12H+13H®24He+01n
or
D + T ® 4He + n
Other fusion reactions:
D + D ® 3He +n
and
D + D ® 3H +p
We can use an earlier table to calculate the energy released in this process:
Some nuclear data
Name Isotope nat abund Mass (amu or u) Half-life neutron * n - 1.008665 12 min. hydrogen H 99.985% 1.007825 - deuterium D 0.015% 2.0140 - tritium * T - 3.01605 12.3 yrs. helium-3 3He 0.00014% 3.01603 - helium-4 4He 99.99986% 4.00260 -
The mass on the left side of the reaction is
The difference 0.018785 u, approximately 17.5 MeV is released in each such
reaction.1
Although this is about ten times smaller than the energy evolved
in the fission of uranium, note that the mass of the ingredients is smaller
by a factor of almost fifty.
To make this reaction occur, the ingredients (on the left side of the
reaction) must be brought within the range of the ``strong'' nuclear force
(say 10-14 m) in spite of the fact that both are positively charged,
and repel each other. To do this requires a combination of high temperature
(or energy) and density (or pressure). Such conditions exist in the inerior
of stars, but are difficult to achieve on earth.
2.0140 u + 3.01605 u = 5.03005 u,
that on the right
4.00260 u + 1.008665 u = 5.011265 u.
1Recall that one atomic mass unit (u) is equivalent to
approximately 930 MeV.