How to calculate the energy of Gamma rays in fusion

[Sebastiaan]

Lets say you have your basic fusion reaction

D + H → He3 + γ + 5.49 MeV

Now exactly how much of the energy is carried away by the gamma ray and how much by the Alfa particle ?

 

[mfb]

You can use conservation of energy and momentum to calculate it. As the photon won't have much momentum, the He-3 nucleus won't move much, which means its energy will be very small. To a good approximation the photon gets the whole energy.

Alpha particles are only He-4.

 

[Sebastiaan]

This is what I feared, but perhaps there are some fundamental laws which limit or mitigate the energy released by Gamma rays in fusion reaction

Do note this answer contradicts what is mention here.

"In the fusion of light elements, the reactions involve a restructuring the nuclei, and in general the resulting energy is 'carried as kinetic energy'."
.

 

[snorkack]

This is what I feared, but perhaps there are some fundamental laws which limit or mitigate the energy released by Gamma rays in fusion reaction

Do note this answer contradicts what is mention here.

There is indeed such a fundamental law.
Strong interaction is strong.
Where it is possible to release energy by restructuring nuclei and carrying energy away by kinetic energy rather than by gamma ray emission, that is usually much faster.

 

[mfb]

This is what I feared, but perhaps there are some fundamental laws which limit or mitigate the energy released by Gamma rays in fusion reaction

You can't violate energy or momentum conservation.

"In the fusion of light elements, the reactions involve a restructuring the nuclei, and in general the resulting energy is 'carried as kinetic energy'."

(Highlight by me) That is correct, and you found one of the exceptions. Most fusion reactions of light elements produce two (sometimes three) nuclei, and then typically the energy is carried away as kinetic energy. Fusion reactions that only produce one nucleus are rare, and in this case the energy is carried away by photons.

 

[snorkack]

(Highlight by me) That is correct, and you found one of the exceptions. Most fusion reactions of light elements produce two (sometimes three) nuclei, and then typically the energy is carried away as kinetic energy. Fusion reactions that only produce one nucleus are rare, and in this case the energy is carried away by photons.

Actually, fusion reactions of light elements in which more than one nucleus is produced and energy is carried away as kinetic energy of nuclei are a minority.
There are also fusion reactions in which energy is carried away as kinetic energy of positron and neutrino, or of neutrino alone.
On average, production of an alpha particle takes 5 fusion reactions. Of which the minority of 1 out of 5 forms more that 1 nuclei - and in that case 3 not 2:
He-3+He-3→α+p+p
The other fusion reactions are:
2 emit gammas:
d+p→He-3+γ
and 2 usually emit positron:
p+p→d+e⁺+ν_e

 

[mfb]

A sample of 5 reactions of your choice is not an argument.
Check this list - all with two reaction products! What does this show? Nothing, because it is a biased list as well.

Fusion reactions with more than two nuclei in the final state can happen via the strong interaction, while reactions with only one need the weaker electromagnetic interaction. That makes the first more likely, especially if both options are possible.

 

[snorkack]

A sample of 5 reactions of your choice is not an argument.
Check this list - all with two reaction products! What does this show? Nothing, because it is a biased list as well.

But is not claiming to be:

Most fusion reactions of light elements

Actually, there are obvious reasons why most fusion reactions must be ones that produce one nucleus.
"Fusion" by definition is formation of fewer, bigger nuclei out of more numerous smaller ones.
Mostly this happens by two nuclei forming one nucleus.
An alternative might be three nuclei forming two. But can you propose an actual common three-nucleus process?
Triple alpha is not, actually. Because its true mechanism is a multistep one:
α+α→Be-8+γ (Be-8 has actual half-life of 10^-16 s - 10⁶ times longer than the duration of nuclear collisions)
Be-8+α→C-12*+γ (Hoyle state)
C-12*→C-12*+γ
C-12*→C-12+γ
Due to selection rules, Hoyle state needs two steps to decay to C-12 ground state.
There IS a somewhat common, but decidedly minority, true three-particle fusion process. But just two are nuclei:
p+p+e→d+ν_e

Fusion reactions with more than two nuclei in the final state can happen via the strong interaction, while reactions with only one need the weaker electromagnetic interaction. That makes the first more likely, especially if both options are possible.

Many fusion reaction cannot proceed via weaker electromagnetic interaction either, and must happen via even weaker weak interaction. Like the simplest way to get 1 α particle out of 4 nuclei, in 3 steps all of which start with 2 particles:
p+p→d+e⁺+ν_e
d+p→He-3+γ
He-3+p→α+e⁺+ν_e
As you see, reactions of "fusion reactions of light elements". None of the 3 is a strong process. Just 1 of the 3 is an electromagnetic interaction, and most of the 3 are weak interaction fusion reactions.

 

[Sebastiaan]

and 2 usually emit positron:
p+p→d+e⁺+ν_e

How much of the kinetic energy is caried by the positron and neutino?

 

[snorkack]

How much of the kinetic energy is caried by the positron and neutino?

Like in beta decay. Most of the energy goes to positron and neutrino, and is freely distributed between them. A small fraction goes to the recoil of the nucleus.

 

[Sebastiaan]

This remind me that whenever nature produces something cool there is a catch. So the only useful thing about this reaction is energenic positron and deuterium core which could be usefull for other fusion reactions.

Question, let's say the energenic positron anihilates when it comes into contact with an electron, what happens with it kenetic energy stored in positron from p-p fusion?, Will the energy leased by Gamma be 1MeV + 0.5 E p-p

 

[snorkack]

This remind me that whenever nature produces something cool there is a catch. So the only useful thing about this reaction is energenic positron and deuterium core which could be usefull for other fusion reactions.

Question, let's say the energenic positron anihilates when it comes into contact with an electron, what happens with it kenetic energy stored in positron from p-p fusion?, Will the energy leased by Gamma be 1MeV + 0.5 E p-p

Yes, roughly that. The energy carried away by the neutrino escapes.
The net energy balance depends on whether the conditions are suitable for further reaction of He-3.

 

[mfb]

Most of the time the positron will lose most of its kinetic energy before it annihilates.

@snorkack: You are wrong with the two/three body discussion, but as it is off-topic anyway I won’t continue that discussion part.

 

[Sebastiaan]

Like in beta decay. Most of the energy goes to positron and neutrino, and is freely distributed between them. A small fraction goes to the recoil of the nucleus.

Wait a second. Exactly what do you mean by "freely distributed between them", the electron-neutrino is much less massive than the positron (which is at least 500000 more massive). Wouldn't this mean it would carry almost all of the 0.42 MeV energy?

 

[snorkack]

Wait a second. Exactly what do you mean by "freely distributed between them", the electron-neutrino is much less massive than the positron (which is at least 500000 more massive). Wouldn't this mean it would carry almost all of the 0.42 MeV energy?

No. It is the nucleus which is the most massive and therefore has to take a tiny fraction of energy. The neutrino and the positron have very different rest masses, but since the momentum conservation is already met by the nucleus, this does not much constrain how the energy is divided between positron and neutrino.

 

[mfb]

The distribution is not symmetric due to the electron mass, but the possible kinetic energy for both (individually) goes from zero to nearly the total released energy minus the electron rest energy.

 

[snorkack]

The distribution is not symmetric due to the electron mass, but the possible kinetic energy for both (individually) goes from zero to nearly the total released energy minus the electron rest energy.

Am I right in guessing that while the distribution of electron and antineutrino energies is not symmetric, the distribution of electron and antineutrino momenta is?

 

[mfb]

That is not symmetric either, but it is a better approximation.
Here are plots.

 

[snorkack]

Fusion of two protons releases an average of 420 keV plus the kinetic energy of protons.

How much of this on average goes to kinetic energy of positron, how much escapes with neutrino?

 

[Sebastiaan]

actually I believe is more than 0.42 MeV which is just the surplus power released, but it first aborbs energy

The first step of the proton–proton chain reaction is a two-stage process; first, two protons fuse to form a diproton:

H + H + 1.25 MeV → ²He

followed by the immediate beta-plus decay of the diproton to deuterium:

²He → D + e+ + ν_e + 1.67 MeV

with the overall formula

H + H→ D + e+ + v_e + 0.42 MeV

So irs not 0.42 MeV but 1.67 MeV which is divided over the positron and neutrino electron. If I applied the lesson leaned until now, I would think most of the energy would be caried away with the nutrino

 

[mfb]

Unless you shoot the two protons together in accelerators, you don’t have this initial 1.25 MeV around. The diproton has to be off-shell, and the total energy released is just 420 keV as required by conservation of energy (add a fee keV from the thermal energy of the protons). As this is smaller than the electron mass most of the energy goes into the neutrino (on average).
Here is a plot, the peak is somewhere around 300 keV.

 

[Sebastiaan]

As this is smaller than the electron mass most of the energy goes into the neutrino (on average).
Here is a plot, the peak is somewhere around 300 keV.

I see there is some probability distribution from which the peak is around 300 keV and it can go up to almost 420 keV. Does this also mean on average its energy is about 300 keV as well?

 

[mfb]

We would have to integrate the distribution to get the average, but it should be something between 250 to 300 keV.