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Mass Defect & Binding Energy (CIE A Level Physics)

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Mass Defect & Binding Energy

  • Experiments into nuclear structure have found that the total mass of a nucleus is less than the sum of the masses of its constituent nucleons
  • This difference in mass is known as the mass defect
  • Mass defect is defined as:

The difference between the mass of a nucleus and the sum of the individual masses of its protons and neutrons

  • The mass defect Δm of a nucleus can be calculated using:

Δm = Zmp + (A – Z)mn – mtotal

  • Where:
    • Z = proton number
    • A = nucleon number
    • mp = mass of a proton (kg)
    • mn = mass of a neutron (kg)
    • mtotal = measured mass of the nucleus (kg)

Mass Defect

Binding Energy, downloadable AS & A Level Physics revision notes

A system of separated nucleons has a greater mass than a system of bound nucleons

  • Due to the equivalence of mass and energy, this decrease in mass implies that energy is released in the process
  • Since nuclei are made up of neutrons and protons, there are forces of repulsion between the positive protons
    • Therefore, it takes energy, ie. the binding energy, to hold nucleons together as a nucleus

  • Binding energy is defined as:

The energy required to break a nucleus into its constituent protons and neutrons

  • Energy and mass are proportional, so, the total energy of a nucleus is less than the sum of the energies of its constituent nucleons
  • The formation of a nucleus from a system of isolated protons and neutrons is therefore an exothermic reaction - meaning that it releases energy
  • This can be calculated using the equation:

E = Δmc2

Exam Tip

Avoid describing the binding energy as the energy stored in the nucleus – this is not correct – it is energy that must be put into the nucleus to pull it apart.

Binding Energy per Nucleon

  • 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

Graph of Binding Energies for Nuclei of Different Masses

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

Worked example

Determine the binding energy per nucleon of iron-56, Fe presubscript 26 presuperscript 56, in MeV.

Mass of a neutron = 1.675×10-27 kg

Mass of a proton = 1.673×10-27 kg

Mass of an iron-56 nucleus = 9.288×10-26 kg

Answer:

Step 1: Calculate the mass defect

  • 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 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

binding space energy space per space nucleon space equals fraction numerator italic space E over denominator A end fraction

E over A space equals space fraction numerator 7.812 cross times 10 to the power of negative 11 end exponent over denominator 56 end fraction space equals space 1.395 cross times 10 to the power of negative 12 end exponent space straight J

Step 4: Convert to MeV

  • J → eV: divide by 1.6 × 10-19
  • eV → MeV: divide by 106

binding space energy space per space nucleon space equals space fraction numerator 1.395 cross times 10 to the power of negative 12 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction space

binding space energy space per space nucleon space equals space 8 space 718 space 750 space eV space equals space 8.7 space MeV

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:

  • You will be expected to draw the best fit curve AND a cross to show the anomaly that is helium
  • 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)

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Leander

Author: Leander

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.