Question

In both D-D reactions in equation (11-18). Two deuterons fuse to produce two particles, a nucleus of$A=3$ and a free nucleon. Mass decreases because the binding energy of theA=3nucleus is greater than the combined binding energies of the two deuterons. The binding energy of helium-4is even greater still. Why can't the deuterons simply fuse into a helium-4nucleus and nothing else? Why must multiple particles be produced?

Short answer

We need a multiple-particle process, so that the extra particle could carry away the excess energy.

Step-by-step solution

Given data

Brass is an alloy of copper and zinc. Given that ${\rho }_{Brass}$ is higher than the ${\rho }_{\text{copper}}$or ${\rho }_{zinc}$ at room temperature and drops much slower with a decrease in temperature.

Concept of Nuclear fission

Nuclear fission, subdivision of a heavy atomic nucleus, such as that of uranium or plutonium, is into two fragments of roughly equal mass.

${\rho }_{\text{Brass}}>{\rho }_{\text{copper}},{\rho }_{zinc}$

Explanation of production of multiple particles in deuteron fusion despite simply

Suppose tile reaction of two deuterons going into single helium-4 did happen, and then choose a reference frame where the single helium-4 particle would be rest.

Initially, the energy equals to the mass energy of the two deuterons and their kinetic energies.

Finally, the energy is only the mass-energy of the helium-4 . Since the mass of two deuterons is bigger than the mass of the helium- 4 , the energy is not conserved.

Thus we need a multiple-particle process, so that the extra particle could carry away the excess energy.