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Questions on mass-energy and binding energy
A nucleus has less mass than its separated nucleons, and that mass difference, converted by E = mc², is the binding energy of the nucleus. One atomic mass unit is equivalent to 931.5 MeV. Binding energy per nucleon peaks at iron-56 near 8.8 MeV, with fusion to the left of the peak and fission to the right.
18 original questions · 48 marks · the mass-energy and binding energy notes · Nuclear physics
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State the equation linking mass and energy, and define the binding energy of a nucleus.
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E = mc2 (1). Binding energy is the energy required to separate a nucleus completely into its individual nucleons (or the energy released when the nucleus forms from separate nucleons) (1).A mass of 1.0 × 10−28 kg is converted entirely into energy. Calculate the energy released.
c = 3.0 × 108 m s−1Mark scheme
E = mc2 = 1.0 × 10−28 × (3.0 × 108)2 (1)
E = 9.0 × 10−12 J (1)State the energy equivalent of one atomic mass unit (u) in MeV.
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1 u is equivalent to 931.5 MeV (1).Define the atomic mass unit, u, and state its value in kilograms.
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One twelfth of the mass of an atom of carbon-12 (1); 1 u = 1.661 × 10−27 kg (1).The water in a kettle is heated, increasing its internal energy. State what happens to the mass of the water, and explain why the change cannot be detected by weighing.
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The mass increases (1). Δm = E/c2, and because c2 is so large the mass gained in any everyday energy transfer is far too small to measure (1).A helium-4 nucleus has a mass of 4.00150 u. The mass of a proton is 1.00728 u and the mass of a neutron is 1.00867 u. Calculate the mass defect of the nucleus and its binding energy in MeV.
1 u = 931.5 MeVMark scheme
Total nucleon mass = 2 × 1.00728 + 2 × 1.00867 = 4.03190 u (1)
Mass defect = 4.03190 − 4.00150 = 0.03040 u (1)
Binding energy = 0.03040 × 931.5 (1)
BE = 28.3 MeV (1)The binding energy of helium-4 is 28.3 MeV. Calculate its binding energy per nucleon.
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BE per nucleon = 28.3/4 (1)
= 7.08 MeV per nucleon (1)Express the binding energy of helium-4 (28.3 MeV) in joules.
1 MeV = 1.60 × 10−13 JMark scheme
E = 28.3 × 1.60 × 10−13 (1)
E = 4.53 × 10−12 J (1)In a nuclear reaction, 2.0 × 10−11 J of energy is released. Calculate the mass converted into energy in the reaction.
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m = E/c2 = (2.0 × 10−11)/(3.0 × 108)2 (1)
m = 2.22 × 10−28 kg (1)The binding energy per nucleon of oxygen-16 is 7.98 MeV. Calculate the mass defect of an oxygen-16 nucleus, in u.
1 u = 931.5 MeVMark scheme
Total binding energy = 7.98 × 16 = 127.7 MeV (1)
Δm = BE/931.5 (1)
Δm = 0.137 u (1)Radium-226 decays by alpha emission: 22688Ra → 22286Rn + 42He. The atomic masses are 226.02540 u, 222.01757 u and 4.00260 u respectively. Calculate the energy released in one decay, in MeV.
1 u = 931.5 MeVMark scheme
Mass after = 222.01757 + 4.00260 = 226.02017 u (1)
Δm = 226.02540 − 226.02017 = 0.00523 u (1)
E = 0.00523 × 931.5 = 4.9 MeV (1)Nucleus X has a total binding energy of 1638 MeV and nucleon number 210. Nucleus Y has a total binding energy of 492 MeV and nucleon number 56. A student claims that X is the more stable nucleus because its binding energy is greater. Deduce whether the student is correct.
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Stability is compared using binding energy per nucleon, not total binding energy (1)
X: 1638/210 = 7.8 MeV per nucleon; Y: 492/56 = 8.8 MeV per nucleon (1)
Y is the more tightly bound per nucleon, so the student is not correct (1)A nucleus with nucleon number 12 has a mass defect of 0.099 u. Calculate its binding energy and its binding energy per nucleon.
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Binding energy = 0.099 × 931.5 (1)
BE = 92.2 MeV (1)
BE per nucleon = 92.2/12 = 7.68 MeV per nucleon (1)Sketch the variation of binding energy per nucleon with nucleon number, and use your sketch to explain why fusion of light nuclei and fission of heavy nuclei both release energy.
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The curve rises steeply for light nuclei to a broad maximum near iron-56 at about 8.8 MeV per nucleon (1); it falls gently for heavier nuclei (1). Any change that moves nuclei towards the peak increases the binding energy per nucleon, releasing energy (1); fusion moves light nuclei up the rising part, and fission moves heavy nuclei back towards the peak, so both release energy (1).Explain the significance of the peak of the binding energy per nucleon curve.
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The peak, near iron-56, corresponds to the most tightly bound and most stable nuclei (1). Nuclei on either side can release energy by changing towards the peak (1); iron itself cannot release energy by fusion or fission, so it is the endpoint of energy-releasing nuclear processes in stars (1).Show that the energy equivalent of one atomic mass unit is about 930 MeV.
1 u = 1.661 × 10−27 kg, c = 3.0 × 108 m s−1, 1 MeV = 1.60 × 10−13 JMark scheme
E = mc2 = 1.661 × 10−27 × (3.0 × 108)2 (1)
E = 1.495 × 10−10 J (1)
E = 1.495 × 10−10/(1.60 × 10−13) = 934 MeV ≈ 930 MeV (1)A nucleus with nucleon number 240 and binding energy per nucleon 7.6 MeV splits into two equal fragments, each with binding energy per nucleon 8.5 MeV. The freed neutrons may be neglected. Estimate the energy released in the fission.
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Each nucleon becomes more tightly bound by 8.5 − 7.6 = 0.9 MeV (1)
Energy released ≈ 0.9 × 240 (1)
E ≈ 216 MeV (1)Calculate the minimum energy, in MeV, needed to remove one neutron from a carbon-13 nucleus, leaving carbon-12.
Nuclear masses: carbon-13 13.00006 u, carbon-12 11.99671 u, neutron 1.00867 u; 1 u = 931.5 MeVMark scheme
Mass after separation = 11.99671 + 1.00867 = 13.00538 u (1)
The products are more massive than carbon-13 by Δm = 13.00538 − 13.00006 = 0.00532 u (1)
This extra mass must be supplied as energy: E = 0.00532 × 931.5 (1)
E = 5.0 MeV (1)
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