Nuclear Fission and Fusion: Reaction, Types, Applications - Embibe
  • Written By Saurav_C
  • Last Modified 24-06-2022
  • Written By Saurav_C
  • Last Modified 24-06-2022

Nuclear Fission and Fusion: Definition, Application and Examples

Nuclear Fission and Fusion: What is the source of this huge amount of energy emitted by the sun and the stars? The answer to this question is a nuclear reaction called fusion. The energy in the core of the sun is produced by the nuclear fusion of hydrogen nuclei into helium at very high temperatures. How does this happen? When two nuclei combine to produce a heavier nucleus or when an atom splits into lighter nuclei, then they produce a huge amount of energy. We will discuss all of this in this article. Due to the increase in energy demand and global warming, the nuclear power plant is being highlighted as one of the clean and reliable sources of energy for all our energy requirements.

The energy produced per unit mass in a nuclear reaction is much larger as compared to any other source of energy, for example, fossil fuels. Since the energy produced in a nuclear reaction is huge, it is quite dangerous too when not handled properly or when used for destructive purposes. We all have heard about the destruction that was caused by the atom bombs which were dropped over Hiroshima and Nagasaki. The bomb which was dropped over Hiroshima released an energy equivalent to that released by the explosion of \(20,000\) tons of TNT (tri-nitro toluene). Since then, more and more powerful (atomic, hydrogen and neutron) bombs have been made whose destructive power is equivalent to several Megatons of TNT.

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Why do Nuclear Reactions Release Tremendous Amounts of Energy?

The mass of an atomic nucleus is always less than the sum of the individual masses of each subatomic particle that constitutes it (protons and neutrons), in their isolated states. This difference in mass is attributed to nuclear binding energy (often referred to as a mass defect). Nuclear binding energy can be defined as the amount of energy required to separate all the constituent particles (protons and neutrons) of a nucleus.

When the nuclear reaction occurs, the difference in mass (mass defect) is converted into energy. The tremendous amount of energy released during this can be calculated by Einstein’s famous equation. It is given by, \(E= mc^{2}\). Here \(m\) represents mass defect and \(c\) is the speed of light.

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Nuclear Reaction

In nuclear reactions, the nuclei of the reactants interact with each other. They result in the formation of new elements. This process is also called the transmutation of nuclei because it involves the change in the identity or characteristics of an atomic nucleus.

There are two types of nuclear reactions as follows:

  1. Nuclear Fission
  2. Nuclear Fusion

Nuclear Fission

Nuclear fission is a process in which the nucleus of an atom is split into two or lighter nuclei (called daughter nuclei). The fission of heavy elements is an exothermic reaction, and huge amounts of energy are released in the process. Nuclear fission occurs with heavier elements, where the electromagnetic force pushing the sub-atomic particles inside the nucleus apart dominates the strong nuclear force holding it together.

To initiate most fission reactions, an atom is bombarded by a neutron to produce an unstable isotope, which undergoes fission. When neutrons are released during the fission process, they can initiate a chain reaction of continuous fission which sustains itself.

nuclear fission

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Types of Nuclear Fission Reaction

Alpha Decay

In \(α\)-particle decay (or alpha decay), the nucleus loses two protons and two neutrons, so the atomic number decreases by two, whereas its mass number decreases by four.

The general equation for alpha decay is: \({}_\rm{Z}^\rm{A} X \to {}_{\left( {\rm{Z} – 2} \right)}^{\left( {\rm{A} – 4} \right)}X’ + {}_2^4\alpha \)

Where A is the mass number and Z is the atomic number. An example of alpha decay is uranium- \(238\) as follows: –

\({}_{92}^{238} \rm{U} \to {}_{{90} }^{ {234} }\rm{Th} + {}_2^4 \rm{He} \)

Beta Decay

Beta decay occurs when a neutron is converted into a proton, which is accompanied by the emission of a beta particle (high-energy electron). In most \(\boldsymbol{\beta}\) -particle decays (or beta decay), either an electron \(\left(\beta^{-}\right)\) or positron \(\left(\beta^{+}\right)\) is emitted by a nucleus.

As an example, the isotope \(\left( _{90}^{234}{\text{Th}} \right)\) is unstable and decays by \(\beta^{-}\) emission with a half-life of \(24\) days. Its decay can be represented as
\(_{90}^{234}{\text{Th}} \to _{{\text{91}}}^{{\text{234}}}{\text{Pa + }}_{ – 1}^0{\text{e + }}\bar \vartheta \)

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Gamma Emission

A nucleus in an excited state can decay to a lower-level state by the emission of a “gamma-ray” photon, and this is known as gamma decay. This is analogous to the de-excitation of an atomic electron. Gamma decay is represented symbolically by:-

\({}_{\text{Z}}^{\text{A}}{\text{X}} \to {}_{\text{Z}}^{\text{A}}{\text{X} + \gamma }\)

An example of gamma emission is the de-excitation of the excited thallium- \(234\) nucleus (which is produced from the alpha decay of uranium- \(238\)). The equation for this nuclear reaction is:

\(^{234}T{h^*}{ \to ^{234}}Th + \gamma \)

Nuclear Fission in Nuclear Power Plants

One of the major applications of a fission reaction is the production of electricity via nuclear power plants. In nuclear power plants, energetic neutrons are directed into a sample of the isotope uranium- \(235\). The equation of the reaction has been given below:

\(U_{92}^{235}+n_{0}^{1} \rightarrow K r_{36}^{92}+B a_{56}^{141}+3 n_{0}^{1}+\text {Energy}\)

The energy from the neutrons can cause the uranium nucleus to break in several different ways. A common fission reaction produces barium- \(141\) and krypton- \(92\). In this particular reaction, one uranium nucleus breaks into a barium nucleus, a krypton nucleus, and two neutrons. These two neutrons can go on to split other uranium nuclei, resulting in a nuclear chain reaction.

Nuclear Fission in Nuclear Power Plants

Nuclear Fusion

The process in which two or more light nuclei combine to form a single heavier nucleus is called nuclear fusion. The nuclear fusion process occurs in elements that have a low atomic number, such as hydrogen. Nuclear Fusion is the opposite of nuclear fission reaction in which heavy elements diffuse and form lighter elements. For a nuclear fusion reaction to occur, it is necessary to bring two nuclei so close that nuclear forces become active and binds the nuclei together. Nuclear forces are small-distance forces and have to act against the electrostatic forces where positively charged nuclei repel each other. This is the reason nuclear fusion reactions occur mostly in highdensity, hightemperature environments. Fusion is the process by which the sun and other stars generate light and heat. The universe is full of instances of nuclear fusion reactions.

Nuclear Fusion

Examples of Nuclear Fusion

1. Deuterium-tritium fusion reaction

The binding energy per nucleon increases as we go from hydrogen to helium. It means that helium is more stable than hydrogen. So, when Nuclear fusion will occur, energy will be released.

Consider the following deuterium-tritium fusion reaction:
We have-

\(D_{1}^{2}+T_{1}^{3} \rightarrow H e_{2}^{4}+n_{0}^{1}\)

Deuterium-tritium fusion reaction

When deuterium and tritium fuse together, their components are recombined to form a helium atom and a fast neutron. As the two heavy isotopes are recombined into a helium atom and a neutron, the leftover’s extra mass is transformed into kinetic energy.

2. Nuclear Fusion in the Sun

Every star in the universe, including the sun, is alive due to nuclear fusion. It is through this process that they produce an enormous amount of heat and energy. The pressure and temperature at the core of any star are tremendously high, and that is where the nuclear fusion reaction occurs. The specific type of fusion that occurs inside of the Sun is known as proton-proton fusion. Inside the core of the Sun, this fusion process begins with protons and through a series of steps, these protons fuse and are turned into helium, and this transformation results in a release of energy that keeps the sun hot.

Nuclear Fission and Fusion – Sample Problems

Q.1. Calculate the energy released in the following spontaneous fission reaction:
\(238_{U} \rightarrow 95_{S r}+140_{X e}+3 n\)
Given the atomic masses to be:-
\(m\left(238_{U}\right)=238.050784\,\text {a.m.u}., m\left(95_{S r}\right)=94.919388\, \text {a.m.u}\).
\(m\left(140_{\mathrm{Xe}}\right)=139.921610\,\text {a.m.u}\). and \(m(n) = 1.008665\,{\text{a}}{\text{.m}}{\text{.u}}.,1~{\text{a}}{\text{.m}}{\text{.u}} = 931.5{\text{MeV}}\)
Sol:- The total mass of products,
\({m_{{\rm{products }}}} = m\left( {{{95}_{Sr}}} \right) + m\left( {{{140}_{Xe}}} \right) = 3{\mkern 1mu} \,{\rm{m}}(n)\)
\(\Rightarrow m_{\text {products }}=(94.919388+139.921610+3 \times 1.008665)\,\text {a.m.u.}\)
\(\Rightarrow m_{\text {products }}=237.866993\,\text {a.m.u.}\)
The mass lost \((\Delta m)\) in the reaction will be,
\(\Delta m=(238.050784-237.866993)\,\text {a.m.u.}\)
\(\Rightarrow \Delta m=0.183791\,\text {a.m.u.}\)
So, the energy released will be,
\(E=(0.183791\,\text {a.m.u.}) \times 931.5\,\text {MeV}\)
\(\Rightarrow E=171.2\,\text {MeV}\)
Hence, the energy released is equal to \(171.2\,\text {MeV}\).

Q.2 Calculate the energy released in the following fusion reaction:-
\(D_{1}^{2}+T_{1}^{3} \rightarrow H e_{2}^{4}+n_{0}^{1}\)
Given the atomic masses to be \(m\left( {D_1^2} \right) = 2.014102{\mkern 1mu} \,{\rm{a}}{\rm{.m}}{\rm{.u}}{\rm{.,}}\,m\left( {T_1^3} \right) = 3.016050{\mkern 1mu} \,{\rm{a}}{\rm{.m}}{\rm{.u}}{\rm{.}}\)
\(m\left(H e_{2}^{4}\right)= 4.002603\,\text {a.m.u.}\) and \(m\left(n_{0}^{1}\right)=1.008665\,\text {a.m.u.}\)
Sol:- 
The total mass of the products is,
\(m_{\text {products }}=m\left(H e_{2}^{4}\right)+m\left(n_{0}^{1}\right)\)
\(\Rightarrow m_{\text {products }}=(4.002603+1.008665)\,\text {a.m.u.}\)
\(\Rightarrow m_{\text {products }}=(5.011268)\,\text {a.m.u.}\)
Now the total mass of  reactants is,
\(m_{\text {reactants }}=m\left(D_{1}^{2}\right)+m\left(T_{1}^{3}\right)\)
\(\Rightarrow m_{\text {reactants }}=(2.014102+3.016050)\,\text {a.m.u.}\)
\(\Rightarrow m_{\text {reactants }}=(5.030152)\,\text {a.m.u.}\)
Then, the mass lost \((\Delta m)\) will be given by,
\(\Delta m=m_{\text {reactants }}-m_{\text {products }}\)
\(\Rightarrow \Delta m=(5.030152-5.011268)\,\text {a.m.u.}\)
\(\Rightarrow \Delta m=(0.018884)\,\text {a.m.u.}\)
So, from the  Einstein mass-energy  equivalence equation, the energy released will be,
\(E=(\Delta m) c^{2}\)
\(\Rightarrow E=(0.018884\,\text {a.m.u.}) \times \frac{1.6605 \times 10^{-27} \mathrm{~kg}}{1 \text { а.m.u. }} \times\left(3 \times 10^{8} \,\text {ms}^{-1}\right)^{2}\)
\(\Rightarrow E=2.82 \times 10^{-12}\,\text {J}\)
\(\Rightarrow E=17.6 \,\text {MeV} \quad\left(1 \mathrm{~J}=6.242 \times 10^{12} \,\mathrm{MeV}\right)\)

Summary

All the nuclear reactions are classified into two categories. These are Nuclear fission and Nuclear fusion. Nuclear fission is a nuclear reaction that splits a heavy atom into two or smaller ones. Fission releases energy when heavy nuclei are split into medium-mass nuclei. The major application of a fission reaction is the production of electricity via nuclear power plants

Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus; energy is released when light nuclei are fused to form medium-mass nuclei. Nuclear fusion explains the reaction between deuterium and tritium that produces a fusion (or hydrogen) bomb; fusion also explains the production of energy in the Sun, the process of nucleosynthesis, and the creation of heavy elements.

Fission reactions on the other hand are the type used in nuclear power plants and can be controlled. Atomic bombs and hydrogen bombs are examples of uncontrolled nuclear reactions.

PRACTICE QUESTIONS RELATED TO NUCLEAR FISSION & FUSION

FAQs on Nuclear Fission and Fusion

Q.1. Why is nuclear fission important?
Ans:
Nuclear fission produces energy for nuclear power and drives the explosion of nuclear weapons. Both uses are possible because certain substances called nuclear fuels undergo fission when struck by fission neutrons, and in turn emit neutrons when they break apart.

Q.2. What are the need and conditions required for nuclear fusion?
Ans: Very high-temperature and high-pressure environment is required. Extremely high energy is required to bring two or more protons close enough that nuclear forces overcome their electrostatic repulsion.

Q.3. What is a real-life example of fission?
Ans: A good example of a fission reaction is the nuclear power plant. In a nuclear power plant, this heat generated during fission is converted to electrical energy for our use in homes and factories.

Q.4. Can we say that nuclear energy is a highly productive source of power?
Ans: Yes, we can say that nuclear energy is a highly productive energy source as a massive amount of energy is produced by triggering one neutron. One kilogram of nuclear fuel can produce energy equivalent to thousands of tons of fossil fuel. Also, nuclear power stations are unaffected by seasonal conditions.

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We hope you find this article on Nuclear Fission and Fusion helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them. 

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