Lesson Explainer: Spontaneous and Stimulated Emission Physics

In this explainer, we will learn how to describe the processes of spontaneous emission of light and stimulated emission of light.

Let’s begin by recalling that electrons in atoms can occupy different energy levels. Given the correct conditions, these electrons can also move between these different energy levels. There are a number of ways in which this can occur.

Spontaneous emission and stimulated emission are both processes by which photons are created as electrons move from one energy level to another. In this explainer, we will be looking at how these two processes work. We will also get to see what the two processes have in common as well as what is different in each case.

Let’s consider a simplified system. Specifically, we will imagine an atom that contains just one electron—in other words, we are considering the hydrogen atom. We will further imagine that there are only two possible energy levels for that electron to occupy. We note that any real atom (even hydrogen) has many more possible energy levels. However, we will see that such a two-level system is sufficient to illustrate all of the concepts presented in this explainer.

We can draw the energy levels of such a two-level atom as follows:

Here, we have drawn a circle on the lower energy level to represent the electron. The only stable state of an atom is that in which all of its electrons are in the lowest energy level that is available to them. For our two-level atom with just a single electron, this means that, in the absence of any external influence, we would expect to find the electron in the lower energy level.

The lowest energy level in an atom is commonly referred to as the ground state, so we have labeled the energy of the lower level as 𝐸, where the 𝑔 stands for ground. We note that the ground state may also be referred to as the relaxed state—these two terms are interchangeable. We have labeled the energy of the upper level as 𝐸, where the 𝑒 stands for excited. Any energy level that is not the ground state is said to be an excited state.

We know that this electron will sit in the ground state if the atom is undisturbed. However, we have also said that electrons can move between energy levels, given the right conditions. So, let’s see how this electron can move from the ground state to the excited one.

Since the electron is getting raised from a lower energy state to a higher energy one, perhaps it is not surprising that the electron will need to have energy transferred to it. What might be less obvious is that the amount of energy needed is actually a very precise quantity. Too much energy, or too little, and the electron will stay put.

It turns out that to move from a lower energy level with energy 𝐸 to a higher energy level with energy 𝐸, the electron needs to be given an amount of energy that is exactly equal to 𝐸−𝐸.

One way for energy to be transferred to the electron is for it to absorb a photon. A photon is an individual packet of light energy. The energy of a photon, which we will call 𝐸p, is given by 𝐸=ℎ𝑓.p

Here, ℎ is a constant known as Plank’s constant (ℎ≈6.626×10⋅/mkgs) and 𝑓 is the frequency of the light.

In order for a photon to excite an electron from a state with energy 𝐸 to a state with energy 𝐸, the energy of the photon must be equal to the energy difference between these states. Mathematically, we can write this as 𝐸=𝐸−𝐸.p

Or, equivalently, we can use our expression for the energy of a photon to write this as ℎ𝑓=𝐸−𝐸.

What this expression is saying is that only a photon of an exact specific frequency is able to be absorbed by the electron, and that frequency is determined by the difference in the energies of the levels that the electron jumps between.

Now, we said that we were considering a simplified model of an atom, specifically one in which the electron only has two levels that it can occupy. Let’s now briefly look at what would happen if we considered an atom with multiple possible energy levels.

We will imagine an electron in the ground state of this more complex atom. We know this electron can only transition from the ground state to one of the higher energy states by absorbing a photon with exactly the same energy as the difference in energy between that higher energy state and the ground state.

So if an incident photon of light has an energy of 𝐸=𝐸−𝐸p, the electron may absorb that photon and move from the ground state to the excited state with energy 𝐸. In this case, all of the other energy levels make no difference to what is going on because the photon energy does not match up to any of the other energy gaps. So, in practice, it is just like the simplified situation with just two levels that we considered in the first place. In this case, those levels are the ground state and the state with energy 𝐸.

Similarly, if the incident photon actually had an energy of 𝐸=𝐸−𝐸p, the system would reduce to a two-level system consisting of the ground state and the excited state with energy 𝐸.

So we can see that because of the limited way in which electrons and photons can interact with each other, our simplified picture with just two levels is indeed sufficient to describe what is going on.

For the rest of this explainer, we will go back to considering just two levels.

With this two-level case, we will imagine that the electron has absorbed a photon of just the right energy to excite it from the lower level to the upper level. We can draw this absorption process as follows:

We can recall that the energy of the photon must have been equal to 𝐸−𝐸.

Now that the electron has absorbed a photon, it is placed in an excited state with a higher energy than the ground state:

We said earlier that the only stable state of an atom is that in which all its electrons are in the lowest possible energy levels available to them. In this current situation, there is clearly a lower energy level available than the excited state that the electron is currently in.

It turns out that if we wait for enough time, the electron will decay back down to this lower level. The electron does not need any external stimulus for this to happen. It just happens as a result of the fact that this excited state is unstable. So we say that this decay occurs spontaneously, and in fact the process we have just described is that of spontaneous decay.

The typical time the electron stays in the excited state before decaying is very short. It varies depending on the particular atom and energy level but is on the order of 10 s or, equivalently, 10 ns.

Even if we know all the details of the atom and energy levels in question, this length of time is still just an average. The decay process really is a spontaneous one, and for a particular electron, we cannot tell exactly when this decay is going to happen. All we know is that, on average, we can expect it to occur roughly 10 s after the electron gets excited up to this excited state. We can think of this length of time as a sort of natural lifetime of the upper energy level: it gives us an idea of how long we expect the electron to remain in this upper level in the absence of any external influence.

When the electron was raised from the lower energy level to the upper energy level, it had to absorb a photon in order to do this. That photon had an energy equal to the difference in energies of the two levels, that is, an energy of 𝐸−𝐸.

When the electron decays back from the upper level to the lower level, the reverse process happens. This is illustrated in the diagram below.

As it decays, the energy of the electron decreases by an amount 𝐸−𝐸. The electron releases this energy by emitting a photon. The energy of this photon, 𝐸p, is equal to the energy lost by the electron. In other words, we know that for the spontaneously emitted photon, 𝐸=𝐸−𝐸.p

We can notice that the energy of the spontaneously emitted photon is equal to the energy of the photon that was originally absorbed by the electron to raise it to the excited state.

We can also recall that, for a photon, we know 𝐸=ℎ𝑓p. This means that the frequency of the emitted photon is the same as the frequency of the absorbed photon.

The spontaneous nature of this process relates not just to when the electron decays, but also to the direction and the phase of the emitted photon. Specifically, these two quantities are random. The photon may be emitted into any direction and with any phase.

Therefore, in general, the direction and the phase of the emitted photon will not be the same as those of the photon that was originally absorbed to excite the electron.

Let’s now look at an example problem.

Example 1: Spontaneous Emission in a Hydrogen Atom

At an instant 𝑡, a hydrogen atom has just absorbed a photon, increasing the energy of its electron to 𝐸. A time interval Δ𝑡≈1µs then elapses, during which no other photons interact with the atom.

  1. How does 𝐸, the energy of the electron at a time Δ𝑡 after 𝑡, compare to 𝐸?
  2. Will any photons have been emitted at a time Δ𝑡 after 𝑡?
  3. Which of the following is the term used for the state of the electron at a time Δ𝑡 after 𝑡?
    1. Spontaneous
    2. Excited
    3. Relaxed
    4. Instantaneous
    5. Stimulated

Answer

Part 1

Let’s begin by considering the scenario that the question describes.

We have a hydrogen atom; that is an atom that has one single electron. We are told that at a particular instant in time, 𝑡, this electron absorbs a photon. We are also told that, by absorbing this photon, the electron increases its energy to a value of 𝐸. In other words, at a time 𝑡, the electron is raised to an excited state.

We can sketch this as follows:

This first part of the question is asking to consider what happens to the electron’s energy following this. Specifically, we are asked to consider a time interval Δ𝑡≈1µs after the time 𝑡 at which the photon was absorbed. We are told in the question that, during this time interval, there are no other photon interactions with this electron.

We can recall that when an electron is in an excited state, it will eventually decay back to the ground state. This ground state, or relaxed state, is lower in energy than the excited state. We know that the electron will do this even in the absence of any other photons interacting with it; the process happens spontaneously.

We can sketch this process of spontaneous emission as follows:

As shown in the sketch, as the electron decays and loses energy, it emits a photon.

Now, this process really is spontaneous, and we cannot predict exactly when it will occur. However, we can recall that a reasonable average lifetime for an electron to stay in an excited state before decaying back down is 10 s.

We are asked to consider a time interval of Δ𝑡≈1µs. We know that 1=10µss. This means that we know Δ𝑡 is 100 times longer than the average time for an excited electron to decay. Therefore, we can be reasonably certain that, after a time Δ𝑡, the electron will already have decayed via spontaneous emission.

With that in mind, we are now in a position to answer the question.

We know that the electron will almost certainly have decayed back down to the lower energy level before the time interval Δ𝑡 is over. Therefore, we can say that the energy of the electron will have decreased. In other words, if the energy of the electron at time 𝑡 (when it is in the excited state) is 𝐸 and the energy of the electron a time Δ𝑡 later is 𝐸, we can say that 𝐸<𝐸.

Part 2

The second part of the question asks whether any photons will have been emitted during the interval Δ𝑡.

We have already deduced that spontaneous emission occurs during this interval. We also know that, during spontaneous emission, the electron emits a photon as it decays. Therefore, our answer to this part of the question is yes, a photon will have been emitted.

Part 3

This final part of the question asks us to identify the term used for the state of the electron after the interval Δ𝑡 has passed.

When the electron absorbs the photon, it gets raised to what we call an excited state. It then decays back down spontaneously.

Now, “excited” and “spontaneous” are both possible choices for our answer. However, we know that the electron decays out of this excited state before the interval Δ𝑡 is over, so this cannot be correct. We also know that “spontaneous” refers to the decay process itself, rather than the final state of the electron, so this answer cannot be correct either.

We can recall that this lower energy state that the electron decays to is referred to either as the ground state or as the relaxed state. Here, one of our possible options—option C—is “relaxed.” So we know that our answer to this final part of the question is that the state of the electron after the time interval Δ𝑡 is relaxed.

In the example above, it was specified that “no other photons interact with the atom” while the electron is in the excited state. It turns out that this condition was important in being able to answer the question.

If we do have other photons that may interact with the atom, then there is another possibility besides spontaneous emission.

Let’s imagine that we have an atom in which an electron has been raised to an excited state. We know that, in the absence of any external influence, after a time of the order of 10 s, we can expect the electron to spontaneously decay back to its ground state.

However, before this decay takes place, it is possible for this excited electron to interact with a photon. Just as we have seen before, this can only happen if the energy of the photon is equal to the difference in energy between the two levels. In other words, the photon has some energy 𝐸=ℎ𝑓p such that 𝐸=𝐸−𝐸p.

In this case, the interaction acts to send the electron back down to its ground state. The process is illustrated in the diagram below:

As shown in the diagram, the electron emits a photon in the process. As in the case of spontaneous emission, the emitted photon has an energy equal to the energy lost by the electron; that is, an energy 𝐸=𝐸−𝐸p.

We should also notice that the original incident photon is not absorbed; it simply interacts with the electron and continues on.

We call this process stimulated emission. The incident photon stimulated the electron, causing it to move from the excited state to the ground state and emit a photon in the process.

We have already seen that the emitted photon must have the same energy as the original photon. Since the energy of a photon is given by 𝐸=ℎ𝑓p, this means that the two photons must also have the same frequency as each other.

It actually turns out that the two photons share some other similarities. They also have the same direction as each other and the same phase as each other. This is in contrast to the case of spontaneous emission, in which the direction and phase of the emitted photon were completely random.

Now let’s look at a couple more example problems involving stimulated emission.

Example 2: Properties of a Photon Produced by Stimulated Emission

The diagram shows the ground state and the excited state of an electron in an atom that is in the path of a stream of photons. Which of the following diagrams most correctly represents the stimulated emission of a photon from the atom due to a change in the energy level of the electron?

Answer

The question tells us that we have an atom in the path of a stream of photons. We are asked to identify which of four possible diagrams most correctly represents the stimulated emission of a photon from this atom.

Let’s begin by recalling what happens in the process of stimulated emission.

For stimulated emission to occur, we know that the electron must be in an excited state. An incident photon interacts with this excited electron and stimulates the emission of another photon. The incident photon is not absorbed by the electron; it continues on following the interaction.

The stimulated emission process causes the electron to fall back from the excited state to the ground state. In order to stimulate this process, the energy of the incident photon must equal the energy difference between the excited state and the ground state.

The emitted photon must have an energy equal to that lost by the electron, that is, the energy difference between the excited state and the ground state.

This means that the incident photon and the emitted photon have the same energy as each other.

The energy, 𝐸p, and frequency, 𝑓, of a photon are related by 𝐸=ℎ𝑓p, where ℎ is Plank’s constant. Since the incident and emitted photon have the same energy as each other, they must also have the same frequency.

We can further recall that the direction and phase of a photon produced by stimulated emission are equal to those of the incident photon that caused the process.

Overall then, we know that after the stimulated emission process there will be two photons: the original incident photon and the emitted one. These photons will have the same energy, frequency, direction, and phase as each other. The process will cause the electron to fall from the excited state to the ground state.

We could therefore illustrate the process of stimulated emission as follows:

Let’s compare this with the diagrams we are given in the question.

Looking at diagram A, we see that we can immediately rule this one out because there is only one photon. We know that we have the emitted photon plus the original photon that does not get absorbed by the electron, and so there should be two photons following the stimulated emission process.

If we now look at diagram B, we see that we do have two photons. However, these photons are clearly different frequencies from each other: one complete cycle of the oscillation of waveform representing the lower photon occurs over a shorter spatial distance than that of the upper photon. So diagram B cannot be correct.

In diagram C, we have two photons with the same frequency as each other, but this time they are traveling in opposite directions. The upper photon is traveling to the right, while the lower one is traveling to the left. We know that the emitted and incident photons must have the same direction as each other, so diagram C cannot be correct.

Finally, looking at diagram D, we see that we have two photons with the same frequency, direction, and phase as each other. This diagram correctly shows the electron falling from the excited state to the ground state. It also shows the incident photon not getting absorbed and the incident and emitted photons having the correct properties.

So our answer to the question is that the process of stimulated emission of a photon from the atom is most correctly represented by diagram D.

Example 3: Stimulated Emission in a Hydrogen Atom

At an instant 𝑡, a photon 𝛾 is absorbed by the electron in a hydrogen atom. The electron is in a relaxed state before 𝑡. At an instant 𝑡, a photon 𝛾 that has energy equal to that of 𝛾 interacts with the electron. The interval between 𝑡 and 𝑡≈0.1ns. A second time interval then elapses, during which a photon 𝛾 is emitted. No photons are absorbed or emitted other than 𝛾 and 𝛾, and no photons other than 𝛾 and 𝛾 interact with the electron.

  1. What is the ratio of the energy of 𝛾 to the energy of 𝛾?
  2. If the direction of travel of 𝛾 just before 𝑡 is defined as being along the 𝑥-axis in the positive direction, which of the following could be the direction of travel of 𝛾 when it is emitted?
    1. Along the 𝑥-axis in the negative direction
    2. Along the 𝑥-axis in the positive direction
    3. Perpendicular to the 𝑥-axis
    4. In a direction between the direction of the 𝑥-axis and a direction perpendicular to the 𝑥-axis
  3. Which of the following is the full range of Φ, the possible phase difference between 𝛾 at 𝑡 and 𝛾 when it is emitted?
    1. 0≤Φ≤𝜋2radrad
    2. 0≤Φ≤2𝜋radrad
    3. 0≤Φ≤𝜋4radrad
    4. 0≤Φ≤𝜋radrad
    5. 0≤Φ≤0radrad
  4. Which of the following is the term used for the type of photon emission that produces 𝛾?
    1. Spontaneous
    2. Relaxed
    3. Excited
    4. Stimulated
    5. Instantaneous

Answer

Part 1

Let’s begin by considering the scenario described in the question so that we may see exactly what is going on.

We have a hydrogen atom. We are told that at some instant in time, 𝑡, a photon 𝛾 is absorbed by the electron in this atom. The electron was in the relaxed state, or ground state, before this occurs.

We know that when the electron absorbs the photon, the electron will be raised to an excited state. This process is illustrated visually in the diagram below:

Let’s say that the ground state has energy 𝐸 and the excited state has energy 𝐸. Then, we know that the energy of photon 𝛾, which we will label 𝐸, must be 𝐸=𝐸−𝐸.

We are told that a second photon 𝛾 interacts with the electron at a time 𝑡. We are also told that the energy of this photon, 𝛾, is the same as the energy of photon 𝛾. If we call the energy of photon 𝛾   𝐸, then we must have that 𝐸=𝐸=𝐸−𝐸. In other words, the energy of this second photon, 𝛾, is also equal to the difference between the two energy levels of the electron.

The fact that photon 𝛾 has this value of energy means that it can also interact with the electron and cause it to move between the two levels. Precisely how this occurs depends on the state of the electron at time 𝑡 that 𝛾 is incident; so let’s work out what that state is.

We know that photon 𝛾 interacts with the electron at a time 𝑡, and we are told that the interval between 𝑡 and 𝑡 is approximately 0.1 ns.

At time 𝑡, the electron was raised to an excited state. We can recall that, if left to its own devices, this electron will eventually return to the ground or relaxed state via spontaneous emission. We can further recall that this happens on a timescale of approximately 10 s, which we may also write as 10 ns.

Comparing this timescale for spontaneous decay with our time interval of 0.1 ns, we see that the time interval between 𝑡 and 𝑡 is 100 times shorter than the expected time before which the electron will decay spontaneously.

This means that the second photon, 𝛾, almost certainly interacts with the electron before it decays, in other words, while it is still in the excited state.

So, we may sketch the situation at time 𝑡 as follows:

We can recall that when a photon interacts with an electron in the excited state, it can cause the electron to fall back to the ground state via stimulated emission. In order for this to occur, the photon must have an energy equal to that of the difference in energies between the excited state and the ground state.

We have already seen that photon 𝛾 has precisely this energy, and so it can cause stimulated emission.

During the process of stimulated emission, we can recall that the electron emits a photon. This photon has an energy equal to that lost by the electron, that is, the difference between the energies of the excited state and the ground state.

In the question, we are told that a photon 𝛾 is emitted following the interaction between photon 𝛾 and the electron.

We can therefore deduce that this photon, 𝛾, is precisely the photon emitted by the electron as it undergoes stimulated emission from the excited state to the ground state. We also know that the energy of photon 𝛾 must be given by 𝐸=𝐸−𝐸.

This process is illustrated in the diagram below:

This first part of the question asks for the ratio of the energy of 𝛾 to the energy of 𝛾. We have labeled these energies as 𝐸 and 𝐸 respectively.

We were told that 𝛾 and 𝛾 have the same energy, and we have seen that this is equal to 𝐸−𝐸. We have also seen that 𝛾, which is emitted via stimulated emission, also has energy 𝐸−𝐸.

Since 𝐸=𝐸−𝐸 and also 𝐸=𝐸−𝐸, the ratio of the two energies is 𝐸𝐸=𝐸−𝐸𝐸−𝐸=1.

Part 2

This second part of the question asks what direction photon 𝛾 must travel in, given that 𝛾 was traveling along the 𝑥-axis in the positive direction immediately before 𝑡.

We know that 𝛾 is the photon that interacts with the excited electron, causing the stimulated emission of photon 𝛾.

We can recall that a photon emitted via stimulated emission must have the same direction as the original photon that caused the emission.

Therefore, we know that photon 𝛾 must also travel along the 𝑥-axis in the positive direction.

Part 3

The third part of the question asks for the range of possible phase differences between 𝛾 at 𝑡 and 𝛾 when it is emitted.

We can recall that the phase of a photon emitted via stimulated emission must be the same as the phase of the photon that caused this emission. In other words, the phase of photon 𝛾 at the instant it is emitted must be equal to the phase of photon 𝛾 at instant 𝑡 at which it interacts with the excited electron.

This means that the phase difference, Φ, between these two photons must be precisely 0 rad. So, our answer to this part of the question is that 0≤Φ≤0radrad.

Part 4

This final part of the question asks us to identify the term used for the type of photon emission producing photon 𝛾.

We have explained already how photon 𝛾 stimulates the electron to fall from the excited state to the relaxed state via stimulated emission and that it is this process that produces photon 𝛾.

So our answer to this final part of the question is that the type of photon emission that produces 𝛾 is stimulated.

Key Points

  • Electrons in atoms can exist in different energy levels. The lowest energy level is referred to as the ground state or relaxed state. Any higher levels are referred to as excited states. Electrons will remain in the lowest available energy level in the absence of external influence.
  • An electron may be excited out of the ground state to a higher energy level by absorbing a photon. In order for this to occur, the energy of the photon must be equal to the difference in energies between the two levels. If the energy of the lower level is 𝐸 and the energy of the higher level is 𝐸, then the energy of the photon must be 𝐸=𝐸−𝐸p.
  • Spontaneous emission is a process by which an electron in an excited state may decay back to a lower energy state. The electron emits a photon in the process. If the higher energy state has energy 𝐸 and the lower energy state has energy 𝐸, then the emitted photon has energy 𝐸=𝐸−𝐸p. The direction and phase of the photon are random. The typical timescale for spontaneous emission is 10 s.
  • Stimulated emission is a process in which an excited electron may interact with a photon, which causes the electron to move to a lower energy state and emit a photon. For this process to occur, the interacting photon must have an energy 𝐸=𝐸−𝐸p. The emitted photon will have this same energy and also have the same frequency, phase, and direction as the interacting photon.

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