# Why doesn't D + D > He4

1. Jan 31, 2009

### Knuckles

The title pretty much says it all. According to my research there are 2 equally probable deuterium-deuterium reactions:

D + D > He3 + n
D + D > T + p

So what prevents D + D > He4 from happening? All these reactions are exothermic, conserve mass, and conserve the number of nucleons. What other rules/conservation laws apply here?

2. Jan 31, 2009

### Morbius

Knuckles,

Conservation of momentum for one. If we had D + D --> He4; look at the reaction in the
center of mass frame of the He4 nucleus. In the center of mass frame of the product He4;
the total momentum is zero. So the total momentum of the incoming deuterons would
also have to be zero in that reference frame. The total kinetic energy after the fusion
reaction would also have to be zero in this reference frame.

You have to be able to conserve BOTH momentum and energy [ including the Q - the
energy released by the reaction ] in that reference frame.

Dr. Gregory Greenman
Physicist

Last edited: Jan 31, 2009
3. Jan 31, 2009

### Knuckles

So you're saying that if we COULD have reactions with only one product, then we would always be free to consider the centre of mass frame of that product and calculate the product's kinetic energy as 0... and this would be a contradiction because the initial kinetic energy and the Q released by the reaction would both, apparently, have vanished? Could we have the extra energy carried away in one or more gamma rays?

4. Jan 31, 2009

### Morbius

Knuckles,

Try it. You know the relationship between energy and momentum for a photon: E = pc
where E is the energy, p is the momentum, and c is the speed of light.

You have to conserve BOTH momentum and energy.

The Q of the reaction that you propose would be just under 24 MeV. [ Use the masses of D and He4 ]

Try to conserve both momentum and energy for this reaction. Both energy and momentum have to
be conserved in ALL frames - but my hint of using the center of mass frame of the product He4 helps
make it simpler since if it is the only product - the momentum in that frame is zero. If you have a
photon too - then the momentum will be non-zero because a photon has momentum in all frames;
but it will be smaller than if you had a nucleus recoiling.

Dr. Gregory Greenman
Physicist

5. Feb 1, 2009

### vanesch

Staff Emeritus
In fact, the problem is that, as Morbius pointed out, that as we have only one particle in the final state, that the energy cannot be kinetic energy, but must be internal energy.

Now, such reactions are a priori not impossible: it could be that the final state is in an excited state with just enough extra excitation energy to be possible.

Now, when we look at the NNDC chart (see http://www.nndc.bnl.gov/chart/reCenter.jsp?z=1&n=1) we see that the mass excess of D is 13.1357 MeV, and that the mass surplus of He-4 is 2.4249 MeV. That means that if the two D nucleae are coming in at very low energy (difficult to do already because of electrostatic repulsion...), we have a Q for this reaction of 2 x 13.1357 - 2.4249 MeV = 23.8465 MeV.
So this means that we should find an excited state of the helium-4 nucleus with an excitation energy of something like 23.8 MeV for it to be possible (and higher if the incoming D energy is higher).
Now, the binding energy of He-4 equals 2 x 7.2890 + 2 x 8.0713 - 2.4249 MeV = 14.591 MeV
as calculated from the mass excess of two protons (H-1) and two neutrons.
This means that the ground state of He-4 is only 14.591 MeV below 0. A bound state with an excitation of 23.8 MeV is hence not possible. That doesn't mean that resonances aren't possible of course. But you cannot have a bound state of He-4 which is excited 23.8 MeV above the ground state, as the ground state itself is only 14.591 MeV deep.