Why there is no reaction Deuterium + Deuterium $=\rm {}^{4}He$?

Question

Why there is no reaction like

$D+D={}^{4}He$

specified here and in other places like this?

Apparently

$2\times2.0141-4.0026=0.0256$

is positive. What is the problem with this reaction?

Answer 1

According to Krane's Introduction to Nuclear Physics (Chapter 14), the reaction $$ D + D \to {}^4\mathrm{He} + \gamma $$ is possible but rare. The gamma ray is necessary because (as you note) there is energy released in the reaction, and it has to go somewhere. However, the $Q$-value for this reaction (23.8 MeV) is greater than either the neutron or proton separation energy for helium-4, and so the reactions $D + D \to {}^3 \mathrm{H} + p$ or $D + D \to {}^3 \mathrm{He} + n$ are more likely.

The likelihood of this reaction actually occurring is discussed in

Wilkinson & Cecil, "²H(d,$\gamma$)⁴He reaction at low energies." Phys. Rev. C31, 2036 (1985)

and references therein. Figure 5 of that article shows branching ratios for the $D + D \to {}^4\mathrm{He} + \gamma$ reaction of between $10^{-4}$ to $10^{-7}$, depending on the energy. (These branching ratios are calculated relative to the $D + D \to {}^3 \mathrm{H} + p$ reaction.) In other words, this reaction is somewhere between ten thousand and ten million times less frequent than the reaction in which a proton is released.

Answer 2

As far as I understand the resulting $^4He$ is highly excited and immediately splits into either $^3He$ and a neutron or $^3H$ and a proton.

Probably this is due to the fact that (in the center of mass frame) the resulting $^4He$ is at rest and thus the excess energy cannot be transferred into kinetic energy and thus has to stay inside the $^4He$ nucleus as excitation energy. By splitting into two parts, which can carry away some of the excess energy in form of kinetic energy, the nuclei can stabilize.