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medium difficulty10 min

Mass-Energy Equivalence and Binding Energy

Mass-energy equivalence states that mass and energy are related by E = mc2. In a nucleus, the mass defect is the difference between the total mass of separated nucleons and the actual nuclear mass, and the corresponding binding energy is the energy needed to separate the nucleus into its nucleons.

In one line: Mass lost is energy locked. Per nucleon tells stability, not just total energy.

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Understand it

  • A stable nucleus has less mass than the sum of its free protons and neutrons.
  • This missing mass is called mass defect and appears as binding energy.
  • Binding energy per nucleon measures average stability per nucleon.
  • The binding energy per nucleon curve rises sharply for light nuclei, reaches a maximum near iron and nickel region, and decreases slowly for heavier nuclei.
  • This graph explains why fusion of light nuclei and fission of heavy nuclei can release energy.

Example

If the mass defect of a nucleus is 0.030 u, then binding energy = 0.030 × 931.5 MeV = 27.945 MeV, using 1 u = 931.5 MeV/c2.

Formula and rule

E = mc2, where E is energy in joule, m is mass in kilogram, and c = 3 × 108 m/s.

For nuclear calculations, E = delta m c2 and 1 u corresponds to about 931.5 MeV.

Binding energy per nucleon = total binding energy/A.

Use these relations when mass defect or nuclear stability is involved.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A nucleus has mass defect 0.050 u and mass number 10.

  1. 1Total binding energy = 0.050 × 931.5 MeV = 46.575 MeV.
  2. 2Binding energy per nucleon = 46.575/10 = 4.6575 MeV per nucleon.
  3. 3This value gives the average stability per nucleon for comparison with other nuclei.

Method you can reuse

  1. 1Write the given masses in consistent units.
  2. 2Find mass defect by subtracting actual nuclear mass from the sum of separated nucleon masses.
  3. 3Convert mass defect to energy using 931.5 MeV per u or E = delta m c2.
  4. 4Divide by A if binding energy per nucleon is required.
  5. 5Interpret stability using the graph trend.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Why is binding energy per nucleon more useful than total binding energy for comparing nuclear stability?

Think first, then show answer
Binding energy per nucleon is more useful because it gives the average binding of each nucleon, so nuclei of different sizes can be compared fairly.

Common wrong answer

A frequent wrong answer is to multiply mass defect in atomic mass unit directly by c2 in SI units without converting u to kilogram. Use either SI units fully or use the MeV shortcut consistently.

Don't mix this up

Students often forget that mass defect must be converted to energy using c2 or the shortcut 1 u = 931.5 MeV/c2. Another common error is calling higher total binding energy more stable without checking binding energy per nucleon.

Exam tip

For stability comparison, use binding energy per nucleon, not total binding energy alone. For energy release, compare binding energy before and after the reaction.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Questions commonly ask for mass defect calculation, binding energy in MeV, binding energy per nucleon, interpretation of the curve, or explanation of energy release in fission and fusion.
  1. 1

    Enough for 1 mark

    Mass-energy equivalence states that mass and energy are related by E = mc2. In a nucleus, the mass defect is the difference between the total mass of separated nucleons and the actual nuclear mass, and the corresponding binding energy is the energy needed to separate the nucleus into its nucleons.

  2. 2

    Add for 2 marks

    Example: If the mass defect of a nucleus is 0.030 u, then binding energy = 0.030 × 931.5 MeV = 27.945 MeV, using 1 u = 931.5 MeV/c2.

  3. 3

    Add for 3 marks

    A stable nucleus has less mass than the sum of its free protons and neutrons.

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