Binding Energy per Nucleon Curve
 In order to compare nuclear stability, it is more useful to look at the binding energy per nucleon
 The binding energy per nucleon is defined as:
The binding energy of a nucleus divided by the number of nucleons in the nucleus
 A higher binding energy per nucleon indicates a higher stability
 In other words, it requires more energy to pull the nucleus apart
 Iron (A = 56) has the highest binding energy per nucleon, which makes it the most stable of all the elements
By plotting a graph of binding energy per nucleon against nucleon number, the stability of elements can be inferred
Key Features of the Graph
 At low values of A:
 Nuclei tend to have a lower binding energy per nucleon, hence, they are generally less stable
 This means the lightest elements have weaker electrostatic forces and are the most likely to undergo fusion
 Helium (^{4}He), carbon (^{12}C) and oxygen (^{16}O) do not fit the trend
 Helium4 is a particularly stable nucleus hence it has a high binding energy per nucleon
 Carbon12 and oxygen16 can be considered to be three and four helium nuclei, respectively, bound together
 At high values of A:
 The general binding energy per nucleon is high and gradually decreases with A
 This means the heaviest elements are the most unstable and likely to undergo fission
Comparing Fusion & Fission
 Fusion occurs at low values of A because:
 Attractive nuclear forces between nucleons dominate over repulsive electrostatic forces between protons
 In fusion, the mass of the nucleus that is created is slightly less than the total mass of the original nuclei
 The mass defect is equal to the binding energy that is released since the nucleus that is formed is more stable
 Fission occurs at high values of A because:
 Repulsive electrostatic forces between protons begin to dominate, and these forces tend to break apart the nucleus rather than hold it together
 In fission, an unstable nucleus is converted into more stable nuclei with a smaller total mass
 This difference in mass, the mass defect, is equal to the binding energy that is released
 Fusion releases much more energy per kg than fission
 The energy released is the difference in binding energy caused by the difference in mass between the reactant and products
 Hence, the greater the increase in binding energy, the greater the energy released
 At small values of A (fusion region), the gradient is much steeper compared to the gradient at large values of A (fission region)
 This corresponds to a larger binding energy per nucleon being released
Worked Example
Step 1: Calculate the mass defect Δm
Number of protons, Z = 26
Number of neutrons, A – Z = 56 – 26 = 30
Mass defect, Δm = Zm_{p} + (A – Z)m_{n} – m_{total}
Δm = (26 × 1.673 × 10^{–27}) + (30 × 1.675 × 10^{–27}) – (9.288 × 10^{–26})
Δm = 8.680 × 10^{–28} kg
Step 2: Calculate the binding energy E of the nucleus
Binding energy, E = Δmc^{2}
E = (8.680 × 10^{–28}) × (3.00 × 10^{8})^{2} = 7.812 × 10^{–11} J
Step 3: Calculate the binding energy per nucleonWorked Example
The equation below represents one possible decay of the induced fission of a nucleus of uranium235.The graph shows the binding energy per nucleon plotted against nucleon number A.Calculate the energy released:
a) By the fission process represented by the equation
b) When 1.0 kg of uranium, containing 3% by mass of U235, undergoes fission
Part (a)
Step 1: Use the graph to identify each isotope’s binding energy per nucleon

 Binding energy per nucleon (U235) = 7.5 MeV
 Binding energy per nucleon (Sr88) = 8.6 MeV
 Binding energy per nucleon (Xe136) = 8.2 MeV
Step 2: Determine the binding energy of each isotope
Binding energy = Binding Energy per Nucleon × Mass Number

 Binding energy of U235 nucleus = (235 × 7.5) = 1763 MeV
 Binding energy of Sr88 = (88 × 8.6) = 757 MeV
 Binding energy of Xe135 = (136 × 8.2) = 1115 MeV
Step 3: Calculate the energy released
Energy released = Binding energy after (Sr + Xe) – Binding energy before (U)
Energy released = (1115 + 757) – 1763 = 109 MeV
Part (b)
Step 1: Calculate the energy released by 1 mol of uranium235

 There are N_{A} (Avogadro’s number) atoms in 1 mol of U235, which is equal to a mass of 235 g
 Energy released by 235 g of U235 = (6 × 10^{23}) × 214 MeV
Step 2: Convert the energy released from MeV to J

 1 MeV = 1.6 × 10^{–13} J
 Energy released = (6 × 10^{23}) × 214 × (1.6 × 10^{–13}) = 2.05 × 10^{13} J
Step 3: Work out the proportion of uranium235 in the sample

 1 kg of uranium which is 3% U235 contains 0.03 kg or 30 g of U235
Step 4: Calculate the energy released by the sample
Exam Tip
Checklist on what to include (and what not to include) in an exam question asking you to draw a graph of binding energy per nucleon against nucleon number:
 Do not begin your curve at A = 0, this is not a nucleus!
 Make sure to correctly label both axes AND units for binding energy per nucleon
 You will be expected to include numbers on the axes, mainly at the peak to show the position of iron (^{56}Fe)
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