Dear Prof. Ghal-Eh,

we can easily get around the range argument by assuming that the electrons or HCPs  are completely stopped in the scintillator. Then let us assume that the incident particles carry identical amounts of energy into the scintillator. Since they are eventually completely stopped they deposit identical amounts of energy into the scintillator. Fluorescence requires that electrons in the scintillating material (in your example PPO) get excited. For each excitation you need at least the excitation energy. On the average there is a certain amount of energy loss of the incident particle necessary to produce an excitation and a resulting photon. If this were all that can happen you just divide the kinetic energy of the incident particle by that average energy loss and you get the average number of photons per incident particle: identical numbers  whether it is an electron or a heavy charged particle.

Now, as I wrote,only a small fraction of the kinetic energy lost by particles in a scintillator is converted to photon production. Most of the energy lost is dissipated without radiation emission. So, electron excitation is not all that can happen. There are also collisions with kinetic energy transfer to atoms/molecules which produces heat.

The dissipation of energy by non-radiative processes  is much more likely for HCP than for electrons. HCP spend most of their energy to heat the material (in your example all three components of the liquid scintillator) and little is left for electronic excitation. Electrons can transfer energy into electronic exciation much more efficiently.

The physical picture behind this is: by the incident particle kinetic energy has to be transferred to a bound electron in the scintillating material to finally produce a photon. An incident electron can ultimately transfer all its energy to another electron in one collision (colliding particles of equal mass in a head-on collision). A heavy particle  can only transfer a fraction of its kinetic energy to an electron just because of kinematic reasons. In a head-on collision of two particles with masses m1 and m2 the (maximum) energy transfer is 4m1 m2 Ekin/ (m1+m2)^2. For incident electrons we have m1=m2=me and the full kinetic energy Ekin can be transferred. If we assume an incident alpha particle the maximum energy that can be transferred to an electron in one collision is approximately Ekin/2000. Hence much of the energy loss of HCP goes into collisions with similarly heavy constituents in the scintillator and these are the atoms that get kinetic energy which means the scintillator is heated. All these arguments are particularly valid at low kinetic energies where nuclear stopping of HCP prevails.

The same arguemnts are basically responsible for the size of atomic excitation cross sections. At kinetic energies of approximately two to  three times the threshold energy electrons have their maximum excitation cross section. At these (low) energies the excitation cross sections  of HCP can almost be negelcted compared to those of the electrons (smaller by many orders of magnitude).

I hope this explanation is sufficient to answer your original question.

Only a small fraction of the kinetic energy lost by particles in a scintillator is converted to photon production. Most of the energy lost is dissipated without radiation emission. Heavy charged particles are much more likely to excite lattice vibrations or heat in the scintillator material than the light electrons. Therefore, a higher fraction of electron kinetic energy can be transferred to electronic excitations and subsequent fluorescence.

Dear Dr. Müller

Thanks for your answer. 

It is known that liquid scintillators are basically made of three components: (1) organic scintillator compounds (e.g., PPO), (2) wavelength-shifters (e.g., POPOP) and (3) solvents (e.g., xylene). Only does the excitation of first component (which are spread over the detector volume) by incident radiation (electrons or heavy charged particles (HCPs)) lead to the production of scintillation light.

I think the reason why electrons produce more lights than HCPs is that since electrons have longer ranges, they can interact with more molecules of component (1) type. HCPs lose most of their energies within a very short pathlengths as a result of interactions mostly with components (2) and (3).

Any comment on this answer (which is regardless of lattice vibration or heat!)?      

Dear Prof. Ghal-Eh,

we can easily get around the range argument by assuming that the electrons or HCPs  are completely stopped in the scintillator. Then let us assume that the incident particles carry identical amounts of energy into the scintillator. Since they are eventually completely stopped they deposit identical amounts of energy into the scintillator. Fluorescence requires that electrons in the scintillating material (in your example PPO) get excited. For each excitation you need at least the excitation energy. On the average there is a certain amount of energy loss of the incident particle necessary to produce an excitation and a resulting photon. If this were all that can happen you just divide the kinetic energy of the incident particle by that average energy loss and you get the average number of photons per incident particle: identical numbers  whether it is an electron or a heavy charged particle.

Now, as I wrote,only a small fraction of the kinetic energy lost by particles in a scintillator is converted to photon production. Most of the energy lost is dissipated without radiation emission. So, electron excitation is not all that can happen. There are also collisions with kinetic energy transfer to atoms/molecules which produces heat.

The dissipation of energy by non-radiative processes  is much more likely for HCP than for electrons. HCP spend most of their energy to heat the material (in your example all three components of the liquid scintillator) and little is left for electronic excitation. Electrons can transfer energy into electronic exciation much more efficiently.

The physical picture behind this is: by the incident particle kinetic energy has to be transferred to a bound electron in the scintillating material to finally produce a photon. An incident electron can ultimately transfer all its energy to another electron in one collision (colliding particles of equal mass in a head-on collision). A heavy particle  can only transfer a fraction of its kinetic energy to an electron just because of kinematic reasons. In a head-on collision of two particles with masses m1 and m2 the (maximum) energy transfer is 4m1 m2 Ekin/ (m1+m2)^2. For incident electrons we have m1=m2=me and the full kinetic energy Ekin can be transferred. If we assume an incident alpha particle the maximum energy that can be transferred to an electron in one collision is approximately Ekin/2000. Hence much of the energy loss of HCP goes into collisions with similarly heavy constituents in the scintillator and these are the atoms that get kinetic energy which means the scintillator is heated. All these arguments are particularly valid at low kinetic energies where nuclear stopping of HCP prevails.

The same arguemnts are basically responsible for the size of atomic excitation cross sections. At kinetic energies of approximately two to  three times the threshold energy electrons have their maximum excitation cross section. At these (low) energies the excitation cross sections  of HCP can almost be negelcted compared to those of the electrons (smaller by many orders of magnitude).

I hope this explanation is sufficient to answer your original question.

Dear Professor Müller,

Thank you very much for your detailed reply. I think I got my answer.

I was thinking about these trajectories.

These trajectories represent just a narrow selection of what can happen

The phenomenon you describe was explored with some detail by Birks, where he used an empirical formulation to describe the results from heavy ions (alpha, protons, etc.) and electrons (and beta particles). Both inorganic and organic scintillators show the effect, and even semiconductors show the effect for low energy ionizing particles.

You can not think of a single particle causing the production of a single ion-pair, except, perhaps, for UV and visible light absorptions. Instead, for higher energy ionizing radiation, energetic liberated electrons from all components in the scintillator, inorganic or organic, can contribute to free electrons. For instance, a liquid scintillator usually has at least three components, being a solvent, a solute (the fluor), and a surfactant emulsifier. Often a second wavelength shifting fluor is added to increase sensitivity to the light collection devices. Usually the total photon yield decreases when a wavelength shifter is added, but the transparency and spectral match are better, ultimately improving light detection. All of these components can become ionized and eject energetic electrons that produce additional ionization in the scintillator. In fact, the solvent is usually chosen with relatively low ionization energy to improve energy conversion. Hence, in the case of charged particles, including electrons, alpha particles and protons, they produce excited and free electrons directly and also indirectly by means of delta rays.

The non-linear scintillation behavior of heavy charged particles is attributed to luminescent center saturation. As heavy charged particles slow down in a medium, the specific ionization density increases (dE/dx), and the local luminescent centers become saturated. Consequently, the excess deposited energy is lost to competing processes, and the light yield becomes almost independent of energy deposition. This same process happens for fission fragments, alpha particles, deuterons, protons, and even electrons of low enough energy. The overall effect is that light charged particles (electrons, beta particles) on average produce more light per unit energy deposited than heavy charged particles.

I recommend that you look over Chapter 6 in The Theory and Practice of Scintillation Counting by J.B. Birks (McMillan, New York, 1964). There you will see in (Fig. 6.3) that all charged particles exhibit this behavior. Because energetic electrons have a much lower dE/dx, the effects are not as noticeable until almost all energy is lost, whereas the dE/dx is much larger for alpha particles, and consequently saturation is more prominent before all energy is lost.