The following is a question and sample answer related to the calculation of loss of mass from fission and, in turn, the calculation of energy released.

In a fission reaction a neutron is absorbed by a uranium-235 nucleus. Barium-139 and krypton-94 nuclei are released as well as some neutrons.

Write a nuclear equation for this reaction. $$^{235}_{92}{\rm U} + {}^1_0{\rm n} \rightarrow {}^{139}_{56}{\rm Ba} + {}^{94}_{36}{\rm Kr} + 3\,{}^1_0{\rm n}$$ Calculate the energy released, in MeV, in this reaction. $$\text{loss in mass}=3.0\times10^{-28}\,\mathrm{kg}\\\begin{align}E&=mc^2\\E&=2.47\times10^{-11}\mathrm J\\E&=171\,\mathrm{MeV.}\end{align}$$

How does the math add up for the calculation of "loss in mass"? The mass numbers of the various particles/atoms are equal on both sides of the equation and thus there doesn't appear to be any loss in mass.

What am I missing?


1 Answer 1


The mass numbers are approximate. If you want to calculate the missing mass you need to look at the actual masses. See: https://wwwndc.jaea.go.jp/NuC/

For example U235 is actually 235.043931368 +/- 0.000001962 u.

  • 1
    $\begingroup$ Note that $ 171\,\mathrm{MeV}/c^2$ corresponds to a little under $0.2\rm\,u$. $\endgroup$
    – rob
    Commented Jun 19, 2021 at 20:19
  • $\begingroup$ Also note that the mass numbers are actually nucleon/baryon number and must be balanced in the reaction. $\endgroup$
    – Bill N
    Commented Jun 20, 2021 at 13:12
  • $\begingroup$ Yes, although pedagogically it would be better if they were described only as baryon numbers and not as mass numbers. $\endgroup$
    – Dale
    Commented Jun 20, 2021 at 13:21

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