# 4.5.3: Energy Released by the Fission of a Single Nucleus


Let's use as an example the reaction that led to the discovery of nuclear fission. The "key" to the discovery was the presence of element Barium with $$Z=56$$. In a fission reaction the $$Z$$ numbers of the daughter nuclei must add up to $$\mathrm{Z}$$ of the parent nucleus, so that the other element must have been $$92-56=36$$, which is the atomic number of the element Krypton. As noted earlier, the fission of U-235 nuclei may produce as many as 150 different combinations of daughter nuclei - so for a single it is not possible to predict how many neutrons there will be in each fragment. Here is one likely "scenario":

$n+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3 n$

Note that the number of protons on both sides are the same: 92 in U, and 56 $$+36=92$$ in Ba and Kr. The same is true for neutrons: left side, 1 incident plus (235-92) in $$U$$ is 144 ; right side: (144-56) in Ba plus (89-36) in Kr plus 3 released also $$=144$$

The calculation of the energy release in the process is pretty straightforward. We need to compare the masses of all objects at both sides of the equation. For that, we need to know the mass defects of all nuclei - we have already got it for U-235, and using the aforementioned Web tool, we readily find: $$1190.23 \mathrm{MeV} / \mathrm{c}^{2}$$ for Ba-144 and $$766.91 \mathrm{MeV}$$ for $$\mathrm{Kr}-89$$. Keeping in mind that the mass of a nucleus is equal to the mass of all protons plus the mass of all neutrons minus the mass defect.

In Table 1, all items needed for calculating the total masses of the elements represented on the left side, and the right side of the Eq. $$4.7$$ are collected. Our objective is to find the difference between the total masses "before" and "after" the fission event.

Table 1. All relevant components arranged in a way making easy "balancing the mass book" in the reaction described by Eq. 4.7.

Left side: $$\quad 1 m_{n}$$
Left side: $$143 m_{n}+92 m_{p}-1783.36 \mathrm{MeV} / c^{2}$$
Right side: $$88 m_{n}+56 m_{p}-1190.23 \mathrm{MeV} / c^{2}$$
Right side: $$53 m_{n}+36 m_{p}-766.91 \mathrm{MeV} / c^{2}$$
Right side: $$\quad 3 m_{n}$$

The way things are arranged in Table 1 facilitates such calculations: it makes it clear immediately that when the "right side" (final) masses are subtracted from the "left side" (initial) ones, all proton and neutron masses cancel out, and the result is:

$\begin{array}{c} M_{\text {total initial }}-M_{\text {total final }}= \\ =[(-1783.36)-(-1190.23)-(-766.91)] \frac{\mathrm{MeV}}{c^{2}}=+173.78 \frac{\mathrm{MeV}}{c^{2}} \end{array}$

The "plus" sign before the final result was added intentionally, in order to emphasize that a significant mass component is missing in the right side of the equation. How comes that the "mass book doesn't balance"? Here the Einstein's famous formula $$E=m c^{2}$$ steps in handy, the missing mass has been converted to $$173.78 \mathrm{MeV}$$ of "pure energy": its main component is the kinetic energy of the two daughter nuclei flying apart (like fragments of the bombshell in an explosion of a bomb), several $$\mathrm{MeV}$$ is the kinetic energy of the fast neutrons released, and some energy may be released in the form of one or more $$Y$$ photons.

Altogether, the total energy released in a single fission event is even higher. Note that the $$N / Z$$ ratio for U-235, $$1.554$$, is considerable higher than that for stable isotopes from the $$Z$$ range close to 46 , i.e., one-half of the Uranium's $$Z$$. Let's check: the element 46 is Palladium, a mixture of stable isotopes with an average $$A$$ of $$106.42$$, meaning an average $$N=(A-Z)$$ of $$60.42$$, and average $$N / Z$$ ratio of $$1.313$$, much less than that of Uranium. Consequently, the fission product nuclei havea "strong surplus" of neutrons. Let's take a look at the nuclei in the Eq. 4.7: for the heavier fragment $$N / Z$$ $$=1.57$$, while for natural Barium it is $$1.452$$; and for the lighter fragment it is $$N / Z=1.472$$, while for natural Krypton gas it is $$1.328$$. It means that both fragments have a huge "surplus" of neutrons, and thus they should be strongly radioactive with very short lifetimes. In fact, sometimes the fission products in real nuclear reactors have lifetimes as short as a small fraction of one second. In some of them the $$N / Z$$ ratio may be so high that they decay through the emission of a neutron - a decay process is not seen in any naturally occurring radioactive material. By the way, such neutrons are called "prompt neutrons" and they play an important role in controlling the chain reaction in nuclear reactors.

The decay of the fission fragments yields additional energy. A detailed list of all processes participating in the overall energy yield in nuclear fission can be found, e.g., in this Hyperphysics Web page. In addition to the immediate energy yield coming from the loss of mass of U-235, the decay of the fission products may contributs extra $$30-40 \mathrm{MeV}$$, so that a reasonable estimate of the average energy release per single fission reaction can be taken as 210-215 $$\mathrm{MeV}^{1} .$$

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1. In fact, the net energy released depends of the timescale because not all decay products of fission reaction fragments are short-lived. For instance, the infamous Caesium-137 that contaminated huge land areas after the Chernobyl disaster has the half-life of over 30 years

4.5.3: Energy Released by the Fission of a Single Nucleus is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.