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Nuclear binding energyEdexcel A-Level Physics: Revision notes

Section 1

Mass deficit

The mass of a nucleus is always less than the total mass of the protons and neutrons that it contains when they are separate. The difference is the mass deficit:

Δm=(Z mp+N mn)−mnucleus\Delta m = (Z\,m_p + N\,m_n) - m_{\text{nucleus}}

where ZZ is the number of protons, NN the number of neutrons and mnucleusm_{\text{nucleus}} the mass of the nucleus (not the atom, which includes electrons).

Worked example: helium-4 has 2 protons and 2 neutrons. Separate nucleons: 2(1.007276)+2(1.008665)=4.0318822(1.007276) + 2(1.008665) = 4.031882 u. The nucleus has a mass of 4.001506 u, so Δm=0.030376\Delta m = 0.030376 u.

Key termsmass deficitnucleon
Common mistake

Use the nuclear mass, not the atomic mass, or the electron masses will be counted wrongly in the deficit.

Section 2

Mass-energy equivalence: E = c²Δm

Einstein showed that mass and energy are equivalent:

E=c2ΔmE = c^2 \Delta m

When nucleons bind together, energy is released, and the system loses mass. The mass deficit is exactly the mass equivalent of the energy released. Mass is not destroyed: it is mass-energy that is conserved.

The mass equivalent of the energy is small in everyday terms, but because c2c^2 is very large (9.00×10169.00 \times 10^{16} m² s⁻²), a tiny mass change corresponds to a large amount of energy per nucleus.

Key termsmass-energy equivalence

Section 3

The atomic mass unit and SI conversion

Nuclear masses are given in the atomic mass unit, u, defined as one twelfth of the mass of a carbon-12 atom:

1 u=1.661×10−27 kg1\text{ u} = 1.661 \times 10^{-27}\text{ kg}

To use E=c2ΔmE = c^2\Delta m in SI units, convert the mass deficit from u to kg by multiplying by 1.661×10−271.661 \times 10^{-27}, then multiply by c2c^2 to get joules.

Energies of nuclear processes are often given in MeV: 1 MeV=1.60×10−131\text{ MeV} = 1.60 \times 10^{-13} J. One atomic mass unit is equivalent to about 931 MeV.

Key termsatomic mass unitMeV
Exam tip

Convert u to kg first, then use c² in m² s⁻². Skipping the conversion is the most common source of a wrong power of ten.

Section 4

Binding energy and binding energy per nucleon

The binding energy of a nucleus is the minimum energy needed to separate it into its individual protons and neutrons. It equals the energy released when the nucleus forms from separate nucleons, and it is the energy equivalent of the mass deficit.

EB=c2ΔmE_B = c^2 \Delta m

The binding energy per nucleon is EB/AE_B / A, where AA is the nucleon number. It measures how tightly bound the average nucleon is, and so how stable the nucleus is.

Key termsbinding energybinding energy per nucleon

Section 5

Worked example: the deuteron

The deuteron has a nuclear mass of 2.013553 u, with proton 1.007276 u and neutron 1.008665 u.

  1. Δm=2.015941−2.013553=0.002388\Delta m = 2.015941 - 2.013553 = 0.002388 u
  2. Δm=0.002388×1.661×10−27=3.97×10−30\Delta m = 0.002388 \times 1.661 \times 10^{-27} = 3.97 \times 10^{-30} kg
  3. E=c2Δm=9.00×1016×3.97×10−30=3.57×10−13E = c^2\Delta m = 9.00 \times 10^{16} \times 3.97 \times 10^{-30} = 3.57 \times 10^{-13} J
  4. In MeV: 3.57×10−13/1.60×10−13=2.233.57 \times 10^{-13} / 1.60 \times 10^{-13} = 2.23 MeV
  5. Per nucleon: 2.23/2=1.122.23 / 2 = 1.12 MeV

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Nuclear binding energy

  1. A student is studying the nucleus of a helium-4 atom, which contains two protons and two neutrons held together by the strong nuclear force. She is comparing the mass of the nucleus with the masses of the particles that it is made from.
    Explain why the mass of the helium-4 nucleus is less than the total mass of two separate protons and two separate neutrons.2 marks
  2. The deuteron, the nucleus of hydrogen-2, contains one proton and one neutron. Nuclear mass of the deuteron = 2.013553 u. Mass of a proton = 1.007276 u. Mass of a neutron = 1.008665 u. Take 1 u=1.661×10−271\text{ u} = 1.661 \times 10^{-27} kg, c=3.00×108c = 3.00 \times 10^{8} m s⁻¹ and e=1.60×10−19e = 1.60 \times 10^{-19} C, so that 1 MeV=1.60×10−131\text{ MeV} = 1.60 \times 10^{-13} J.
    Use your answer to (b) to calculate the binding energy per nucleon of the deuteron, in MeV.2 marks
  3. The Sun transfers energy to space at a rate (power) of 3.8×10263.8 \times 10^{26} W, and this energy comes from nuclear reactions in which mass is converted to energy. The mass of the Sun is 2.0×10302.0 \times 10^{30} kg, its age is 4.6×1094.6 \times 10^{9} years and 1 year = 3.16×1073.16 \times 10^{7} s. Take c=3.00×108c = 3.00 \times 10^{8} m s⁻¹ and assume that the power has been constant.
    Calculate the mass that the Sun converts to energy each second.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).