2. 1© Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion Nuclear binding energy Nuclear fission Nuclear fusion

3. 2 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 233) Nuclear binding energy The masses of proton, neutron and electron can be used to find the total mass of the particles that constitute an atom. The difference between the mass of an atom and the total mass of the particles in the atom taken separately is known as the mass defect.

4. 3 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 234) Nuclear binding energy 1. Binding energy The binding energy (E b ) of a nucleus is the energy released when an atom is formed from its constituent particles. Conversely, it is also the energy furnished to a nucleus when it is torn apart into its individual nucleons. Note: The mass (m) must be converted from u to kg when using the equation E = mc 2.

5. 4 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 235) Nuclear binding energy 2. Binding energy per nucleon - it is the binding energy of nucleus divided by the nucleon number of nucleus - use it to compare the binding energies of different isotopes Note: The binding energy per nucleon of a nucleus is a measure of the nucleus stability. The higher binding energy per nucleon, the more stable of the nucleus.

6. 5 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 235) Nuclear binding energy 2. Binding energy per nucleon Fe-56 has the largest binding energy per nucleon (smallest nuclear PE)

7. 6 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 236) Nuclear binding energy 2. Binding energy per nucleon (a) light nuclei of small E b /A combine to form heavier nucleus of large E b /A, binding energy increases (energy is released) (fusion)

8. 7 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 236) Nuclear binding energy 2. Binding energy per nucleon (b) heavier nuclei of small E b /A breaks up into two lighter nuclei of large E b /A, binding energy increases (energy is released) (fission) Go to Example 3 Example 3 Go to Example 4 Example 4

9. 8 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 239) Nuclear fission Nuclear fission is the splitting of a large nucleus into smaller nuclei.

10. 9 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 239) Nuclear fission 1. Nuclear fission in uranium - when a nucleus of U-235 captures a neutron, it breaks down by one of many possible reaction routes, e.g.

11. 10 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 240) Nuclear fission 1. Nuclear fission in uranium Chain reaction

12. 11 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 240) Nuclear fission 2. Nuclear fission reactor

13. 12 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 241) Nuclear fission 2. Nuclear fission reactor (a) Uranium fuel - To maintain the chain reactions, the uranium fuel rods must be increased the concentration of U-235 to about 3.2%.

14. 13 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 241) Nuclear fission 2. Nuclear fission reactor (b) Moderator - surrounding the fuel rods to reduce the speed of fast fission neurons - the slow neutron generally has high probability of being captured by the uranium fuel - water, sometimes graphite

15. 14 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 241) Nuclear fission 2. Nuclear fission reactor (c) Coolant - carry heat away from fuel rods - water (d) Control rod - neutron-absorbing rods - absorb secondary neutrons to reduce reaction rate - Cadmium or boron

16. 15 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 242) Nuclear fission 3. Advantages and disadvantages of nuclear power (a) Operating cost (b) Pollution (c) Radioactive waste (d) Nuclear weapons (e) Nuclear accidents Go to Example 5 Example 5

17. 16 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 243) Nuclear fusion Large nucleus splits into two smaller nuclei whose binding energy per nucleon is higher

18. 17 © Manhattan Press (H.K.) Ltd. 25.2 Energy release in fission and fusion (SB p. 244) Nuclear fusion The joining together of two small nuclei to form a larger nucleus is known as nuclear fusion. Go to Example 6 Example 6

19. 18 © Manhattan Press (H.K.) Ltd. End

20. 19 © Manhattan Press (H.K.) Ltd. Q: Q: The mass of an atom of is exactly 12 u. (a) Find the sum of the rest masses of the constituent particles of the atom in terms of u. (b) Why is your answer not exactly 12 u? (1 u = 1.66 × 10 –27 kg, rest mass of electron (m e ) = 9.0 × 10 –31 kg, rest mass of proton (m p ) = 1.672 6 × 10 –27 kg, rest mass of neutron (m n ) = 1.675 0 × 10 –27 kg) Solution 25.2 Energy release in fission and fusion (SB p. 236)

21. 20 © Manhattan Press (H.K.) Ltd. Solution: (a) Total mass of the constituent particles in a atom = 6 m p + 6 m n + 6 m e = 6 (1.672 6 + 1.675 0 + 9.0 ×10 –4 )10 –27 = 20.091 ×10 –27 kg = 12.10 u (b) The total mass is greater than 12 u because in the nucleus, the protons and neutrons are in a lower energy state than when they were separated. The energy equivalent of the mass difference is the binding energy of the nucleus. Return to Text 25.2 Energy release in fission and fusion (SB p. 236)

22. 21 © Manhattan Press (H.K.) Ltd. Q: Q: A nuclide is represented by the symbol. (a) Write down (i) the number of protons, and (ii) the number of neutrons in its nucleus. (b) The mass of the nucleus is 80.897 1 u. If the mass of a proton is 1.007 3 u and mass of a neutron is 1.008 7 u, deduce the nuclear binding energy per nucleon. (c) Draw a graph to show how the binding energy per nucleon changes with the mass number, and show the approximate position for on your graph. (d) Explain the implication to the stability of a nuclide which has a low value for the binding energy per nucleon. Solution 25.2 Energy release in fission and fusion (SB p. 236)

23. 22 © Manhattan Press (H.K.) Ltd. Solution: (a) (i) Number of protons in = 35 (ii) Number of neutrons in = 81 – 35 = 46 (b) Total mass of protons and neutrons in nucleus = 35 m p + 46 m n = 35 (1.007 3) u + 46 (1.008 7) u = 81.655 7 u Mass defect (Δm) = (81.655 7 – 80.897 1) u = 0.758 6 u = 0.758 6 × (1.66 ×10 –27 ) = 1.259 3 ×10 –27 kg Hence nuclear binding energy = mc 2 = (1.259 3 ×10 –27 ) × (3.0 ×10 8 ) 2 = 1.13 ×10 –10 J Binding energy per nucleon = = 1.40 ×10 –12 J (d) Nuclides with a lower binding energy per nucleon are less stable because the nucleons in the nuclei are in a higher energy state than those in a nucleus possessing a higher binding energy per nucleon. Return to Text (c) 25.2 Energy release in fission and fusion (SB p. 237)

24. 23 © Manhattan Press (H.K.) Ltd. Q: Q: (a) Explain what is meant by the statement: “ 235 U is made to undergo fission by thermal neutrons.” (b) Outline the process by which a nuclear chain reaction may be maintained in natural uranium. Summarize the factors that make such a reaction unlikely to take place in nature. (c) Assuming that each fission of a 235 U atom releases 3 × 10 –11 J of energy and eventually results in the formation of one 239 Pu atom, calculate, to the nearest month, how long a reactor of total output 100 MW would take to produce 10 kg of plutonium, Pu. (Avogadro constant = 6.02 × 10 23 mol –1, 1 month = 2.6 × 10 6 s) Solution 25.2 Energy release in fission and fusion (SB p. 242)

25. 24 © Manhattan Press (H.K.) Ltd. Solution: (a) Thermal neutrons are neutrons which are slowed down by setting them in equilibrium with the surroundings which are maintained at a constant temperature. When a 235 U nucleus is bombarded by a thermal neutron and the neutron is captured by the nucleus, it becomes unstable, splits into two smaller nuclei and produces three secondary neutrons. (b) A nuclear chain reaction may be maintained in natural uranium if the secondary neutrons from the fission of a uranium-235 nucleus can be slowed down sufficiently to strike other U-235 nuclei. This happens only if the concentration of U-235 nuclei is sufficiently high and the sample of uranium-235 is greater than the critical size. Such a chain reaction is unlikely to take place in nature because: (i) In uranium ore, U-235 constitutes less than 1%. A large percentage of it is U-238. (ii) Most of the secondary neutrons are captured by U-238 nuclei. (iii) Most of the other secondary neutrons escape from the uranium ore. (iv) The concentration of U-235 nuclei is too low for a nuclear chain reaction to take place. 25.2 Energy release in fission and fusion (SB p. 243)

26. 25 © Manhattan Press (H.K.) Ltd. Solution (cont’d): (c) The fission of 1 atom of U-235 releases 3 ×10 –11 J of energy and forms 1 atom of Pu-239. The number of Pu-239 atoms in 1 mole or 239 ×10 –3 kg = 6.02 ×10 23. ∴ In 10 kg of Pu-239, the number of atoms The energy released = (2.52 ×10 25 ) ×(3 ×10 –11 ) = 7.56 ×10 14 J Hence time taken for the formation of 10 kg of Pu-239 Return to Text 25.2 Energy release in fission and fusion (SB p. 243)

27. 26 © Manhattan Press (H.K.) Ltd. Q: Q: (a) Identify the numbers a, b, c and d, and the symbols X and Y in the reaction (b) The product particles each has a greater mass than at rest. Account for this and calculate the overall difference. (c) If 100 kg of the mixed materials, X and H were used each year as fuel in a power station with a conversion efficiency of 20%, estimate (i) the electrical power output and (ii) the waste heat produced. Solution 25.2 Energy release in fission and fusion (SB p. 245)

28. 27 © Manhattan Press (H.K.) Ltd. Solution: (a) X = H, its atomic number is 1. a = 1, the atomic number of H is 1. b = 2, the atomic number of He is 2. Hence c = 0, d = 1 and Y = n (neutron). (b) When a particle is moving and has kinetic energy E, according to Einstein’s theory of relativity, the particle mass is greater than its rest mass by: where c = speed of light. The energy equivalent of the overall mass difference, E = 17.5 MeV = (17.5  10 6 )  (1.60  10 –19 ) = 2.80  10 –12 J 25.2 Energy release in fission and fusion (SB p. 245)

29. 28 © Manhattan Press (H.K.) Ltd. Solution (cont’d): (c) (i) From the mass numbers of and, a mass of (3 + 2) u of the mixture released 2.80 ×10 –12 J of energy. Therefore, in one year, the energy released by 100 kg of mixture Hence the rate of release of energy 20% of the energy released per second is converted into electrical power. ∴ Electrical power output = 20/100 ×(1.07 ×10 9 ) = 2.14 ×10 8 W (ii) Also, the rate at which heat is wasted = 80/100 × (1.07 ×10 9 ) = 8.56 ×10 8 W Return to Text 25.2 Energy release in fission and fusion (SB p. 245)