interval.
Here then are five diagrams that make up our five answer options (A), (B), (C), (D), and (E). In each diagram, including the original one, we see three electron energy states. There is the ground state, where electrons tend to go and remain unless energy is transferred to them. Then above that, there is a state called the metastable state. Energetically, this is just below the third state called the excited state.
Our problem statement tells us that these states represent electron energy levels in atoms in the active medium of a laser. Our initial condition, we could call it, is to have one, two, three, four, five, six, seven electrons, where three of those electrons are in the excited state, one in the metastable state, and three are in the ground state.
Given this distribution of electrons, we’re told that a very short amount of time Δ𝑡 passes and that this time is the smallest amount of time during which an electron could change energy states. Our five answer options show us possibilities for what those changes might be. Notice that none of our five answer options look exactly like our original diagram.
To start narrowing down our answers, it’s helpful to think in terms of a lifetime of an electron in these three different states. That term just refers to how long an electron is likely to stay in that given energy state. We can start with the ground state. This, as we’ve seen, is where an electron is likely to remain unless external energy is provided to it. We’re explicitly told though that no energy is supplied to the system during our time interval Δ𝑡. That tells us that these three electrons, originally in the ground state, will not be able to transfer out of that state over the time interval Δ𝑡. We expect our final answer then to have at least three electrons in the ground state.
Next, let’s consider the lifetime of an electron that’s in the excited state. On average, an electron will spend very little time in that state. Over a time on the order of 10 to the negative eight seconds, that’s less than one one millionth of a second, electrons in the excited state are likely to go through a process called spontaneous decay. When this happens, such electrons drop down to a lower energy state. Because of the position of the metastable state, energetically right below the excited state, when the electrons in the excited state do decay spontaneously, they’ll likely decay down to the metastable state.
Like the excited state, the metastable state is also above ground level. That is, electrons in this state possess greater energy than electrons in the ground state. However, the name metastable state gives us a clue that this electron energy level is different compared to the excited state. An electron in the metastable state is capable of spontaneously decaying just like an electron in the excited state. However, the average lifetime of an electron in the metastable state is about 10,000 times longer than that of an electron in the excited state.
We would therefore expect that over the shortest possible interval of time during which an electron could transition to another energy level, that transition would not be likely to include a metastable state transition down to the ground state. Rather, it’s much more probable that that transition would be an excited state electron down to the metastable state.
By the way, this relatively long lifetime of an electron in the metastable state is what allows this three-energy-level system within an atom in the gain medium of a laser to achieve what’s called a population inversion, where more electrons are out of the ground state than are in it.
Knowing then that after a time interval of Δ𝑡 we expect our metastable state electron, right here, to still be in that state, let’s consider our answer choices. And note that in answer option (A), option (D), and option (E) that electron is not in the metastable state any longer but rather has decayed down to the ground state.
We’ve said though that this transition is highly unlikely over our time interval of Δ𝑡. It’s not impossible that this would happen. But looking to choose our best answer, we can safely eliminate options (A), (D), and (E) from consideration. Note that answer options (B) and (C) do have this electron still in the metastable state. The difference between these two answers though is that in answer option (B) all the excited state electrons over a time interval of Δ𝑡 have decayed down to the metastable state. In option (C) on the other hand, they’ve decayed from the excited state down to the ground state.
Energetically, the more likely of these two transitions is the one that requires the least change. It is easier, we could say, for an electron to decay from the excited state to the metastable state rather than all the way down to the ground state. Therefore, option (B) will be our final answer. This is the diagram that most correctly represents the population of the three different energy levels after a time interval Δ𝑡. Note that this is the only one of our answer options that also demonstrates to us a population inversion. This is the mechanism that allows for lasing.
