1. P.1 Book 5 Section 3.2 Mass-energy relationship Missing mass Mass-energy equivalence Energy released in nuclear reactions Check-point 3 3.2Mass-energy relationship

2. P.2 Book 5 Section 3.2 Mass-energy relationship Missing mass In the fission of a U-235 nucleus, the total mass of the products is lighter than the reactants. Where does the mass go? Becomes the energy released in the reaction. From the energy of the  radiation. On the other hand, particles can be created by extremely energetic  radiation. Where does the mass of the particles come from? paths of particles created

3. P.3 Book 5 Section 3.2 Mass-energy relationship 1 Mass-energy equivalence In 1905, Albert Einstein proposed the theory of special relativity: ‘Mass and energy are equivalent.’ This means:  mass and energy are different forms of the same thing  mass can be converted into energy and vice versa

4. P.4 Book 5 Section 3.2 Mass-energy relationship 1 Mass-energy equivalence The amount of energy released is given by the mass-energy relationship:  E =  mc 2  E : increase in energy (unit: J)  m : decrease in mass (unit: kg) c : the speed of light, i.e. 3  10 8 m s –1  huge amount of energy from a tiny mass The mass-energy relationship Example 4

5. P.5 Book 5 Section 3.2 Mass-energy relationship Example 4 The mass-energy relationship Find the amount of energy released if a Hong Kong 10-cent coin (mass 1.85 g) were converted completely into energy. Energy released=  mc 2 = 1.85  10 –3  (3  10 8 ) 2 = 1.67  10 14 J

6. P.6 Book 5 Section 3.2 Mass-energy relationship 2 Energy released in nuclear reactions a Atomic mass unit atomic mass unit (u)  small unit of mass  Suitable for measuring mass in atomic scale = 1.66  10 –27 kg 1 u =  mass of a carbon-12 atom 1 12

7. P.7 Book 5 Section 3.2 Mass-energy relationship a Atomic mass unit Mass of some particles in atomic mass unit: ParticleMass / u Proton1.007 276 Neutron1.008 665 Electron0.000 549 Hydrogen-2 (deuteron) nucleus2.013 553 Hydrogen-3 (tritium) nucleus3.015 500 Helium-4 nucleus4.001 506 Atomic mass unit and kilogram Example 5

8. P.8 Book 5 Section 3.2 Mass-energy relationship Example 5 Atomic mass unit and kilogram (a) Express the mass of a uranium-235 atom in kg. (Mass of an uranium-235 atom = 235.043 923 u) Mass = 235.043 923  1.66  10 –27 = 3.90  10 –25 kg (b) Express the mass of a helium nucleus in u. (Mass of a helium nucleus = 6.64  10 –27 kg) = 4.00 u Mass = 6.64  10 –27 1.66  10 –27

9. P.9 Book 5 Section 3.2 Mass-energy relationship 2 Energy released in nuclear reactions b Mass difference and energy In nuclear fission and fusion, total mass of the particles after the reaction is usually smaller than before. Loss in mass  energy Consider the nuclear fission:

10. P.10 Book 5 Section 3.2 Mass-energy relationship b Mass difference and energy Total mass before the reaction = 236.052 588 u Total mass after the reaction = 235.859 799 u Loss in mass (0.192 789 u) is converted into energy. ∴ Energy released in the reaction =  mc 2 = (0.192 789  1.66  10 –27 )  (3  10 8 ) 2 = 2.88  10 –11 J (or 0.192 789 u) The atomic mass unit can be used as a unit of energy. The energy released in nuclear fusion Example 6

11. P.11 Book 5 Section 3.2 Mass-energy relationship Total mass before reaction = mass of H-2 nucleus + mass of H-3 nucleus = 2.013 553 + 3.015 500 Total mass after reaction = mass of He-4 nucleus + mass of neutron = 4.001 506 + 1.008 665 Example 6 The energy released in nuclear fusion (a) Total mass of the particles before and after the reaction = ? = 5.029 053 u = 5.010 171 u A fusion reaction: H + H  He + n 2121 3131 4242 1010

12. P.12 Book 5 Section 3.2 Mass-energy relationship Example 6 The energy released in nuclear fusion (b)Find the energy released in (i) atomic mass units; (ii) joules. (1 u = 1.66  10 –27 kg) (i) Energy released = 5.029 053 – 5.010 171 = 0.018 882 u (ii) Energy released =  mc 2 = 0.018 882  1.66  10 –27  (3  10 8 ) 2 = 2.82  10 –12 J

13. P.13 Book 5 Section 3.2 Mass-energy relationship b Mass difference and energy Exploring the moon Example 7

14. P.14 Book 5 Section 3.2 Mass-energy relationship Example 7 Exploring the moon On October 24, 2007, China launched its first lunar orbiting spacecraft, Chang’e 1. One of its objectives was to probe the amount of He-3 on the moon. Scientists estimate that about 1.1  10 9 kg of He-3 have been deposited on the moon by solar wind. He-3 is a light, non-radioactive isotope of helium which is rare on earth. It is a highly efficient nuclear fusion fuel for power generation. The electricity generated by 3000 kg of He-3 can satisfy China’s energy consumption for a whole year!

15. P.15 Book 5 Section 3.2 Mass-energy relationship Example 7 Exploring the moon (a) The fusion of He is represented as follows: H + He  He + Z Find x and y. Hence, state what Z is. Consider the atomic numbers: 1 + 2 = 2 + x  x = 1 Consider the mass numbers: 2 + 3 = 4 + y  y = 1 ∴ Z is a hydrogen nucleus (or a proton). 3232 2121 3232 4242 yxyx

16. P.16 Book 5 Section 3.2 Mass-energy relationship Example 7 Exploring the moon (b)Energy released in the reaction = 2.94  10 –12 J What is the difference between the total mass of the reactants and products? By  E =  mc 2, = 3.27  10 –29 kg = 0.0197 u  m =  E c 2 = 2.94  10 –12 (3  10 8 ) 2

17. P.17 Book 5 Section 3.2 Mass-energy relationship Example 7 Exploring the moon (c)Give one advantage and one disadvantage of using He-3 as a fuel for fusion. Advantage: highly efficient (or non-radioactive) Disadvantage: hard to obtain

18. P.18 Book 5 Section 3.2 Mass-energy relationship b Mass difference and energy Nuclear power Example 8

19. P.19 Book 5 Section 3.2 Mass-energy relationship Example 8 Nuclear power (a) Consider the following fission reaction: (i) Find x and y. Consider the mass numbers: 235 + 1 = x + 92 + 3  1  x = 141 Consider the atomic numbers: 92 + 0 = 56 + y + 3  0  y = 36

20. P.20 Book 5 Section 3.2 Mass-energy relationship Total mass of the particles before reaction = 235.043 923 + 1.008 665 = 236.052 588 u Example 8 Nuclear power (a) (ii) Find the energy (in u) released in the reaction.

21. P.21 Book 5 Section 3.2 Mass-energy relationship Total mass of the particles after reaction = 140.914 411 + 91.926 156 + 3  1.008 665 = 235.866 562 u Energy released= loss in mass = 236.052 588 – 235.866 562 = 0.186 026 u Example 8 Nuclear power

22. P.22 Book 5 Section 3.2 Mass-energy relationship Example 8 Nuclear power (b)A nuclear reactor consumes 1.39 g of fuel every second. Of the consumed fuel, 0.003% of mass is converted into energy. If the capacity of the reactor is 700 MW, efficiency = ? Energy released by the fuel in each second =  mc 2 = 1.39  10 –3  0.003%  (3  10 8 ) 2 = 3.75  10 9 J

23. P.23 Book 5 Section 3.2 Mass-energy relationship Example 8 Nuclear power energy supplied by the reactor energy released by the fuel Efficiency =  100% 700  10 6 3.75  10 9 = = 18.7% (b)A nuclear reactor consumes 1.39 g of fuel every second. Of the consumed fuel, 0.003% of mass is converted into energy. If the capacity of the reactor is 700 MW, efficiency = ?

24. P.24 Book 5 Section 3.2 Mass-energy relationship Example 8 Nuclear power (c)Give one advantage and one disadvantage of using nuclear power. Advantage: It helps solve the future energy shortage crisis. Disadvantage: Waste products are highly radioactive.

25. P.25 Book 5 Section 3.2 Mass-energy relationship A 3  10 8 J B 6  10 16 J C 9  10 16 J D It cannot be determined since the type of element is unknown. Check-point 3 – Q1 What is the amount of energy produced if 1 kg of a certain element completely changes into energy?

26. P.26 Book 5 Section 3.2 Mass-energy relationship Check-point 3 – Q2 ParticleMass / kgMass / u Carbon-12 nuclide12.0000 Carbon-13 nuclide 2.1586  10 –26 Carbon-14 nuclide14.0032 Uranium-234 nuclide 3.8851  10 –25 Roentgenium-272 nuclide 272.1535 1.99  10 –26 2.32  10 –26 13.0 4.52  10 –25 234

27. P.27 Book 5 Section 3.2 Mass-energy relationship Check-point 3 – Q3 A nuclear fission: U + n  Cs + Rb + 2 n Mass of the particles involved in the fission: 235 92 1010 138 55 96 37 1010 235.043 923 u 137.911 017 u 95.934 273 u 1.008 665 u MassParticle 1010 96 37 138 55 235 92 n Rb Cs U

28. P.28 Book 5 Section 3.2 Mass-energy relationship Check-point 3 – Q3 Find the total energy released during the reaction in joule. Total mass before reaction = 235.043 923 + 1.008 665 = 236.052 588 u Total mass after reaction = 137.911 017 + 95.934 273 + 2  1.008 665 = 235.862 620 u Energy released=  mc 2 = 2.84  10 –11 J 235.043 923 u 137.911 017 u 95.934 273 u 1.008 665 u MassParticle

29. P.29 Book 5 Section 3.2 Mass-energy relationship The End