A-level Physics
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    8.4 Nuclear energy and E = mc² — Practise with help

    Work each problem first. Open a hint only if you are stuck, and the full working only to check. Decide first whether the mass is in kilograms or in atomic mass units, because the two routes use different constants.

    Warm-up: remember

    1. Write an energy of 2.0 MeV in joules.
    2. A mass of 1.0×10−30 kg1.0 \times 10^{-30}\,\text{kg} is converted completely into energy. Find the energy released.
    3. Find the energy of a photon of energy 0.511 MeV in joules.
    Answers
    1. 2.0×106×1.60×10−19=3.2×10−13 J2.0 \times 10^{6} \times 1.60 \times 10^{-19} = 3.2 \times 10^{-13}\,\text{J}.
    2. E=mc2=1.0×10−30×(3.00×108)2=9.0×10−14 JE = mc^2 = 1.0 \times 10^{-30} \times (3.00 \times 10^{8})^2 = 9.0 \times 10^{-14}\,\text{J}.
    3. 0.511×106×1.60×10−19=8.18×10−14 J0.511 \times 10^{6} \times 1.60 \times 10^{-19} = 8.18 \times 10^{-14}\,\text{J}.

    If question 1 goes wrong, one megaelectronvolt is 1.60×10−13 J1.60 \times 10^{-13}\,\text{J}. If question 2 goes wrong, use E=mc2E = mc^2 with the speed of light squared. If question 3 goes wrong, change mega to 10610^{6} and multiply by the charge of the electron.

    Practise with help

    1. A reaction converts a mass of 3.0×10−29 kg3.0 \times 10^{-29}\,\text{kg} into energy. Find the energy released.

    Hint

    Use the mass-energy relation with the speed of light squared.

    Next step

    Write E=mc2E = mc^2 with m=3.0×10−29 kgm = 3.0 \times 10^{-29}\,\text{kg} and c=3.00×108 m s−1c = 3.00 \times 10^{8}\,\text{m}\,\text{s}^{-1}.

    Full working E=mc2=3.0×10−29×(3.00×108)2=3.0×10−29×9.00×1016=2.7×10−12 J\begin{aligned} E &= mc^2 = 3.0 \times 10^{-29} \times (3.00 \times 10^{8})^2 \\ &= 3.0 \times 10^{-29} \times 9.00 \times 10^{16} \\ &= 2.7 \times 10^{-12}\,\text{J} \end{aligned}

    The energy released is 2.7×10−12 J2.7 \times 10^{-12}\,\text{J}.

    2. A nucleus has a mass defect of 0.050 u. Find its binding energy in MeV.

    Hint

    Use the energy equivalent of one atomic mass unit from the data sheet.

    Next step

    Multiply the mass defect by 931.5 MeV per u.

    Full working E=0.050×931.5=46.6 MeVE = 0.050 \times 931.5 = 46.6\,\text{MeV}

    The binding energy is 46.6 MeV.

    3. A helium-4 nucleus has a mass defect of 0.0304 u. Find its binding energy in MeV and its binding energy per nucleon.

    Hint

    Find the total binding energy first, then divide by the number of nucleons.

    Next step

    Multiply 0.0304 by 931.5, then divide by 4.

    Full working E=0.0304×931.5=28.3 MeVper nucleon=28.34=7.08 MeV\begin{aligned} E &= 0.0304 \times 931.5 = 28.3\,\text{MeV} \\ \text{per nucleon} &= \frac{28.3}{4} = 7.08\,\text{MeV} \end{aligned}

    The binding energy is 28.3 MeV, or 7.08 MeV per nucleon.

    4. A fission reaction releases 200 MeV. Find the mass defect in kilograms.

    Hint

    Change the energy to joules, then divide by c2c^2.

    Next step

    Write E=200×106×1.60×10−19 JE = 200 \times 10^{6} \times 1.60 \times 10^{-19}\,\text{J}, then use Δm=Ec2\Delta m = \dfrac{E}{c^2}.

    Full working E=200×106×1.60×10−19=3.20×10−11 JΔm=3.20×10−11(3.00×108)2=3.20×10−119.00×1016=3.56×10−28 kg\begin{aligned} E &= 200 \times 10^{6} \times 1.60 \times 10^{-19} = 3.20 \times 10^{-11}\,\text{J} \\ \Delta m &= \frac{3.20 \times 10^{-11}}{(3.00 \times 10^{8})^2} = \frac{3.20 \times 10^{-11}}{9.00 \times 10^{16}} \\ &= 3.56 \times 10^{-28}\,\text{kg} \end{aligned}

    The mass defect is 3.56×10−28 kg3.56 \times 10^{-28}\,\text{kg}.

    5. A positron and an electron annihilate. Each has a rest energy of 0.511 MeV. Find the total energy released and the energy of each photon, in joules.

    Hint

    The two rest energies add. Two photons share the total energy equally.

    Next step

    Find 2×0.511 MeV2 \times 0.511\,\text{MeV} in joules, then halve it.

    Full working Etotal=2×0.511=1.022 MeV=1.022×1.60×10−13=1.64×10−13 JEphoton=1.64×10−132=8.2×10−14 J\begin{aligned} E_{\text{total}} &= 2 \times 0.511 = 1.022\,\text{MeV} \\ &= 1.022 \times 1.60 \times 10^{-13} = 1.64 \times 10^{-13}\,\text{J} \\ E_{\text{photon}} &= \frac{1.64 \times 10^{-13}}{2} = 8.2 \times 10^{-14}\,\text{J} \end{aligned}

    Each photon carries 8.2×10−14 J8.2 \times 10^{-14}\,\text{J}, or 0.511 MeV.

    6. Explain why energy is released when a very heavy nucleus undergoes fission.

    Hint

    Compare the binding energy per nucleon of the heavy nucleus with that of the products.

    Next step

    Say where the heavy nucleus sits on the binding energy curve, and where the products sit. Then link a larger binding energy per nucleon to a smaller mass.

    Full working

    A very heavy nucleus sits on the right-hand side of the binding energy curve, where the binding energy per nucleon is lower than at the peak. Its fission products are nearer the peak, so they have a larger binding energy per nucleon. A more tightly bound set of nucleons has less mass. The missing mass is released as the kinetic energy of the products, through E=mc2E = mc^2.

    7. Explain why energy is released when two light nuclei fuse.

    Hint

    Use the same curve, but start at the far left.

    Next step

    Compare the binding energy per nucleon of the light nuclei with that of the heavier product.

    Full working

    Two light nuclei sit on the steep left-hand part of the binding energy curve, where the binding energy per nucleon is small. The heavier product is closer to the peak, so its binding energy per nucleon is larger. The nucleons are more tightly bound in the product, so the product has less mass than the two nuclei together. The mass difference is released as energy, through E=mc2E = mc^2.

    8. The Sun converts about 4.3×109 kg4.3 \times 10^{9}\,\text{kg} of mass into energy each second. Find the energy released each second.

    Hint

    Use E=mc2E = mc^2 with the mass converted each second.

    Next step

    Multiply 4.3×1094.3 \times 10^{9} by (3.00×108)2(3.00 \times 10^{8})^2.

    Full working E=mc2=4.3×109×(3.00×108)2=4.3×109×9.00×1016=3.9×1026 J\begin{aligned} E &= mc^2 = 4.3 \times 10^{9} \times (3.00 \times 10^{8})^2 \\ &= 4.3 \times 10^{9} \times 9.00 \times 10^{16} \\ &= 3.9 \times 10^{26}\,\text{J} \end{aligned}

    The Sun releases about 3.9×1026 J3.9 \times 10^{26}\,\text{J} each second.

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