Why is the D-D nuclear fusion reaction unlikely despite having a higher Q value?

[Incand]

I'm having trouble understand a passage in our book.
The author claims that the reaction
$^2H+^2H \to ^4He + \gamma$
is unlikely since the $Q$ value is large ($23.8$ MeV) which happens to be greater than both the neutron and proton separation energies.

This seem very counter intuitive to me. Shouldn't a large $Q$ value rather make the reaction more probably?

We also have the D-T reaction
$^2H+^3H \to ^4He +n$ with $Q=17.6$
that for some reason is more probable but I really don't see the difference.

 

[mfb]

3He + n and 3H + p are more likely as they don't involve the emission of a photon and work via the strong interaction only. They should also have a larger phase space.

 

[Incand]

3He + n and 3H + p are more likely as they don't involve the emission of a photon and work via the strong interaction only.

That makes some sense but then I wonder why
$^2H+ \; ^1H \to \; ^3He + \gamma$
and
$2 \; ^3 He \to \; ^4He +2 \; ^1H+ \gamma$
are part of the p-p chain as opposed to for example
$^2H+ \; ^1H \to \; ^3H+p$
Shouldn't these by the same reasoning be unlikely?

 

[Vanadium 50]

Your last equation creates a neutron out of nowhere.

 

[Incand]

Your last equation creates a neutron out of nowhere.

That's embarrassing,I tried to come up with an example.

But for the other two reactions is the reasoning here that they are indeed unlikely but there just isn't anything more probably around either? Not that gives a reaction chain with enough released energy to be a significant part of the powering of a star in any case.

 

[Vanadium 50]

Your first two equations involve the weak interaction. This makes them rare.

 

[mfb]

@Vanadium 50: None of the reactions involve the weak interaction.

2H + 1H -> 3He + photon is the only reaction that is possible with those hydrogen nuclei.
3He + 3He -> 4He + 1H + 1H can work without photons.

 

[Vanadium 50]

You're right. I looked at 3He and thought "tritium". Dumb of me.