# Why is the photopeak at a higher energy than the Compton edge?

Why does the photoeffect deposit more energy than interactions via Compton scattering? Or the other way around: Why is the photopeak right (at a higher energy) than the Compton edge? Compton edge (Wikipedia)

I know that the interactions vary with the incident photon energy (from photoelectric effect to Compton effect to pair production). Therefore, I thought that Compton and pair production deposits more energy?!

A scintillator crystal (and photomultiplier tube) are designed to take a single high-energy gamma ray and turn it into a current pulse proportional to the photon's energy. On a microscopic level, this happens via a multi-step process:

1. The photon enters the scintillator crystal and strikes an electron.
2. The electron receives all of the energy of the initial gamma ray. This can happen via an analog of the photoelectric effect.
3. The electron then bounces around inside the scintillator, losing energy to photons (usually in the UV-visible range) with each interaction. By conservation of energy, these UV/vis photons have the same total energy as the initial gamma ray.
4. These UV/vis photons are then amplified by a photomultiplier tube, which uses a series of electrodes to turn a single photon into an electric pulse.

If everything goes perfectly, we get what's called the "photopeak" in the gamma-ray spectrum.

However, sometimes this process fails to capture all of the energy of the incoming gamma ray. This can happen in one of two main ways:

• Instead of the electron receiving all of the energy from the incoming gamma ray, the gamma ray can bounce off of it via Compton scattering. In this case, it transfers some, but not all, of its energy to the electron. (In other words, we have $e + \gamma \to e + \gamma$ instead of $e + \gamma \to e$.) If the outgoing photon then interacts with another electron as above, then the books balance out, and the current pulse will be the same as though the initial gamma ray had just deposited all its energy into one electron. However, sometimes the outgoing photon then escapes the scintillator crystal, taking some of the original photon's energy with it. Following the logic through, we see that current pulses that result from Compton scattering will be smaller than those in the photopeak.

• If an incoming photon has energy greater than $2 m_e c^2$ (where $m_e$ is the electron mass), it can undergo pair production: $\gamma \to \gamma + e^+ + e^-$. The outgoing gamma ray can then undergo the same steps 1–4 above, while the electron-positron pair can undergo steps 2–4. If all three of these things happen, then the books balance out, and the current pulse from the photomultiplier will be the same as if the photopeak. But if one of the electron-positron pair escapes the scintillator without any further interaction, the total energy deposited in the scintillator crystal will be less than the energy of the original photon by the amount $m_e c^2$; and if both escape, then the energy deposited in the scintillator will be less than the original gamma-ray energy by $2m_e c^2$. You'll therefore get peaks at $E_\gamma - m_e c^2$ and $E_\gamma - 2 m_e c^2$; these are the "single escape" and "double escape" peaks respectively.

In both cases, the energy deposited in the scintillator crystal is less than the energy of the original particle, since one or more particles have escaped the crystal, taking some energy with them.

[A technical aside: the interactions $e^- + \gamma \to e^-$ and $\gamma \to e^- + e^+ + \gamma$ have to occur in the presence of another particle to make the books balance out for momentum conservation. Usually this other particle is a nucleus, and it's not too hard to show that since a nucleus is much heavier than either an electron or photon, it only ends up with a small fraction of the energy of the original photon.]

In short, the photopeak is formed in the case of complete absorption of the gamma ray's energy in the scintillator or detector, while the Compton edge is the maximum amount of energy absorbed by the scintillator in the process of Compton scattering, where there is an incomplete absorption of the gamma ray's energy as it scatters off of the detector.

Let's call the energy of the incident gamma ray $E.$ The photopeak occurs when the amount of energy transferred to the scintillator or detector, $E_{T},$ is equal the the energy of the incident gamma ray, $$E_{T}=E.\tag{1}\label{1}$$ In a Compton scattering process, the amount of energy exchanged by the gamma ray and an electron in a material depends on the angle that the gamma ray is scattered through, and is given by the formula $$\frac{1}{E'}-\frac{1}{E}=\frac{1}{m_{e}c^{2}}(1-cos\theta),$$ where $E$ is still the energy of the incident gamma ray, $E'$ is the energy of the scattered gamma ray, $m_{e}$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the angle that the gamma ray is scattered through. This can also be written $$E'=\frac{E}{1+\frac{E}{m_{e}c^{2}}(1-cos\theta)}.$$ The amount of energy exchanged is $$E_{T}=E-E'.\tag{2}\label{2}$$ We see that the amount of energy exchanged is maximized when $\theta$ approaches $180$ degrees, at which point the energy transferred is $$E_{T}=E(1-\frac{1}{1+\frac{2E}{m_{e}c^{2}}}).$$ We call this maximum amount of energy absorbed during Compton scattering the Compton edge.

Since the energy of the scattered gamma ray must be positive, we see that the energy transferred to the detector through Compton scattering $\eqref{2},$ which has its maximum at the Compton edge, is necessarily smaller than the energy transferred when all of the gamma ray's energy is absorbed by the detector $\eqref{1}.$

• Thanks a lot for the explanation! Just for a final understanding: Will the photopeak always be higher than the Compton peak independent from the photon's energy?
– Ben
Jan 9, 2018 at 14:14
• Yes, the photopeak will always be at a higher energy than the Compton edge, regardless of the photon's energy. This is because the energy of the Compton edge is smaller than the energy of the incoming photon (due to the photon scattering off of the electron with an incomplete transfer of its energy), while the energy of the photopeak is equal to the energy of the incoming photon (the photon completely transfers its energy to the electron). Jan 9, 2018 at 16:02

You said:

I know that the interactions vary with the incident photon energy (from photoelectric effect to Compton effect to pair production).

What you need to understand is that the probabilities of the interactions vary with the incident photon energy. As the energy of the incident photon, $E$, increases, the probability of a photoelectric effect decreases and Compton scattering increases. PE dominates below 500 keV and depends on the material according to $Z^5$, where $Z$ is the atomic number.

Between 500 keV and 1100 keV, Compton scattering dominates the interaction probability with a material dependence of $Z$. There are a few other effects such as Raleigh scattering, electron resonance and Thomson scattering, but these are fairly minor in the the overall detector process.

As the excellent answers by Michael Seifert and Jared Popowski state, a single Compton event will always deposit less energy in the detector than the incident photon has. So, while PE does dominate at low energies, it is possible for a Compton event to occur, and this will produce a small Compton-edge in the spectrum even for low energy photons. Photons with $E>500$ keV are expected to have substantial Compton-edges, again associated with a full-energy peak, but lower than $E$.

Occasionally, the secondary photon will interact in the crystal by a subsequent photoelectric event within the time resolution of the detector system and the full energy $E$ might be "counted," but this isn't certain. So for mid-range gamma-energy photons, the Compton events will always show up at a lower energy than the associated full-energy peak.

OTOH, the probability of a large Compton edge increases with $E$, so the Peak:Compton ratio usually decreases with increasing $E$, as illustrated in the graphic you posted.

When $E>2m_e c^2$, the probability of pair production increases and probability of Compton scattering will decrease. Pair production has a broad maximum between 10-30 MeV. And as pointed out by the other answers, a single escape or double-escape maximum could occur due to the escape of the annihilation photons which follow pair-production.

• When the photopeak has a peak at higher energies, this means ultimately the following: Though the photopeak has a lower probability for high energetic photons, it will occur and when it does, the photons transfer all their energy to the electron. Is this true? If so, I wouldn't expect the counts of the photopeak at a similar level as the compton continuum. Otherwise, when the entire compton regime is piled up, the counts are much more than the photopeak.
– Ben
Aug 21, 2019 at 19:09
• @Ben Don't confuse the height of the photopeak with the area. Yes, for higher energy primary photons, the area under the compton energies will most likely be larger than the area under the primary photopeak. Aug 21, 2019 at 19:39