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Binding energy is defined as the energy corresponding to a mass defect. A nucleus resembles an inflexible spherical ball formed by assembling a large number of tiny spherical balls known as nucleons. A glueing agent is needed to bind the nucleons that have been gathered. To transfer the energy resulting in a mass defect, each nucleon must contribute some of its mass. The energy required to break down a nucleus into its constituent nucleons is known as its Binding Energy.
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Key Terms: Binding Energy, Mass Defect, Nucleons, Atoms, Neutrons, Protons, Nucleus, Binding Energy per Nucleon, Subatomic Particles
What is Binding Energy?
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Binding energy is the amount of energy necessary to break down or split a nucleus into its constituent nucleons. The protons, neutrons, and other nuclear particles that make up the nucleus of each atom are known as nucleons. The strong nuclear forces keep the nucleons together, and the more tightly the nucleus components are linked, the more binding energy is required to separate them.
Nucleons consist of protons, neutrons, and other nuclear particles that make up an atom's nucleus. The nucleons are bound together by forces known as strong nuclear forces. Similarly, the higher the strength of forces, the higher the binding energy required to separate them. In most cases, the binding energy is always positive. This is because it takes energy to move these nucleons away from each other, which are attracted to one another by a strong nuclear force.
According to Einstein's equation E = mc2, the mass of an atomic nucleus is smaller than the sum of the individual masses of the unbound component protons and neutrons. This missing mass is referred to as a Mass Defect since it was discharged when the nucleus was created.
Also Check:
| Concepts Related to Binding Energy Formula | ||
|---|---|---|
| Nuclei | Mass-Energy Equivalence | Radioactive Decay Formula |
| Deuteron Mass | Geiger Counter | Electron Mass |
Main Features of Binding Energy Curve
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The binding energy per nucleon is the average energy per nucleon required to break a nucleus into its component nucleons.
- For nuclei with a middle mass number of 30 < A < 170, the binding energy per nucleon (Ebn) is virtually constant and is almost independent of the atomic number. The curve has a maximum value of around 8.75 MeV for A = 56 and a value of 7.6 MeV for A = 238.
- Both light nuclei (A<30) and heavy nuclei (A>170) have lower binding energy per nucleon.
Based on the above two observations, the following conclusions can be drawn:
- The force is attractive and strong enough to create binding energy of a few MeV per nucleon.
- The range of binding energy constancy is from 30 to 170, which is due to the short-range nature of nuclear force.
- In comparison, a heavy nucleus A = 240 has a lower binding energy per nucleon than an A = 120 nucleus; hence, if an A = 240 nucleus splits into two A = 120 nuclei, the nucleons become more securely bonded, implying that energy is liberated.
Binding Energy Formula
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Binding Energy formula can be denoted as:
| Binding Energy = (mass defect).(c2) = [(Zmp + Nmn) – mtot] c2 |
Where,
- c = speed of light in vacuum = 2.9979 x 108 m/s.
The above equation defines the difference in mass after the nucleus splits, which is referred to be a Mass Defect.
Because Z is the number of protons and N is the number of neutrons, the nucleus mass must be the sum of both, which is Zmp + Nmn. The resulting mass defect is this sum minus the total mass when the particles collide (mtot).
Applications of Binding Energy
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Binding energy is used widely in the realm of Nuclear Physics. It's mostly helpful in two fields:
Both of these fields look at how light nuclei fuse or divide. It is also used to generate energy and as a nuclear weapon.
Types of Binding Energy
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Previous Year Questions
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Things to Remember
- The units of Binding energy are reported to be Joules or eV per nucleus, which are energy units.
- The binding energy is the amount of energy necessary to break down or split a nucleus into its constituent nucleons.
- It's worth noting that when the nucleons are all together, the overall mass of the nucleus is lower than when the particles are separated. For all atoms, it is constant.
- If the binding energy is positive or zero, the nucleus will split and escape into space; hence, it is negative.
- The protons and neutrons in a nucleus are bound together by strong nuclear force and contain potential energy (and negligible kinetic energy).
- Electron binding energy is produced by the electron's electromagnetic interaction with the nucleus and other electrons of the atom, which is mediated by photons.
- The atomic binding energy is the amount of energy necessary to deconstruct an atom into free electrons and nuclei.
- When protons and neutrons react to form bonds, the nuclear binding energy is released. The binding energy gives away a considerable quantity of energy as the number of nucleons grows.
Also Check:
Sample Questions
Ques. Calculate the per-nucleon binding energy for an alpha particle with a mass defect of 0.0292amu. (5 marks)
Ans. Given: Mass defect = 0.0292amu
(1 amu = 1.6606 x 10-27 kg)
Mass defect = (0.0292).(1.6606 x 10-27) = 0.04848 x 10-27 kg/nucleus
Using, DE = Dmc2 to convert this mass to energy, using c = 2.9979 x 108 m/s.
E = (0.04848 x 10-27).(2.9979 x 108 )2
= 0.4357 x 10-11 J/nucleus
Converting energy from kJ/mole to kJ/mole (1 kJ = 1000 J)
Convert to mole by multiplying with the Avogadro number (6.022 x 1023 nuclei/mol)
Therefore, E = (0.4357 x 10-11) (6.022 x 1023)/1000
Binding Energy, E = 2.62378 x 109 kJ/mole
Ques. Calculate the binding energy of a beryllium-4 nucleus with a mass of 9.012182 u. (3 marks)
Ans. Calculate the mass defect of beryllium as your first step.
There are 4 protons and 5 neutrons in this atom. One proton has a mass of 1.00728 amu, while each neutron has a mass of 1.00867 amu:
[5 neutrons(1.00867 u) + 4 protons(1.00728 u)] – 9.012182 u
= 0.060288 u 1.6606 10-27 kg/amu
= 1.00114 10-28 kg/nucleus
Thus, the binding energy is BE = (m) c2 = 0.060288 u (2.9979 × 108 m/s) 2
= 8.9976 × 10-12 J/nucleus.
Ques. Atomic mass of 8O16 is 16. (5 marks)
Mass of one neutron =1.00893 amu
Mass of one proton =1.00757 amu
Mass of one electron =0.0005486 amu
Calculate its mass defect & binding energy?
(8O16 have 8p, 8n & 8e)
Ans. The mass of the nucleus is equal to the sum of the masses of 8p and 8n.
= 8 × 1.00757 + 8 × 1.00893
= 8.06056 + 8.07144
= 16.1320 a.m.u
Mass of nucleus is equal to the mass of 8e subtracted from the atomic mass
= 16 - (8×0.0005486)
= 16 - 0.0043888
= 15.9956112 amu
Mass defect (Δm) = 16.1320 - 15.9956
(Δm) = 0.1364 amu
The Binding energy (B) is Δm × 931 MeV = 0.1364×931
B = 126.988 MeV
Ques. Calculate 33As75 ‘s Δm & Binding energy of 33As75. One proton, one neutron, and one electron have a mass of 1.0073 amu, 1.0087 amu, and 0.0055 amu, respectively, while the atomic mass of 33As75 is 74.9216 amu. (5 marks)
Ans. 33As75 = 33e, 33p & 42n
Mass of nucleus is the sum of the mass of 33p and mass of 42n
= 33×1.0073+42×1.0087
= 33.2409 +42.3654
= 75.6063 amu
Mass of nucleus is Mass of 33e subtracted from atomic mass.
= 74.9216 –33×0.00055
= 74.9216- 0.01815
= 74.90345 amu
Δm = 75.6063–74.90345
Δm = 0.70285 amu
Binding Energy is calculated by Δm×931 MeV = 0.70285×931
B = 654.35 MeV Ans
Ques. Calculate 2He4’s Δm & binding energy. The atomic mass of this object is 4.0039 amu. One n and one p combined have a mass of 2.0165 amu. 0.0005486amu is the mass of one electron. (5 marks)
Ans. 2He4= 2e, 2p&2n
Mass of Nucleus is the atomic mass - the mass of 2e = 4.0039–2×0.0005486
= 4.0028 amu
Mass of nucleus is the sum of mass of 2p and mass of 2n = 2(mass of P + mass of n)
= 2×2.0165
= 4.033 amu
Mass defect (Δm) is 0.0302 amu & Binding Energy is Δm×931 MeV = 0.0302 × 931
= 28.12 MeV.
Ques. An element's binding energy per nucleon is 7.14 MeV. Calculate the amount of nucleons in the nucleus if the element's binding energy is 28.6MeV. (2 marks)
Ans. No. of nucleons =?
B = 28.6MeV
Binding Energy per Nucleon = 7.14MeV
No. of nucleons =4
Ques. An element's binding energy is 64 MeV. The nucleon's binding energy is 6.39 MeV. How many protons and neutrons are there in the nucleus? (2 marks)
Ans. No. of nucleons is the sum of no. of protons and no. of neutrons
B= 64MeV
No. of Nucleons =10
Total no. of neutrons & protons = 10
Ques. 2He4 has a binding energy of 28.8 MeV. Calculate the nucleon's binding energy. (2 marks)
Ans. B = 28.8 MeV
No. of Nucleons in 2He4 = 4
Binding energy per nucleon = 7.2 MeV
Ques. Determine the binding energy per nucleon for 3Li7 if its mass is 7.01653 amu. (3 marks)
Ans. E = ΔE/A = Δm × 931/ A MeV
Δm = (3mp + 4mn) – mass of Li7
= (3×1.00759 + 4×1.008898) - 7.01653
= 0.04216
ΔE =((0.04216×931)/7)
= 39.25 / 7
= 5.6 MeV
Ques. How much energy is released in the following reaction? (5 marks)
1H2 + 1H2 = 2He4
if the Binding Energy/Nucleon of 1H2 and 2He4 are 1.123 MeV and 7.2 MeV respectively.
Ans. B. E. of 1H2, ΔE = 1.125
E = A × ΔE
E = 2 × 1.125
= 2.25 MeV
B.E. of two 1H2 = 2.25
Ed = 4.5 MeV
Binding Energy of an α-particle = 4 × 7.2
Ea = 28.8
Energy released ER = Ea – Ed
Energy Released = 28.8 - 4.5
= 24.3 MeV
Ques. Calculate the deuteron's mass defect and binding energy. The mass of the deuteron is mD=3.34359×10-27 kg or 1875.61MeV/c2 (2 marks)
Ans. Δm =mp+mn-mD =938.28MeV/c2+939.57MeV/c2-1875.61MeV/c2 =2.24MeV/c2
The binding energy of the deuteron:
Eb=(Δm)c2=(2.24MeV/c2) (c2)=2.24MeV
Ques. Calculate 4He( particle)’s binding energy per nucleon. (3 marks)
Ans. For 4He, we have Z=N=2. The total binding energy is
BE= {[2mp+2mn] – m (4He)} c2
These masses are m (4He) =4.002602u,mp=1.007825u, and mn=1.008665u. Thus we have,
BE=(0.030378u)c2
Noting that 1u=931.5MeV/c2, we find
BE =(0.030378)(931.5MeV/c2)c2 =28.3MeV
Since A=4, the total nucleon binding energy is
Binding Energy=7.07MeV/ nucleon.
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