What happens to the matter that is consumed in nuclear fusion?

Answer 1

The matter that is "consumed" is converted into energy, according the the equation E=mc2.

The original matter mostly becomes the final matter, though a small amount is converted into heat. The amount converted into heat is small enough however, that the larger subatomic particles are all accounted for.

We could take as an example a fusion reaction in which a deuterium atom and a tritium atom are fused into helium. Deuterium and tritium are both isotopes of hydrogen, 2H and 3H respectively. The 2H nucleus consists of one proton and one neutron. The 3H nucleus consists of one proton and two neutrons. Each atom also has one electron. The total before fusion is two protons, two electrons, and three neutrons.

After the fusion takes place, the product is one helium atom, of the isotope 4He, plus one free neutron. The 4He atom has two protons and two neutrons, plus two electrons. Thus, the total of particles after fusion is two protons, two electrons, and three neutrons. In this case, however, the helium atom and the neutron are both very, very hot.

So the number of protons, neutrons, and electrons is the same after the reaction as it was before.

The equation on converting between energy and mass is E=mc2, as you know. The amount of energy released in the fusion example above is the difference between the mc2 before the reaction and the mc2 after the reaction. While this difference in mass is so small that it is not reflected in the counts of large subatomic particles, it is nonetheless there.

The masses, in Atomic Mass units, of the atoms and the neutron are:

at the beginning

2H - 2.014102

3H - 3.016049

which add to 5.030151

at the end

n - 1.008665

4He - 4.002602

which add to 5.011267

so the difference between the masses before and after the reaction is 0.018884 Atomic Mass units, which represents the amount of mass converted into energy in the reaction.

Answer 2

The matter that is "consumed" is converted into energy, according the the equation E=mc2.

The original matter mostly becomes the final matter, though a small amount is converted into heat. The amount converted into heat is small enough however, that the larger subatomic particles are all accounted for.

We could take as an example a fusion reaction in which a deuterium atom and a tritium atom are fused into helium. Deuterium and tritium are both isotopes of hydrogen, 2H and 3H respectively. The 2H nucleus consists of one proton and one neutron. The 3H nucleus consists of one proton and two neutrons. Each atom also has one electron. The total before fusion is two protons, two electrons, and three neutrons.

After the fusion takes place, the product is one helium atom, of the isotope 4He, plus one free neutron. The 4He atom has two protons and two neutrons, plus two electrons. Thus, the total of particles after fusion is two protons, two electrons, and three neutrons. In this case, however, the helium atom and the neutron are both very, very hot.

So the number of protons, neutrons, and electrons is the same after the reaction as it was before.

The equation on converting between energy and mass is E=mc2, as you know. The amount of energy released in the fusion example above is the difference between the mc2 before the reaction and the mc2 after the reaction. While this difference in mass is so small that it is not reflected in the counts of large subatomic particles, it is nonetheless there.

The masses, in atomic mass units, of the atoms and the neutron are:

at the beginning

2H - 2.014102

3H - 3.016049

which add to 5.030151

at the end

n - 1.008665

4He - 4.002602

which add to 5.011267

so the difference between the masses before and after the reaction is 0.018884 atomic mass units, which represents the amount of mass converted into energy in the reaction.