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Mass-energy equivalence explains how mass and energy are two forms of the same physical quantity. This cheat sheet helps students solve problems using E=mc2E = mc^2, including conversions between kilograms, joules, electronvolts, and atomic mass units. Worked examples are especially important because the numbers are very large or very small, so unit control matters.

Key Facts

  • Mass-energy equivalence is given by E=mc2E = mc^2, where EE is energy in joules, mm is mass in kilograms, and c=3.00×108 m/sc = 3.00 \times 10^8\ \text{m/s}.
  • A mass change Δm\Delta m corresponds to an energy change ΔE=Δmc2\Delta E = \Delta m c^2.
  • Because c2=9.00×1016 m2/s2c^2 = 9.00 \times 10^{16}\ \text{m}^2/\text{s}^2, even a tiny mass can correspond to a large amount of energy.
  • The joule conversion for electronvolts is 1 eV=1.602×1019 J1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.
  • The atomic mass unit energy equivalent is 1 u931.5 MeV/c21\ \text{u} \approx 931.5\ \text{MeV}/c^2, so ΔE in MeV=Δm in u×931.5\Delta E\text{ in MeV} = \Delta m\text{ in u} \times 931.5.
  • Mass defect in a nucleus is Δm=mseparate particlesmnucleus\Delta m = m_{\text{separate particles}} - m_{\text{nucleus}}.
  • Binding energy is Eb=Δmc2E_b = \Delta m c^2, and it represents the energy required to separate a nucleus into its nucleons.
  • In annihilation, the total rest mass converted to energy follows Etotal=mtotalc2E_{\text{total}} = m_{\text{total}}c^2.

Vocabulary

Mass-energy equivalence
The principle that mass and energy are related by E=mc2E = mc^2 and can be converted into each other.
Rest energy
The energy an object has because of its mass when it is not moving, given by E0=mc2E_0 = mc^2.
Mass defect
The missing mass Δm\Delta m between separate particles and the bound system they form.
Binding energy
The energy Eb=Δmc2E_b = \Delta m c^2 needed to break a bound system, such as a nucleus, into separate parts.
Electronvolt
A small energy unit used in atomic and nuclear physics, where 1 eV=1.602×1019 J1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.
Atomic mass unit
A mass unit used for atoms and nuclei, with 1 u1.6605×1027 kg1\ \text{u} \approx 1.6605 \times 10^{-27}\ \text{kg}.

Common Mistakes to Avoid

  • Using grams instead of kilograms in E=mc2E = mc^2 is wrong because the SI unit of mass must be kilograms when energy is calculated in joules.
  • Forgetting to square the speed of light is wrong because the formula uses c2c^2, not cc, and this changes the answer by a factor of about 3.00×1083.00 \times 10^8.
  • Mixing joules and electronvolts without conversion is wrong because J\text{J} and eV\text{eV} are different energy units related by 1 eV=1.602×1019 J1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.
  • Using the final mass instead of the mass defect is wrong in nuclear binding problems because the released or required energy depends on Δm\Delta m, not the full nuclear mass.
  • Ignoring signs for energy changes is wrong because a decrease in mass usually means energy is released, while an increase in mass means energy must be supplied.

Practice Questions

  1. 1 A reaction converts 2.50×106 kg2.50 \times 10^{-6}\ \text{kg} of mass into energy. Calculate EE using c=3.00×108 m/sc = 3.00 \times 10^8\ \text{m/s}.
  2. 2 A nucleus has a mass defect of 0.0450 u0.0450\ \text{u}. Find its binding energy in MeV\text{MeV} using 1 u=931.5 MeV/c21\ \text{u} = 931.5\ \text{MeV}/c^2.
  3. 3 An electron and a positron annihilate, with total mass 1.82×1030 kg1.82 \times 10^{-30}\ \text{kg}. Calculate the total energy released in joules.
  4. 4 Explain why a small mass defect in a nucleus can correspond to a large binding energy.