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Mass Defect & Binding Energy (CIE A Level Physics)

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Mass defect & binding energy

Mass defect

  • 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)mnmtotal

  • 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 of carbon-12

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

  • 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
    • This means that it releases energy
  • This energy can be calculated using the equation:

E = Δmc2

  • In a typical nucleus, binding energies are usually measured in MeV
    • This is considerably larger than the few eV associated with the binding energy of electrons in the atom
  • Nuclear reactions involve changes in the nuclear binding energy whereas chemical reactions involve changes in the electron binding energy
    • This is why nuclear reactions produce much more energy than chemical reactions

Worked example

Calculate the binding energy per nucleon, in MeV, for the radioactive isotope potassium-40 (19K).

  • Nuclear mass of potassium-40 = 39.953 548 u
  • Mass of one neutron = 1.008 665 u
  • Mass of one proton = 1.007 276 u

Answer:

Step 1: Identify the number of protons and neutrons in potassium-40

  • Proton number, Z = 19
  • Neutron number, N = 40 – 19 = 21

Step 2: Calculate the mass defect, Δm

  • Proton mass, mp = 1.007 276 u
  • Neutron mass, mn = 1.008 665 u
  • Mass of K-40, mtotal = 39.953 548 u

Δm = Zmp + Nmnmtotal

Δm = (19 × 1.007276) + (21 × 1.008665) – 39.953 548

Δm = 0.36666 u

Step 3: Convert from u to kg

  • 1 u = 1.661 × 10–27 kg

Δm = 0.36666 × (1.661 × 10–27) = 6.090 × 10–28 kg

Step 4: Write down the equation for mass-energy equivalence

E = Δmc2

  • Where c = speed of light

Step 5: Calculate the binding energy, E

E = 6.090 × 10–28 × (3.0 × 108)2 = 5.5 × 10–11 J

Step 6: Determine the binding energy per nucleon and convert J to MeV

  • Take the binding energy and divide it by the number of nucleons
  • 1 MeV = 1.6 × 10–13 J

binding energy per nucleon = (5.5 × 10–11)/40 = 1.375 × 10–12 J

binding energy per nucleon = (1.375 × 10–12)/(1.6 × 10–13) = 8.594 MeV

Examiner Tip

The terms binding energy and mass defect can cause students confusion, so be careful when using them in your explanations.

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 separate all the nucleons.

The same goes for the term mass defect, make sure to only use this when all the nucleons are separated and not to describe the decrease in mass which occurs during radioactive decay.

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, more energy is required to separate the nucleons contained within a nucleus

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 have lower binding energies per nucleon than at large values of A, but they tend to be stable when N = Z
    • This means light nuclei have weaker electrostatic forces and will undergo fusion
    • The gradient is much steeper compared to the gradient at large values of A
    • This means that fusion reactions release a greater binding energy than fission reactions
  • At high values of A:
    • Nuclei have generally higher binding energies per nucleon, but this gradually decreases with A
    • This means the heaviest elements are the most unstable and will undergo fission
    • The gradient is less steep compared to the gradient at low values of A
    • This means that fission reactions release less binding energy than fission reactions
  • Iron (A = 56) has the highest binding energy per nucleon, which makes it the most stable of all the elements
  • 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

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

Examiner 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

Expertise: Physics

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.