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Theory: What is binding energy?
Using Mass Spectroscopy we can measure the mass of individual isotopes of an element, the observed mass of an atom.
We can use the rest mass of protons, neutrons and electrons given in the table below:
Subatomic Particle 
Rest Mass atomic mass units 
Rest Mass kg 
proton 
1.007276 
1.673 × 10^{27} 
neutron 
1.008665 
1.675 × 10^{27} 
electron 
0.000549 
9.109 × 10^{31} 
to calculate what the mass of an atom should be, the predicted mass of an atom:
predicted mass = mass of all protons + mass of all neutrons + mass of all electrons
We find that the observed mass of an atom is less than the mass we predict:
predicted mass > observed mass
This difference in mass is known as the mass defect (Δm):
mass defect = predicted mass  observed mass
Mass cannot just disappear, this would be contrary to the Law of Mass Conservation, but it can be converted into energy.
We can use Einstein's famous equation for massenergy equivalence, E = mc^{2},
E = energy (J)
m = mass (kg)
c = speed of light = 3 × 10^{8} m s^{1} (vacuum)
to convert the mass defect, Δm in kilograms (kg), into an energy term in Joules (J):
E = Δm × c^{2}
= Δm × 3 × 10^{8}
= Δm × 9 × 10^{16}
The binding energy must be related to the energy released when the subatomic particles come together to form an atom:
We could think of this like a chemical equation:
protons + neutrons → nucleus + energy
The binding energy of a nucleus is in reality defined the other way around, that is, as the energy required to completely separate the individual components of the nucleus:
If we were to write a chemical equation for this it would look like the one below:
nucleus + binding energy → protons + neutrons
We would expect that the binding energy of a nucleus would increase as the number of nucleons (numbers of protons and neutrons) in the nucleus increases, that is, because there are more particles in the nucleus (nucleons) it will take more energy to separate them completely.
So a better way of comparing the binding energy of different nuclei is to compare the binding energy per nucleon:
binding energy per nucleon = binding energy for the atom ÷ number of nucleons
binding energy per nucleon = binding energy for the atom ÷ (no. protons + no. neutrons)
Since the number of protons and neutrons in the nucleus of an atom is given by its mass number (A), we can write:
binding energy per nucleon = binding energy for the atom ÷ mass number
binding energy per nucleon = binding energy for the atom ÷ A
You will sometimes see the binding energy given the symbol B (or B.E.) in which case we can write:
binding energy per nucleon = B ÷ A
A nucleus that is very stable will require a lot of energy to break it up, so it will have a large value for the binding energy per nucleon.
A nucleus that is not very stable will require much less energy to break it up, so it will have a smaller value for the binding energy per nucleon.
Binding Energy Calculations in Joules
Steps to solve binding energy problems in SI units:
Step 1: Calculate the mass defect (Δm) in kilograms (kg)
Step 2: Convert this mass defect into energy in joules (J)
E = Δm × c^{2}
E = binding energy in joules (J)
Δm = mass defect in kilograms (kg)
c = speed of light in metres per second (m s^{1}) = 3 × 10^{8} m s^{1}
Example: Calculate the binding energy for helium4 using SI units
 Step 1: Calculate the mass defect in SI units of kilograms (kg)
We can predict the mass of an atom of helium4 using the SI units for the rest mass of protons, neutrons and electrons given above.
No. protons = atomic number = Z = 2 (from Periodic Table)
No. electrons = no. protons = 2 (for a neutral atom number of protons = number of electrons)
No. neutrons = mass number (A)  atomic number (Z) = 4  2 = 2 (mass number given in name)
predicted mass 
= 
(no. protons × mass proton) 
+ 
(no. neutrons × mass neutron) 
+ 
(no. electrons × mass electron) 

= 
(2 × 1.673 × 10^{27}) 
+ 
(2 × 1.675 × 10^{27}) 
+ 
(2 × 9.109 × 10^{31}) 

= 
(3.346 × 10^{27}) 
+ 
(3.35 × 10^{27}) 
+ 
(1.8218 × 10^{30}) 

= 
6.6978 × 10^{27} kg 




Using Mass Spectroscopy, the mass of an atom of helium4 is found to be 6.646 × 10^{27} kg
Mass defect 
= 
predicted mass 
 
observed mass 
Δm 
= 
6.6978 × 10^{27} 
 
6.646 × 10^{27} 

= 
5.18 × 10^{29} kg 


 Step 2: Convert this mass defect (kg) into energy (J)
E 
= 
Δm 
× 
c^{2} 

= 
5.18 × 10^{29} 
× 
(3 × 10^{8})^{2} 

= 
5.18 × 10^{29} 
× 
9 × 10^{16} 

= 
4.662 × 10^{12} J 


Calcuating the binding energy for helium4 using atomic mass units (u)
 Step 1: Calculate the mass defect in kilograms (kg)
We can predict the mass of an atom of helium4 using the nonSI units of "atomic mass units" (u) for the rest mass of protons, neutrons and electrons given above.
No. protons = atomic number = Z = 2 (from Periodic Table)
No. electrons = no. protons = 2 (for a neutral atom number of protons = number of electrons)
No. neutrons = mass number (A)  atomic number (Z) = 4  2 = 2 (mass number given in name)
predicted mass 
= 
(no. protons × mass proton) 
+ 
(no. neutrons × mass neutron) 
+ 
(no. electrons × mass electron) 

= 
(2 × 1.007276) 
+ 
(2 × 1.008665) 
+ 
(2 × 0.000549) 

= 
2.014552 
+ 
2.01733 
+ 
0.001098 

= 
4.03298 u 




Using Mass Spectroscopy, the mass of an atom of helium4 is found to be 4.003 u
Mass defect 
= 
predicted mass 
 
observed mass 
Δm 
= 
4.03298 
 
4.003 

= 
0.02998 u 


Convert this mass in atomic mass units (u) to a mass in kilograms (kg):
1 atomic mass unit = ^{1}/_{12} × mass carbon12 atom in kg
1 atomic mass unit = ^{1}/_{12} × 1.993 × 10^{26} = 1.6608 × 10^{27} kg
0.02998 u = 0.02998 × 1.6608 × 10^{27} = 4.980 × 10^{29} kg
(Note there is a difference between the mass defect in kg calculate here and the one above. This is due to rounding errors in the rest masses of particles given in kg in the table)
 Step 2: Convert this mass defect (kg) into energy (J)
E 
= 
Δm 
× 
c^{2} 

= 
4.980 × 10^{29} 
× 
(3 × 10^{8})^{2} 

= 
4.980 × 10^{29} 
× 
9 × 10^{16} 

= 
4.482 × 10^{12} J 


Energy Conversions: J and MeV
Physicists often use the electronvolt (eV) as a unit of energy.
1 electronvolt (1 eV) is the energy gained by an electron accelerated through a potential difference of 1 volt (1 V).
1 eV = 1.602 × 10^{19} J
1 megaelectronvolt (1 MeV) = 1,000,000 eV
1 MeV = 1,000,000 × 1.602 × 10^{19} = 1.602 × 10^{13} J
In order to convert the binding energy in joules (J) to megaelectronvolts (Mev), we need to divide the energy in J by 1.602 × 10^{13} J/MeV:
energy (MeV) = energy (J) ÷ 1.602 × 10^{13} (J/MeV)
Energy Conversion Example: Convert binding energy of helium4 in J to MeV
binding energy of helium4 = 4.482 × 10^{12} J (calculated above)
binding energy of helium4 in MeV = (4.482 × 10^{12}) ÷ (1.602 × 10^{13}) = 27.98 MeV
Binding Energy per Nucleon
Binding energy per nucleon can be calculated using the binding energy of the nucleus (as above) and the number of protons and neutrons in the nucleus of the atom (number of nucleons):
binding energy per nucleon = binding energy for the nucleus ÷ number of nucleons
binding energy per nucleon = binding energy for the nucleus ÷ (number of protons + number of neutrons)
binding energy per nucleon = binding energy for the nucleus ÷ mass number
Calculating Binding Energy per Nucleon for Helium4
Binding energy per nucleon tells us how much energy per nucleon we need to apply in order to completely separate the components of a nucleus.
The binding energy for helium4 was calculate above to be 4.482 × 10^{12} J (or 27.98 MeV).
The number of nucleons = the number of protons + the number of neutrons = mass number = A = 4
binding energy per nucleon (J/nucleon) = 4.482 × 10^{12} J ÷ 4 = 1.1205 × 10^{12} J
binding energy per nucleon (MeV/nucleon) = 27.98 MeV ÷ 4 = 6.995 MeV
Binding Energy per Nucleon and Isotope Stability
Carbon12 is known to be a stable isotope of carbon, whereas carbon14 (used for radiocarbon dating) is an unstable isotope of carbon.
The mass of the carbon12 atom is defined as being exactly 12.000 u, whereas the mass of the carbon14 atom has been found to be 14.003 u.
Let us calculate, and compare, the binding energy per nucleon for each of these isotopes of carbon.

carbon12 
carbon14 
no. protons = Z = 
6 
6 
no. neutrons = A  Z = 
12 6 = 6 
14  6 = 8 
no. electrons = no. protons = 
6 
6 
predicted mass (u) = 
(6×1.007276)+(6×1.008665)+(6×0.000549) 
(6×1.007276)+(8×1.008665)+(6×0.000549) 
= 12.09894 u 
= 14.11627 u 
observed mass (u) = 
12.000 
14.003 
mass defect (u) = 
12.09894  12.000 
14.11627  14.003 
= 0.09894 u 
= 0.11327 u 
mass defect (kg) 
0.09894 × 1.6608 × 10^{27} 
0.11327 × 1.6608 × 10^{27} 
= 1.643 × 10^{28} kg 
= 1.881 × 10^{28} kg 
binding energy (J) 
1.643 × 10^{28} × (3 × 10^{8})^{2} 
1.881 × 10^{28} × (3 × 10^{8})^{2} 
= 1.4787 × 10^{11} J 
= 1.6929 × 10^{11} J 
binding energy per nucleon (J) 
1.4787 × 10^{11} J ÷ 12 
1.6929 × 10^{11} J ÷ 14 
= 1.23225 × 10^{12} J 
= 1.20921 × 10^{12} J 
binding energy per nucleon (MeV) 
1.23225 × 10^{12} J ÷ 1.602 × 10^{13} J/MeV 
1.20921 × 10^{12} J ÷ 1.602 × 10^{13} J/MeV 
= 7.692 MeV 
= 7.548 MeV 
We can see that the more stable carbon isotope, carbon12, has a higher binding energy per nucleon (7.692 MeV) than the unstable (radioactive) carbon14 isotope (7.548 MeV).
Drill
Quick Question 1
Calculate the nuclear binding energy per nucleon in MeV for silicon29 which has an observed mass of 28.976 u.
Mass of a proton = 1.007276 u
Mass of a neutron = 1.008665 u
mass of an electron = 0.000549 u