# Energy released in Nuclear Reactions

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**Einstein’s Equation Relating Mass and Energy**

** According to Albert Einstein, mass can be converted into energy and vice versa. His famous equation relating mass and energy is:

where E = energy ; m = mass and c = velocity of light.

** In nuclear reactions, a change in mass, Δm, is accompanied by release of energy, ΔE. Thus equation may be written as:

** If we substitute the value 3.00 × 10

^{10}cm/sec for the velocity of light, the equation (2) directly gives the relation between the energy change in ergs and the mass change in grams:
** Making use of the conversion factor 1 erg = 2.39 × 10

^{–11}kcal, we can express the energy change in k cals.
** Very often, in a nuclear reaction, the mass of the products is less than that of the reactants. The mass difference is converted into energy. Therefore by using equation (4), we can calculate the amount of energy released in a particular reaction.

** For example, in the equation:

The atomic mass difference between the reactants and products is 0.0186 gram. Using equation (4):

**Mass Defect**

** We know that atomic nucleus consists of protons and neutrons; collectively known as nucleons.

** It is found that the measured mass of nucleus is always less than the sum of the masses of the individual protons and neutrons which make it up.

** Let us take the example of helium,

^{4}He_{2}. It consists of two protons and two neutrons. Its mass may be calculated as:
** However, the experimental mass of the helium nucleus is only 4.00388. This is less by 0.03040 amu than that calculated above. This is called the mass defect of helium nucleus.

** The difference between the experimental and calculated masses of the nucleus is called the Mass defect or Mass deficit.

**(Experimental mass of nucleus) – (Mass of protons + Mass of neutrons) = Mass defect**

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**Nuclear Binding Energy**

** Atomic nucleus is made of protons and neutrons closely packed in a small volume. Although there exist intensive repulsive forces between the component protons, the nucleus is not split apart. This is so because the nucleons are bound to one another by very powerful forces. The energy that binds the nucleons together in the nucleus is called the Nuclear binding energy.

** Nuclear binding energy is The energy that binds the nucleons together in the nucleus.

** When a nucleus is formed from individual protons and neutrons, there occurs a loss of mass (mass defect). According to Einstein’s theory, it is this mass defect which is converted into binding energy. Hence binding energy is the energy equivalent of the mass defect. The various nuclei have different binding energies.

** Binding energy is a measure of the force that holds the nucleons together. Hence an energy equivalent to the binding energy is required to disrupt a nucleus into its constituent protons and neutrons. Since the nuclear energy is of an extremely high order, it is not easy to fission a nucleus.

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**Calculation of Binding Energy**

** The binding energy of a nucleus can be calculated from its mass defect by using Einstein’s

equation, ΔE = Δm × c

^{2}.**Solved Problem (1): What is the binding energy for**

^{11}B_{5}nucleus if its mass defect is 0.08181 amu?**Solution:**

Substituting values in Einstein’s equation,

No. of nuclei in one mole is 6.02 × 10

^{23}(Avogadro’s Law).
∴ Binding energy
for

^{11}B_{5}nucleus, ΔE, may be expressed as:####
**Binding energy per nucleon**

** It can be calculated by dividing the total binding energy by the sum of the number of protons and neutrons present in the nucleus:

** By plotting the binding energy per nucleon against the mass number, we get the graph shown in the following Figure:

** Figure shows the relative stability of the various nuclei.

** The greater the binding energy per nucleon the more stable is the nucleus. Thus the nuclei of about 60 atomic mass having maximum energy per nucleon are most stable e.g.,

^{56}Fe.
** The nuclei that are heavier or lighter than this have lower binding energies per nucleon and are less stable. Thus

^{235}U undergoes fission into lighter and more stable isotopes as^{139}Ba and 94Kr with the release of energy.
** Similarly two or more lighter nuclei (

^{2}H,^{3}H) with lower binding energy per nucleon combine or fuse together into a heavier and more stable nucleus. This is also accompanied by release of energy.####
**Equivalence of amu and Energy**

Since 1 amu is exactly equal to1/12 th of the mass of C

^{12}atom, therefore**Solved Problem (2): Calculate the binding energy per nucleon (in Mev) in He atom 42 He which has a mass of 4.00260 amu. Mass of an electron = 1.008655 amu and mass of 1 hydrogen atom = 1.007825 mass.**

**Solution:**

In Helium atom there are 2 electrons, 2 protons and 2 neutrons.

1 amu = 931.5 Mev

Binding energy per nucleon:

*Reference:**Essentials of Physical Chemistry /Arun Bahl, B.S Bahl and G.D. Tuli / multicolour edition.*

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