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I have a question about calculating the energy release from a fusion reaction. With parameters I have been supplied the liquid drop model semi-empirical binding energy formula gives the wrong answer. Does anybody know what the reason for this is?

The fusion of two tritium nuclei (T-T fusion) is given by the equation:

(3,1)H +(3,1)H =(4,2)He + 2n

We need to find the energy release, Q. For given constants in the semi-empirical binding energy we can calculate Q as: B = 14.0A − 13.1A2/3 − 0.584*Z(Z − 1)/A1/3− 19.4(A − 2Z)2/A+(±, 0) 34/A3/4,

where A is the Atomic Number and Z is the number of protons.

I get a value of 17.71MeV. Why is this different to the actual value of 11.33 MeV?

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closed as off-topic by Emilio Pisanty, Jon Custer, Kyle Kanos, Daniel Griscom, Mitchell Nov 8 '17 at 15:31

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The real value depends on quantum effects that are missing from the semi-empirical formula (SEF). The SEF works best for larger isotopes while your example is from the other end of the periodic table.

The SEF was motivated by the liquid drop model before the nuclear shell model (with a large spin-orbit force) was found to explain the magic numbers. The parameters of the SEF were found by curve fitting the binding energies over the entire periodic table. This averaging procedure glossed over the quantum effects associated with the shell structure.

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  • $\begingroup$ Awesome this tallies but in more coherent fashion than with what I have found subsequently!! $\endgroup$ – Tom Weisner Nov 5 '17 at 17:23
  • $\begingroup$ My thesis advisor (Alex Green) was the guy who did the curve fitting. $\endgroup$ – Lewis Miller Nov 5 '17 at 19:09

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