Positron cross section with H,D,T

[Salman2]

I have a question about cross section probability for a positron (e+) to interact with the electron (e-) that would be bound to three different atomic systems (1) hydrogen atom: H(e-), (2) deuterium atom: D(e-), (3) tritium atom: T(e-).

My hypothesis is that a positron (e+) would have the highest cross section interaction probability with H(e-) and lowest with T(e-), with D(e-) intermediate. My thinking is that the addition of neutrons in the D and T systems would serve to slightly shield the positron (e+) and (e-) wavefunctions from forming unstable positronium, thus lower cross section probability of interaction between the (e+) and (e-). Would my hypothesis be correct ?

 

[mfb]

I don't see why the neutrons would "shield" anything.

The positron needs some minimal energy, as the positronium bond is weaker than the electron/proton bond.
Due to the isotope effect, the energy levels of deuterium and tritium are a bit lower than those for hydrogen. If you get the right positron energy, the reaction might be possible with some isotopes, but not with heavier ones.

 

[Salman2]

...due to the isotope effect, the energy levels of deuterium and tritium are a bit lower than those for hydrogen...

OK, thanks, we can forget about any shielding effect.

So, are you saying that, due to the 'isotope effect' you mention, the positronium formation cross section potential would be highest for H(e-) isotope, and lowest for tritium T(e-) isotope, because the electron energy density would be a bit lower in tritium, thus less possibility for (e-) and (e+) wavefunctions to overlap to form positronium ?

 

[mfb]

No, I did not say this.

The binding energy for H+e in the ground state is -13.606 eV.
The binding energy for D+e in the ground state is something like -13.611 eV (rough estimate).
The binding energy for positronium in the ground state is -6.803 eV.

The reaction $e^+ + {}^1H \to (e^+e^-) + {}^1H^+$ needs additional 6.803 eV of energy.
The reaction $e^+ + {}^2H \to (e^+e^-) + {}^2H^+$ needs additional 6.808 eV of energy.

If your positron has 6.805 eV, it can react with H, but not with D.

 

[Salman2]

No, I did not say this.

The binding energy for H+e in the ground state is -13.606 eV.
The binding energy for D+e in the ground state is something like -13.611 eV (rough estimate).
The binding energy for positronium in the ground state is -6.803 eV.

The reaction $e^+ + {}^1H \to (e^+e^-) + {}^1H^+$ needs additional 6.803 eV of energy.
The reaction $e^+ + {}^2H \to (e^+e^-) + {}^2H^+$ needs additional 6.808 eV of energy.

If your positron has 6.805 eV, it can react with H, but not with D.

Ok, very helpful, thank you.

I have a few more questions.

Your estimate in binding energy between H and D has a delta of 0.005 eV, so does that mean there would be a similar delta between D and T predicted, that is, we would predict the binding energy for a T+e in ground state to be about -13.616 eV, given the added neutron isotope effect ? So, we would predict this:

The reaction $e^+ + {}^3H \to (e^+e^-) + {}^3H^+$ needs additional 6.813 eV of energy.

If yes above, it seems reasonable that we predict that a positron (e+) with energy between 6.809 and 6.812 eV would react with either a H and D system, but not with T=tritium ? Would this be correct ? Are you aware if this experiment has ever been conducted to confirm the prediction ?

One final question. Suppose a positron (e+) with an energy of 6.820 eV. Would it be predicted to have a higher cross section reaction probability with a D+e(-) isotope in ground state, or a T+e(-) isotope, or would the cross section be the same for the two ? Again, are you aware if this experiment has ever been conducted ?

Again, thanks for help with these questions.

 

[snorkack]

I do not see why positron needs to form positronium in order to be unstable.

Positron is itself a somewhat special isotope of hydrogen. Very special isotope, though, due to its having such a small reduced mass.

Just like two hydrogen nuclei and one electron can form a dihydrogen cation, a positron and one hydrogen atom can form a molecular ion. Which would constantly have an overlap between positron and electron wavefunctions.

 

[mfb]

I do not see why positron needs to form positronium in order to be unstable.

It does not have to, and I would guess that direct annihilation is the most probable reaction* (and it is the only possible reaction* for low-energetic positrons).

*ignoring elastic collisions

 

[snorkack]

direct annihilation is the most probable reaction* (and it is the only possible reaction* for low-energetic positrons).

*ignoring elastic collisions

What do you mean low-energetic?

Energetic positron can undergo direct annihilation, or lose energy by elastic collisions, or lose energy by inelastic collisions, or undergo direct capture into positronium - or undergo direct capture into positronium molecules.

So what could be the further fate of positronium molecules and molecular ions?

 

[mfb]

What do you mean low-energetic?

Below the 6.8 eV I calculated.

So what could be the further fate of positronium molecules and molecular ions?

Positronium decays (unless it hits something and disintegrates before that happens), and ions usually capture an electron after a while, but that depends on your setup.

 

[snorkack]

Positronium decays (unless it hits something and disintegrates before that happens), and ions usually capture an electron after a while, but that depends on your setup.

Cannot positronium molecular ions also annihilate?