# Deuterium fusion into Helium-4

## Main Question or Discussion Point

Hi All

Just wondering why does deuterium fused with itself not make helium-4 - instead it makes a triton or a helion and a proton or a neutron. These products could then fuse with deuterium to make helium-4.

What stops deuterium from fusing straight into alpha particles? Is the kinetic energy too high for it to stick together as a whole or is it driven by a more exotic process?

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Hi there,

I could be mistaking, but the reason for deuterium not fusing into helium-4 come from the nuclear spin. Deuterium has a nuclear spin of +1, and helium-4 has a spin of 0. Then again, both combination of tritium (+1/2) + proton (+1/2) or helium-3 (+1/2) + neutron (+1/2) fit well.

Cheers

malawi_glenn
Homework Helper
but spin 1 coupled to spin 1 can give spin 2,1 or 0 as final state.

clem
In order to conserve both energy and momentum, fusion can only occur with two particles going to two (or more) particles.

Staff Emeritus
2019 Award
What stops deuterium from fusing straight into alpha particles?
What makes you think the reaction $d+d \rightarrow {\rm{^4He}} + \gamma$ doesn't occur?

QuantumPion
Gold Member
In order to conserve both energy and momentum, fusion can only occur with two particles going to two (or more) particles.
How do you explain neutron capture then? E.g. Au-197 + n -> Au*-198

clem
What makes you think the reaction $d+d \rightarrow {\rm{^4He}} + \gamma$ doesn't occur?
Emitting a photon brings in a factor of alpha=1/137 compared topurely strong processes.

clem
How do you explain neutron capture then? E.g. Au-197 + n -> Au*-198
The key there is the star. When states of finite width are involved there can be absorption of neutrons with energies within the width of the excited state. That can't happen with
d+d-->He, all of which are stable.

I believe clem is correct, there's nothing to carry away extra energy unless you go electromagnetic (d + d -> He4 + gamma) and that process is suppressed by the factor of 1/137 compared to d + d -> He3 + n and d + d -> t + p.

Well that makes sense. The case of muonium fusion doesn't change that because the muon is another particle in the mix.

What's the relative fusion probabilities of d+d and d+t and d+He3 then? Obviously it depends on the abundance of d, but how much easier or harder are the other two reactions when both the t and He3 are being made from the d+d reaction?

I'm wondering just how hot the core of a brown dwarf star gets in order to burn deuterium, which must be pretty easy to do (inside brown dwarfs I mean) since it's only 1/6000th of the hydrogen available.

A recent paper on the Anthropic Principle (do you want me to drag out a reference?) discussed alternative Universes with higher levels of deuterium. What if a star was mostly deuterium? How light would it be and still achieve fusion?

I believe clem is correct, there's nothing to carry away extra energy unless you go electromagnetic (d + d -> He4 + gamma) and that process is suppressed by the factor of 1/137 compared to d + d -> He3 + n and d + d -> t + p.

Staff Emeritus