Binding Energy

1. Binding Energy Nuclear Physics Lesson 13

2. Learning Objectives • To define binding energy. • To define mass defect. • To know which are the most stable nuclei. • Explain why energy is released in nuclear fission and nuclear fusion.

3. The Atomic Mass Unit • The atomic mass unit (u) is far more convenient to use with nuclear masses.  • It uses carbon-12 as a reference and is defined as: • 1 atomic mass unit (u) = 1.661 × 10-27 kg. Exactly 1/12th the mass of a carbon 12 atom.

4. The table shows particle masses in atomic mass units.  • Note that the numbers are expressed to a large number of significant figures as the changes are quite subtle.

5. Be Careful! • Remember to distinguish between the atomic mass and the nuclear mass. • The atomic mass is the mass of an atom complete with its electrons. • The nuclear mass is the mass of the nucleus alone.  To get the nuclear mass we need to take away the mass of the electrons.

6. Binding Energy • If you want to remove a nucleon from the nucleus of an atom, then you have to do work to overcome the strong nuclear force. • Definition from specification book:- • The removed nucleons gain potential energy. • The binding energy of a nucleus is the work that must be done to separate a nucleus into its constituent neutrons and protons.

7. Binding Energy II • If we run that process in reverse, then the binding energy can also be defined as the energy released when a nucleus is assembled from its constituent nucleons. • This means that the mass of the nucleus is less than the mass of the separate nucleons because energy has been released…

8. Mass Defect • Definition from the specification book:- • The mass defect Δm of a nucleus is defined as the difference between the mass of the separated nucleons and the mass of the nucleus.

9. Calculating Mass Defect • For a nucleus of an isotope ZAX, composed of Z protons and (A-Z) neutrons with mass MNUC, the mass defect, Δm, is given by:-

10. Question 1 • What is the mass defect of a helium atom?

11. Answer 1 • What is the mass defect of a helium atom? • The atomic mass of a helium atom is 4.002603 u • Therefore mass defect, Δm=0.030377 u

12. Calculating Binding Energy • The mass defect Δm exists because energy is released when the constituent nucleons bind together to form a nucleus. • The energy released is equal to the binding energy of the nucleus:- • Note that Δm must be in kg to get energy in J.

13. Question 2 • What is the binding energy of the helium atom whose mass defect is 0.030377 u?   Express your answer in MeV. • m = 0.030377 u × 1.661 × 10-27 kg = 5.046 × 10-27 kg • E = mc2 = 5.046 × 10-27 kg × (3 × 108)2 = 4.541 × 10-12 J • E = 4.541 × 10-12 J = 28.38 × 106 eV= 28.38 MeV • Note: 1 u = 931.3 MeV

14. Binding Energy Per Nucleon • If we know the binding energy in a nucleus, and the number of nucleons, we can work out the binding energy per nucleon, which is the work done needed to remove each nucleon.  • The higher the binding energy per nucleon, the more stable is the nucleus.  For helium (4He) the binding energy per nucleon is: • Binding energy per nucleon = 28.38 MeV/4 = 7.1 MeV

15. Question 3 • What is the mass defect in atomic mass units (u) and in kilograms for the lithium nucleus which has 7 nucleons, and a proton number of 3?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 7.014353 u.  

16. Answer 3 • Add them together to get 7.056488 u • Now take away the nuclear mass from the number above to get the mass deficit.7.056488 u - 7.014353 u = 0.042135 u   • Now convert to kilograms: 1 u = 1.661 ´ 10-27 kg0.042135 u × 1.661 ´ 10-27 kg  = 6.9986235 × 10-29 kg     • Now use E = mc2 to work out the binding energy: E = 6.9986235 × 10-29 kg × (3 × 108 m/s)2 =  6.3 × 10-12 J • In electron volts, this is 6.3 × 10-12 J ÷ 1.6 × 10-19 eV/J = 3.9 × 107 eV = 39 MeV.   • There are 7 nucleons so the binding energy per nucleon = 3.9 × 107 eV ÷ 7 = 5.6 × 106 eV  

17. Question 4 • What is the mass defect in atomic mass units (u) and in kilograms for the copper nucleus which has 63 nucleons, and a proton number of 29?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 62.91367 u.

18. Answer 4 • Number of protons = 29; number of neutrons = 63 – 29 = 34  • Mass of protons = 29 ´ 1.007276 = 29.211004 u  • Mass of neutrons = 34 ´ 1.008665 = 34.29461 u • Total mass = 29.211004 u + 34.29461 u = 63.505614 u • Mass defect = 63.505614 u – 62.91367 u = 0.591944 u • Mass defect in kg = 0.591944 ´ 1.661 ´ 10-27 = 9.83218 ´ 10-28 kg • Binding energy = mc2 = 9.83218 ´ 10-28 kg ´ (3 ´ 108 m/s)2 = 8.85 ´ 10-11 J • Binding energy in eV = 8.85 ´ 10-11 J ¸ 1.6 ´ 10-19 J/eV = 5.53 ´ 108 eV • Binding energy per nucleon =  5.53 ´ 108 eV ¸ 63 = 8.78 ´ 106 eV

19. Binding Energy per Nucleon • We can plot a graph of binding energy per nucleon against nucleon number.

20. From this graph we can see that • The maximum value is about binding energy of 8.7 MeV per nucleon and iron is the most stable nuclide. • Helium has a particularly high value of binding energy per nucleon, much higher than the light isotopes of hydrogen. • There is a trend for nuclides of nucleon numbers in multiples of 4 to be particularly stable (i.e. have a high binding energy). • The largest nuclides tend to be less stable, with slightly lower binding energies per nucleon.

21. The Most Stable Nucleus • Iron has the highest binding energy per nucleon so is the most stable nucleus.  • If we look at large nuclei (greater than iron), we find that the further to the right (greater nucleon number) the less stable the nuclei.  • This is because the binding energy per nucleon is getting less. • The explanation for this observation lies in that the strong nuclear force that binds the nucleus together has a very limited range, and there is a limit to the number of nucleons that can be crammed into a particular space. 

22. Nuclear Fission • A large unstable nucleus splits into two fragments which are more stable than the original nucleus. • The binding energy per nucleon increases in this process and energy is released. • The change in binding energy per nucleon is about 0.5 MeV in a fission reaction.

23. Nuclear Fusion • Small nuclei fuse together to form a larger nucleus. • The product nucleus has a higher binding per nucleon as long as A is no greater than ~50. • The change in binding energy per nucleon can be more than 10 times greater in a fusion reaction than a fission reaction.

24. Summary • Atomic Mass Unit: 1/12th the mass of a carbon atom • Mass defect: Difference between the mass of nucleons separately and together within a nucleus. Difference between the two sides of a nuclear interaction equation. Energy worked out by E = mc2.                                                     • Binding Energy: Energy equivalent of the mass defect in a nucleus.  Binding energy per nucleon increases in more stable nuclei. • Fission Splitting of a nucleus.  Rarely spontaneous.  Occurs after the nucleus has been tickled with a neutron • FusionJoining together of two light nuclei to make a heavier nucleus.