Binding Energy, as we know, is the energy released in the processes in which the light nuclei fuse or the heavy nuclei split to form a transmuted nucleus. This energy is made available as nuclear energy. Let’s learn about the nuclear fission in section below.

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While studying about Binding energy we observed that the binding energy per nucleon (E_{bn}) is nearly constant (around 8 MeV) for mass numbers in the range 30 < A < 170. Also, for lighter nuclei (A < 30) and heavier nuclei (A > 170), E_{bn} is less than 8 MeV. And, the nuclei in the range 30 < A < 170 are more tightly bound than those below or above the range specified.

Further, in conventional energy sources like coal or petroleum, energy is released through chemical reactions. However, for the same quantity of matter, nuclear sources provide million times larger energy than these conventional sources. For example, one kilogram of coal gives 107J of energy on burning. On the other hand, one kilogram of uranium creates 1014J of energy on fission.

**Browse more Topics under Nuclei**

- Atomic Mass and Composition of Nucleus
- Mass-Energy and Nuclear Binding Energy
- Nuclear Force
- Nuclear Energy – Nuclear Fusion
- Radioactivity – Law of Radioactive Decay
- Radioactivity – Types of Radioactive Decay

## Nuclear Fission

Post the discovery of the neutron, Enrico Fermi found that on bombarding a neutron different elements produced new radioactive elements. However, on bombarding a uranium target, the nucleus broke into two nearly equal fragments and released a great amount of energy. An example of the same is:

^{1}_{0}n + ^{235}_{92}U → ^{236}_{92}U → ^{144}_{56}Ba + ^{89}_{36}Kr + 3^{1}_{0}n

Fission does not always produce Barium and Krypton. Here is another example:

^{1}_{0}n + ^{235}_{92}U → ^{236}_{92}U → ^{133}_{51}Sb + ^{99}_{41}Nb + 4^{1}_{0}n

One more example:

^{1}_{0}n + ^{235}_{92}U → ^{236}_{92}U → ^{140}_{54}Xe + ^{94}_{38}Sr + 2^{1}_{0}n

All the fragmented nuclei produced in fission are neutron-rich and unstable. Also, they are radioactive and emit beta articles until they reach a stable end-product.

## Nuclear Energy Released in a Fission Reaction

In the example above, the nuclear energy released is of the order of 200 MeV per nucleus undergoing a fission. Here is how it is calculated: Imagine a heavy nucleus having A = 240. Now, this nucleus breaks into two nuclei with A = 120 each. Therefore,

E_{bn} (A = 240 nucleus) = 7.6 MeV

E_{bn} (each of A = 120 nuclei) = 8.5 MeV

Hence, the gain in binding energy per nucleon is about 0.9 MeV (8.5 – 7.6). Therefore, the total gain in binding energy is 240×0.9 = 216 MeV.

The disintegration energy in fission events initially appears as the kinetic energy of the fragments and neutrons. Subsequently, the surrounding matter starts heating up. The source of energy in nuclear reactors, which produce electricity, is nuclear fission. The enormous energy released in an atom bomb comes from uncontrolled nuclear fission.

## Nuclear Reactor

When ^{235}_{92}U undergoes a fission after being bombarded by a neutron, it splits into two nuclei and releases a neutron. This extra neutron now initiates the fission of another ^{235}_{92}U nucleus. For that matter, 2.5 neutrons are released per fission of a uranium nucleus. Also, fission produces more neutrons than what can be consumed.

This increases the chances of a chain reaction with each neutron that is produced, triggering another fission. If this chain reaction is uncontrolled, then it can lead to destruction (like a nuclear bomb). On the other hand, in a controlled manner, it can be harnessed to generate electric power.

However, there was a small problem. The neutrons generated in fission were highly energetic. They would escape rather than trigger another fission reaction. Also, it was observed that slow neutrons have a higher possibility of inducing fission in ^{235}_{92}U than their faster counterparts.

Now, the energy of a neutron produced in fission of ^{235}_{92}U is around 2 MeV. Unless these neutrons are slowed down, they tend to escape without inducing fission. This simply means that we need a lot of fissionable material to sustain the chain reaction.

### Slowing Down Fast Neutrons in a Nuclear Reactor to Generate Nuclear Energy

These fast neutrons are slowed down by elastic scattering with a light nuclei. Chadwick, in his experiments, showed that in an elastic collision with hydrogen, these neutrons almost come to rest and the protons carry away the energy.

*To understand this, imagine two marbles. One at rest and one travelling fast. When the fast-moving marble hits the stationary marble head-on, it comes to rest.*

Therefore, in nuclear reactors, light nuclei are provided along with the fissionable nuclei to slow down the fast neutrons. These light nuclei are the ‘*moderators’*. The most commonly used moderators are water, heavy water (D_{2}O) and graphite.

### Moderators

Moderators can ensure that the number of neutrons generated by a given generation is greater than those produced by the preceding generation. Hence, the multiplication factor ‘K’ or the measure of the growth rate of neutrons in the reactor can be in the range 1 < K < 1. Here is what it means:

- K < 1 means that the neutrons are not increasing in number over the previous generations. Or, they are escaping without inducing fission.
- K> 1 means that the reaction rate and nuclear power in the reactor is increasing exponentially and can even explode. It needs to be brought down to as close to unity as possible.
- K = 1 means that the reactor will be able to generate steady power.

This control is managed by using cadmium rods which can absorb neutrons. So, there is a dual control on a nuclear fission reaction.

- Slowing down the fast neutrons so that the induce fission and start a chain reaction.
- Introducing neutron absorbing rods in the reactor to ensure that the value of ‘K’ stays as close to 1 as possible.

Another interesting isotope is ^{238}_{92}U. This isotope is available in abundance and does not fission. Also, it forms plutonium by capturing neutrons. Here’s how:

^{238}_{92}U + ^{1}_{0}n → ^{239}_{92}U → ^{239}_{93}Np + e^{–} + v^{–}

^{239}_{93}Np → ^{239}_{94}Pu + e^{–} + v^{–}

Plutonium is highly radioactive and can also undergo nuclear fission under bombardment by slow neutrons.

## A Nuclear Power Plant

The image above outlines a typical nuclear power plant based on a pressurized water reactor. In such reactors, water is used as a moderator and heat transfer medium. In the primary-loop, water is circulated through the reactor vessel and transfers energy at high temperature and pressure (at about 600 K and 150 atm) to the steam generator, which is part of the secondary loop.

In the steam generator, evaporation provides high-pressure steam to operate the turbine that drives the electric generator. The low-pressure steam from the turbine is cooled and condensed to water and forced back into the steam generator.

Since the energy released in nuclear reactors is very high, they need very less fuel. However, in these reactors, highly radioactive elements are produced continually. Hence, this radioactive waste needs to be accumulated regularly. This includes both fission products and heavy transuranic elements like plutonium and americium.

## Solved Question for You

Question: The fission properties of ^{239}_{94}Pu are very similar to those of ^{235}_{92}U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ^{239}_{94}Pu undergo fission?

Solution:

Average energy released per fission of ^{239}_{94}Pu, E_{av} = 180 MeV

Amount of pure ^{239}_{94}Pu, m = 1 kg = 1000 grams

Avogadro’s number = N_{A} = 6.023 x 10^{23}Mass number of ^{239}_{94}Pu = 239

Now, 1 mole of ^{239}_{94}Pu contains N_{A} atoms. Therefore, ‘m’ grams of ^{239}_{94}Pu will contain {(N_{A}/Mass number) x m} number of atoms,

= [(6.023 x 10^{23})/239] x 1000 = 2.52 x 10^{24} atoms.

Hence, total energy released during the fission of 1 kg of ^{239}_{94}Pu is:

E = E_{av} x 2.52 x 10^{24}= 180 x 2.52 x 10^{24} = 4.536 x 10^{26} MeV.