# IB DP Physics: SL复习笔记7.2.4 Binding Energy per Nucleon Curve

### 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 (4He), carbon (12C) and oxygen (16O) do not fit the trend
• Helium-4 is a particularly stable nucleus hence it has a high binding energy per nucleon
• Carbon-12 and oxygen-16 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 = Zmp + (A – Z)mn – mtotal

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 = Δmc2

E = (8.680 × 10–28) × (3.00 × 108)2 = 7.812 × 10–11 J

Step 3: Calculate the binding energy per nucleon Worked Example

The equation below represents one possible decay of the induced fission of a nucleus of uranium-235. 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 U-235, undergoes fission

Part (a)

Step 1: Use the graph to identify each isotope’s binding energy per nucleon • Binding energy per nucleon (U-235) = 7.5 MeV
• Binding energy per nucleon (Sr-88) = 8.6 MeV
• Binding energy per nucleon (Xe-136) = 8.2 MeV

Step 2: Determine the binding energy of each isotope

Binding energy = Binding Energy per Nucleon × Mass Number

• Binding energy of U-235 nucleus = (235 × 7.5) = 1763 MeV
• Binding energy of Sr-88 = (88 × 8.6) = 757 MeV
• Binding energy of Xe-135 = (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 uranium-235

• There are NA (Avogadro’s number) atoms in 1 mol of U-235, which is equal to a mass of 235 g
• Energy released by 235 g of U-235 = (6 × 1023) × 214 MeV

Step 2: Convert the energy released from MeV to J

• 1 MeV = 1.6 × 10–13 J
• Energy released = (6 × 1023) × 214 × (1.6 × 10–13) = 2.05 × 1013 J

Step 3: Work out the proportion of uranium-235 in the sample

• 1 kg of uranium which is 3% U-235 contains 0.03 kg or 30 g of U-235

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 (56Fe) 