# Nuclear Binding Energy

Nuclear binding energy

It may be defined as the energy released when the given number of protons and neutrons combine to form its nucleus. In other words, it is the energy required to disrupt its nucleus into its constituents. This released energy corresponds to mass defect $(\Delta m)$, e.e., the difference between the expected mass and the actual mass of the atoms. It may be written as:

Binding energy = [mass of neutrons + mass of protons + mass of electron] of the atom – actual mass of the atom

= mass defect a.m.u.

931.5 × Mass defect (MeV)

(since 1 a.m.u. = 931.5 MeV)
Aston expressed the variation of isotopic mass from mass number in terms of packing fraction for each isotope as: $\text{pacing fraction} = \dfrac{\text{Isotopic mass- Mass number}}{\text{Mass number}} \times 10^4$
It is evident that the smaller value of packing fraction shows grater stability of the atom and vice versa.

Example 1: the fission fragments in the thermal neutron fission of $^{325}_{92}U$ found to be $^{98}_{42}Mo \text{and} ^{136}_{54}Xe$ what are the elementary particles and energy, released in the reaction? (Isotopic mass of $^{325}U ^{98}Mo, ^{136}Xe \text{and} ^1_0n$ are 235.044, 97.906, 135.907 and 1.0086 amu, respectively).

Solution:

According to the example, $^{325}_{92}U + ^1_0n \to ^{98}_{42}Mo + ^{136}_{54}Xe$

On balancing the equation by elementary particles the above equation many be written as: $^{235}_{92}U + ^1_0n \to ^{98}_{42}Mo + ^{136}_{54}Xe + 2 ^1_0n + 4 ^1_{-1}e\\ \therefore$elementary particles released are neutrons and electrons.
Mass of the reactants = 235.044 + 236.0526

Mass of the products

= 97.906 + 135.907 + 2(1.0086) + 4(0.0005486) = 235.8324

Therefore mass defect $(\Delta m)$ = 235.8324 – 236.0526 = -0.2202 amu

Since the value of $(\Delta m)$ is negative, it means energy is released during the process.

Energy = 0.2202 × 931 MeV

Example 5: calculate the loss in mass accompanying combustion of one mole of fuel to release 1800KJ heat energy

Solution: since $E = \Delta m c^2$

Where E = released energy, $\Delta m$ and c = velocity of ligh

Hence $\Delta m = \dfrac{E}{c^2} = \dfrac{1800 \times 10^3 J}{(3 \times 10^8 m s^{-1})^2} \\ 2 \times 10^{-11} kg (\text{since} J = kg m^2s^{-2})$

Henry Danlos use Ra for the first time in the treatment of Tuber Culons Skin Lesin disease.
CO – 60 $\to$ in treatment of Cancer.
I-131 $\to$in treatment of Thyroid Gland.
P-32 $\to$ in treatment of Leukemia.
Na-24 $\to$ to trace flow of blood
I-123 $\to$ for imaging the brain.
Au-198 $\to$ blood cancer treatment.
Technetium-99M $\to$ for scanning bones.

Other uses $^3_1H$ is used as a Tracer in absorption of water by plant roots.

P-32 $\to$ TO Truce plant growth etc.

O-18 $\to$ in photosynthesis. $6CO_2 + 12H_2O^{18} \to C_6H_{12}O_6 + O_2^{18} + 6 H_2O$

Na-24 to find leakage pipes.

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