Physics · Ch 13 — Nuclei

Fission

13.7.1

Fission

The Discovery of Nuclear Fission

When we move beyond the natural radioactive decays that happen spontaneously, a whole new world opens up: nuclear reactions. These are processes where we deliberately bombard a nucleus with another particle — a proton, a neutron, an alpha particle, and so on. Among all such reactions, one stands out for its sheer energy release and its practical importance: nuclear fission.

The classic example is the fission of uranium-235. When a slow neutron is absorbed by a $^{235}_{92}\text{U}$ nucleus, the compound nucleus $^{236}_{92}\text{U}$ is formed. This highly excited nucleus is unstable and splits almost instantly into two medium-mass fragments, along with a few neutrons. One such reaction is:

{}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{236}_{92}\text{U} \rightarrow {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n}

This is not the only possible outcome. The same reaction can produce different pairs of fragments. For instance:

{}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{236}_{92}\text{U} \rightarrow {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4\,{}^{1}_{0}\text{n}

Or, as another example:

{}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \rightarrow {}^{140}_{54}\text{Xe} + {}^{94}_{38}\text{Sr} + 2\,{}^{1}_{0}\text{n}

Note

The fragment products — like $^{144}_{56}\text{Ba}$ , $^{89}_{36}\text{Kr}$ , $^{133}_{51}\text{Sb}$ , etc. — are all radioactive. They are neutron-rich and unstable. They achieve stability by undergoing a series of beta-minus ( $\beta^-$ ) decays, each converting a neutron into a proton and emitting an electron and an antineutrino. This is why fission products are a major source of radioactive waste.

The Energy Released in Fission: A Q-Value of ~200 MeV

The truly staggering fact is that the energy released — the Q-value — in the fission of a single uranium nucleus is about 200 MeV. To put that in perspective, a typical chemical reaction (like burning a carbon atom) releases only a few eV. Fission releases roughly a hundred million times more energy per atom.

How do we arrive at this number? The textbook gives a clear, approximate calculation based on the binding energy per nucleon curve.

›Proof

Step 1: The starting point.

Consider a heavy nucleus with mass number $A = 240$ (like the compound nucleus formed in fission). From the binding energy per nucleon curve, the binding energy per nucleon for $A=240$ is about $E_{bn} \approx 7.6\ \text{MeV}$ .

Step 2: The fragments.

This nucleus splits into two roughly equal fragments, each with $A \approx 120$ . For $A=120$ , the binding energy per nucleon is higher: $E_{bn} \approx 8.5\ \text{MeV}$ .

Step 3: The gain per nucleon.

The gain in binding energy per nucleon is:

$8.5\ \text{MeV} - 7.6\ \text{MeV} = 0.9\ \text{MeV}$

Step 4: The total gain.

Since there are 240 nucleons in total, the total gain in binding energy is:

$240 \times 0.9\ \text{MeV} = 216\ \text{MeV}$

This 216 MeV is the energy released. The textbook rounds this to "of the order of 200 MeV" — a very good approximation. The slight difference comes from the fact that the fragments are not exactly $A=120$ , and the binding energy values are approximate.

Watch out

Do not confuse binding energy with binding energy per nucleon. The total binding energy of the original $A=240$ nucleus is $240 \times 7.6 \approx 1824\ \text{MeV}$ . The total binding energy of the two fragments together is $2 \times (120 \times 8.5) = 2040\ \text{MeV}$ . The difference, $2040 - 1824 = 216\ \text{MeV}$ , is the energy released. The gain comes from the fact that the fragments are more tightly bound (higher $E_{bn}$ ) than the original heavy nucleus.

Where Does This Energy Go? …

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