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- 1. Dr Ahmad Taufek Abdul Rahman School of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan Studies, Sciences, is d fi d i defined as a physical process in which there is a change in identity of an atomic nucleus. nucleus CHAPTER 9: Nuclear reaction (2 Hours) Four types of nuclear reaction: 1
- 2. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Learning Outcome: L i O t 9.1 Nuclear reaction (1 hour) At the end of this chapter, students should be able to: State the conservation of charge (Z) and nucleon number (A) in a nuclear reaction. Write and complete the equation of nuclear reaction. reaction Calculate the energy liberated in the process of nuclear reaction 2
- 3. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.1 Nuclear reaction 9.1.1 Conservation of nuclear reaction Any nuclear reaction must obeyed conservation laws stated below: Conservation of relativistic energy (kinetic and rest energies): ∑ ∑ relativistic energy = gy before reaction relativistic energy gy after reaction Conservation of linear momentum: ∑ ∑ linear momentum = before reaction linear momentum after reaction Conservation of angular momentum: ∑ ∑ angular momentum = before b f reaction ti angular momentum after reaction ft ti 3
- 4. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Conservation of charge (atomic number Z): ∑ atomic number Z before reaction =∑ atomic number Z after reaction Conservation of mass number A: ∑ mass number A before reaction =∑ mass number A after reaction However, it is very hard to obey all the conservation laws. Note: N t The most important of conservation laws should be obeyed by every nuclear reaction are conservation of charge (atomic atomic number )and of mass number number. 4
- 5. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.1.2 Reaction energy (Q) Energy is released (liberated) in a nuclear reaction in the form of kinetic energy of the particle emitted the kinetic energy of energ emitted, energ the daughter nucleus and the energy of the gamma-ray gammaphoton that may accompany the reaction. The energy is called the reaction OR disintegration energy (Q). It may be calculated by finding the mass defect of the reaction where Mass defect =∑ mass of nucleus before reaction ∆m = mi − mf −∑ products after reaction mass of nucleus (9.1) (9.1) The reaction energy Q is the energy equivalent to the mass defect ∆m of the reaction, thus Q = (∆m )c 2 (9.2 (9 2) (9.2) 2) Speed of light in vacuum 5
- 6. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Note: If the value of ∆m OR Q is positive the reaction is called positive, exothermic (exoergic) in which the energy released in the form the kinetic energy of the product. If the value of ∆m OR Q is negative, the reaction is called g endothermic (endoergic) in which the energy need to be absorbed for the reaction occurred. 9.1.3 Radioactivity decay is defined as the phenomenon in which an unstable nucleus disintegrates to acquire a more stable nucleus without absorb an external energy energy. The disintegration is spontaneous and most commonly involves the emission of an alpha particle (α OR 4 He ), a beta 0 0 2 particle (β OR −1 e ) and gamma-ray (γ OR γ ). It also 0 release an energy Q k l known as di i t disintegration energy ti energy. 6
- 7. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 1 : Polonium nucleus decays by alpha emission to lead nucleus can be represented by the equation below: 212 Po→208 Pb+ 4 He + Q 84 82 2 Calculate a. the energy Q released in MeV. b. the wavelength of the g g gamma-ray p y produced. (Given mass of Po-212, mPo=211.98885 u ; mass of Pb-208, mPb=207.97664 u and mass of α particle , mα=4.0026 u) 212 Solution : α decay Po→208 Pb+ 4 He + Q 84 82 2 before decay ∑Z = ∑Z i f after decay and ∑A =∑A i f 7
- 8. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : a. The mass defect (difference) of the reaction is given by ∆m = mi − mf = mPo − (mPb + mα ) = 211.98885 − (207.97664 + 4.0026) ∆m = 9.61× 10−3 u The energy released in the decay reaction can be calculated by gy y y using two method: 2 st method: 1 u = 1.66 ×10−27 kg 1 Q = ∆m c ( ) ( in kg )( ) ∆m = 9.61 × 10 −3 1.66 × 10 −27 −29 = 1.5953 × 10 kg − 29 8 Q = 1.5953 × 10 3.00 ×10 Q = 1.436 × 10 −12 J ( )( ) 2 8
- 9. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : a. Thus the energy released in MeV is 1.436 × 10 −12 Q= −13 1.60 × 10 Q = 8.98 MeV 2nd method: Q = (∆m )c 2 1 MeV = 1.60 ×10−13 J 1 u = 931.5 MeV/c 2 in u ⎡ ⎛ 931.5 MeV/ c 2 ⎞⎤ 2 = ⎢∆m⎜ ⎥c ⎜ 1u ⎠⎦ ⎣ ⎝ ⎡ ⎛ 931 .5 MeV/c 2 ⎞⎤ 2 −3 ⎟⎥ c = ⎢ 9.61 × 10 u ⎜ ⎜ ⎟ 1u ⎢ ⎝ ⎠⎥ ⎣ ⎦ Q = 8.95 MeV ( ) 9
- 10. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : b. The reaction energy Q is released in form of gamma-ray where its wavelength can be calculated b appl ing the Planck’s a elength calc lated by applying hc quantum theory: E= =Q λ hc λ= Q (6.63 ×10 )(3.00 ×10 ) = −34 Note: 1.436 × 10 −13 λ = 1.39 × 10 m 8 −12 The radioactive decay only occurred when the value of ∆m OR Q is positive positive. 10
- 11. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 2 : ( ) A nickel-66 nucleus 66 Ni decays to a new nucleus by emitting a 28 beta particle. particle a. Write an equation to represent the nuclear reaction. p (a) b. If the new nucleus found in part ( ) has the atomic mass of 65.9284 u and the atomic mass for nickel-66 is 65.9291 u, what is the maximum kinetic energy of the emitted electron? (Given mass of electron, me =5.49 × 10−4 u and c =3.00 × 108 m s−1) Solution : a. N l Nuclear reaction equation must obey th conservation of atomic ti ti t b the ti f t i number and the conservation of mass number. 66 66 0 28 Ni→ 29 X + −1 e + Q β decay 11
- 12. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : b. Given m Ni = 65 .9291 u ; mX = 65 .9284 u The mass defect (difference) of the reaction is given by ∆m = mi − mf = mNi − (mX + me ) N = 65.9291 − 65.9284 + 5.49 × 10−4 −4 ∆m = 1.51× 10 u ( ) If the reaction energy is completely convert into the kinetic energy of emitted electron, therefore the maximum kinetic energy of the emitted electron is given by K max = Q 2 = (∆m )c ( K max −4 )( = 1.51 × 10 1.66 × 10 = 2.26 × 10 −14 J − 27 )(3.00 ×10 ) 8 2 12
- 13. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 3 : Table 9.1 shows the value of masses for several nuclides. Nuclide 4 2 He 23 11 Na 27 13 Al Mass (u) 4.0026 22.9898 22 9898 26.9815 Table 9.1 Discuss whether i i possible f Di h h it is ibl for 27 Al to emit spontaneously an i l 13 alpha particle. Solution : If by 27 13 Al emits an alpha particle, the α decay would be represented 27 13 Al 26.9815 u → 23 11 Na 22.9898 u + 4 2 He 4.0026 u 26.9924 u Since the total mass after the reaction is greater than that before 13 the reaction, therefore the reaction does not occur.
- 14. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.1.4 Bombardment with energetic particles is defined as an induced nuclear reaction that does not occur spontaneously; occ r spontaneo sl it is ca sed b a collision bet een a caused by between nucleus and an energetic particles such as proton, neutron, alpha particle or photon photon. Consider a bombardment reaction in which a target nucleus X is bombarded by a particle x, resulting in a daughter nucleus Y, an emitted particle y and reaction energy Q: X + x → y + Y+ Q sometimes this reaction is written in the more compact form: target (parent) nucleus X(x, y )Y daughter nucleus bombarding emitted particle p particle The calculation of reaction energy Q has been discussed in section 9.1.2. 14
- 15. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Examples of bombardment reaction: (α , p )17 O 8 7 Li+1 H →2 4 H + Q OR 7 Li(p,α )4 He 3 1 2 He 3 2 10 1 B+ 0 n → 7 Li+ 4 He + Q OR 10 B(n,α )7 Li 5 3 2 5 3 14 4 17 1 7 N + 2 He→ 8 O+1 H + Q Example 4 : OR 14 N 7 17 A nitrogen nucleus 14 N is converted into an oxygen nucleus 8 O 7 and a proton if it is bombarded by an alpha particle carrying certain amount of energy. a. Write down an expression for this nuclear reaction, showing the atomic number and the mass number of each nuclide and particle emitted. b. b Calculate the minimum energy of the alpha particle for this reaction to take place. (Given mp =0.16735×10−26 kg; mα =0.66466 ×10−26 kg ; mass of 0.16735 10 0.66466 10 nitrogen nucleus , mN =2.32530×10−26 kg; mass of oxygen nucleus, 15 mO =2.82282×10−26 kg ; c =3.00×108 m s−1)
- 16. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : a. The expression represents the nuclear reaction is 14 7 +4 H → 17 O + 1 H + Q N 2 He 8 1 b. The mass defect of the reaction is ∆m = mi − mf = (mN + mHe ) − (mO + mH ) ( −26 −26 ) = 2.32530 ×10 + 0.66466 ×10 − (2.82282 ×10−26 + 0.16735 ×10−26 ) ∆m = −2.1×10 −30 kg k Therefore the minimum energy of the alpha particle for this reaction to take place is K min = Q K min = (∆m )c 8 −30 = 2.1×10 3.00 ×10 −13 K min = 1.89 ×10 J 2 ( )( ) 2 16
- 17. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Exercise 9 1 : E i 9.1 Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u, 1. 1 Complete the following radioactive decay equations : a. 8 Be→4 He + 4 2 [ ] b. 240 Po→97 Sr +139 Ba + 94 38 56 ( )+ [ c. 236 U→131 I + 3 1 n 92 53 d. 11 Na →−1 e + 29 0 [ e. 47 Sc∗ →47 Sc + 21 21 f. 19 f 40 K →40 C + 20 Ca [ 0 [ [ ] ] ] ] ] 17
- 18. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Exercise 9 1 : E i 9.1 2. Calculate the energy released in the alpha decay below: 238 92 U→ Th + He + Q 234 90 4 2 (Given mass of U 238 mU=238.050786 u ; mass of Th 234 (Gi f U-238, 238 050786 f Th-234, mTh=234.043583 u and mass of α particle , mα=4.002603 u) ANS. : 6.87×10−13 J ANS 6 87× 6.87 87× 3. The following nuclear reaction is obtained : 14 7 N + n → C+ H + 0.55 MeV 1 0 14 6 1 1 14 6 Determine the mass of C in atomic mass unit (u). (Given the mass of nitrogen nucleus is 14.003074 u) (Gi th f it l i 14 003074 ) ANS. : 14.003872 u 18
- 19. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Learning Outcome: L i O t 9.2 Nuclear fission and fusion (1 hour) At the end of this chapter, students should be able to: Distinguish th processes of nuclear fi i and fusion. Di ti i h the f l fission d f i Explain the occurrence of fission and fusion in the form of graph of binding energy per nucleon. g p g gy p Explain chain reaction in nuclear fission of a nuclear reactor. Describe th process of nuclear fusion in the sun. D ib the f l f i i th 19
- 20. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.2 Nuclear fission and fusion 9.2.1 9 2 1 Nuclear fission is defined as a nuclear reaction in which a heavy nucleus splits into two lighter nuclei that are almost equal in mass p g q with the emission of neutrons and energy energy. Nuclear fission releases an amount of energy that is greater than the energy released in chemical reaction reaction. Energy is released because the average binding energy per nucleon of the fission products is greater than that of the parent. parent It can be divided into two types: spontaneous fission – very rarely occur. occur induced fission – bombarding a heavy nucleus with slow neutrons or thermal neutrons of low energy (about 10−2 eV). This Thi type of fission i the i f fi i is h important process i the energy in h production. 20
- 21. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION For F example, consider th b b d l id the bombardment of 235 U b slow t f 92 by l neutrons. One of the possible reaction is 235 1 U + 0 n →236 U∗ →85 Br +148 La 92 92 35 57 1 +30 n + Q Nucleus in the excited state. The reaction can also be represented by the diagram in Figure 9.1. 91 85 35 Br 1 0n 1 0n 1 0n 235 92 U 1 0n 236 ∗ 92 U Figure 9.1 91 Other possible reactions are: 235 1 U + 0 n →236 U∗ →89 Kr +144 Ba 92 92 36 56 148 57 La 1 +30 n + Q 235 1 1 U + 0 n →236 U∗ →94 Sr +139 Xe +30 n 92 92 38 54 +Q 21
- 22. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Most f th fi i f M t of the fission fragments (daughter nuclei) of the uraniumt (d ht l i) f th i 235 have mass numbers from 90 to 100 and from 135 to 145 as shown in Figure 9.2. Figure 9.2 22
- 23. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 5 : Calculate the energy released in MeV when 20 kg of uranium-235 undergoes fission according to 235 1 U + 0 n →89 Kr +144 Ba 92 36 56 1 +30 n + Q (Given the mass of U-235 =235.04393 u, mass of neutron =1.00867 u, mass of Kr-89 =88.91756 u, mass of Ba-144 =143.92273 u and NA =6.02×1023 mol−1) 143.92273 6.02 10 Solution : The mass defect (difference) of fission reaction for one nucleus U235 is ∆m = mi − mf = (mU + mn ) − (mKr + mBa + 3mn ) = (235.04393 + 1.00867) − (88.91756 + 143.92273 + 3 ×1.00867) ∆m = 0.1863 u 23
- 24. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : The energy released corresponds to the mass defect of one U-235 is s Q = ∆m c 2 ( ) ⎡ ⎛ 931 .5 MeV/c 2 ⎞⎤ 2 ⎟⎥ c = ⎢(0.1863 u )⎜ ⎜ ⎟ 1u ⎢ ⎝ ⎠⎥ ⎣ ⎦ Q = 174 MeV 235 × 10−3 kg of uranium-235 contains of 6.02 × 1023 nuclei 20 ⎞ 20 kg of urainum-235 contains of ⎛ 6.02 × 10 23 ⎜ −3 ⎟ ⎝ 235 × 10 ⎠ ( ) = 5.12 × 10 25 nuclei Therefore Energy released by 20 kg U-235 ( ) = 5.12 × 10 25 (174 ) 27 = 8.91 × 10 MeV 24
- 25. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 6 : A uranium-235 nucleus undergoes fission reaction by bombarding it with a slow neutron The reaction produces a strontium 90 nucleus neutron. strontium-90 90 Sr , a nucleus X and three fast neutrons. ( ) 38 a. Write down the expression represents the fission reaction. p p b. If the energy released is 210 MeV, calculate the atomic mass of nucleus X. (Given the mass of U 235 =235.04393 u, mass of neutron (Gi h f U-235 235 04393 f =1.00867 u and mass of Sr-90 =89.90775 u) Solution : a. The expression represents the fission reaction is 235 1 U + 0 n →90 Sr +143X 92 38 54 1 + 30 n + Q 25
- 26. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : The energy released of 210 MeV equivalent to the mass defect for U 35 s U-235 is 2 Q = (∆m )c ⎡ ⎛ 931 .5 MeV/c 2 ⎞⎤ 2 ⎟⎥ c 210 = ⎢(∆m )⎜ ⎜ ⎟ 1u ⎢ ⎝ ⎠⎥ ⎣ ⎦ ∆m = 0.22544 u Therefore the atomic mass of the nucleus X is given by ∆m = mi − mf ∆m = (mU + mn ) − (mSr + mX + 3mn ) 0.22544 = (235.04393 + 1.00867) − (89.90775 + mX + 3 × 1.00867) mX = 142 .8934 u 26
- 27. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.2.2 9 2 2 Chain reaction is defined as a nuclear reaction that is self- sustaining as a selfresult of the products of one fission reaction initiating a subsequent fission reaction. b t fi i reaction ti Figure 9.3 shows a schematic diagram of the chain reaction. Stimulation 9.1 Figure 9.3 27
- 28. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION From Figure 9.3, one neutron initially causes one fission of a uranium-235 nucleus, the two or three neutrons released can go on to cause additional fissions, so the process multiples. p p This reaction obviously occurred in nuclear reactor. Conditions to achieve chain reaction in a nuclear reactor : Slow neutrons are better at causing fission – so uranium are mixed with a material that does not absorb neutrons but slows them down. The fissile material must has a critical size which is defined as the minimum mass of fissile material that will sustain a nuclear chain reaction where the number of neutrons produced in fission reactions should balance the number of neutron escape from the reactor core core. The Th uncontrolled chain reactions are used i nuclear weapons – ll d h i i d in l atomic bomb (Figure 9.4). The controlled chain reactions take place in nuclear reactors p (Figure 9.5) and release energy at a steady rate. 28
- 29. DR.ATAR @ UiTM.NS Figure 9.4 94 PHY310 NUCLEAR REACTION Figure 9.5 95 29
- 30. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.2.3 Nuclear fusion is defined as a type of nuclear reaction in which two light nuclei f se n clei fuse to form a heavier n cle s with the release of hea ier nucleus ith large amounts of energy energy. The energy released in this reaction is called thermonuclear gy energy. Examples of fusion reaction releases the energy are 2 3 1 H + 2 H→ 2 He+ 0 n + Q 1 1 2 H + 2 H→3 H +1 H + Q 1 1 1 1 The two reacting nuclei in f fusion reaction above themselves have to be brought into collision. As bo nuclei a e pos e y c a ged there is a s o g s both uc e are positively charged e e s strong repulsive force between them, which can only be overcome if the reacting nuclei have very high kinetic energies. These high kinetic energies imply temperatures of the order of 108 K. 30
- 31. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION At these elevated temperatures however fusion reactions temperatures, are self sustaining and the reactants are in form of a plasma (i.e. nuclei and free electron) with the nuclei possessing sufficient energy to overcome electrostatic repulsion forces forces. The nuclear fusion reaction can occur in fusion bomb and in the core of a star. Deuterium-tritium f i i other example of f i reaction D i ii fusion is h l f fusion i where it can be represented by the diagram in Figure 9.6. Deuterium Tritium 2 1H Figure 9.6 Alpha p p particle 4 2 He 3 1H Fusion reaction Neutron N t 1 0n 2 1 H+ 3 H→4 He+ 0 n 1 1 2 +Q Stimulation 9.2 31
- 32. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Example 7 : A fusion reaction is represented by the equation below: 2 H + 2 H→3 H +1 H 1 1 1 1 Calculate a. the energy in MeV released from this fusion reaction, b. the energy released from fusion of 1.0 kg deuterium, (Given mass of proton =1.007825 u, mass of tritium =3.016049 u (Gi f 1 007825 f ii 3 016049 and mass of deuterium =2.014102 u) Solution : a. The mass defect of the fusion reaction for 2 deuterium nuclei is ∆m = mi − mf = (mD + mD ) − (mT + mp ) = (2.014102 + 2.014102) − (3.016049 + 1.007825) −3 ∆m = 4.33 × 10 u 32
- 33. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Solution : a. Therefore the energy released in MeV is Q = (∆m )c 2 ⎡ ⎛ 931 .5 MeV/c 2 ⎞⎤ 2 −3 ⎟⎥ c = ⎢ 4.33 × 10 u ⎜ ⎜ ⎟ 1u ⎢ ⎝ ⎠⎥ ⎣ ⎦ Q = 4.03 MeV ( ) b. The mass of 2 deuterium nuclei is 4 ×10−3 kg. 4 × 10−3 kg of deuterium contains of 6.02 × 1023 nuclei 1.0 kg of deuterium contains of ⎛ 1.0 ⎞ 6.02 × 10 23 ⎜ −3 ⎟ ⎝ 4 × 10 ⎠ ( ) = 1.505 × 10 26 nuclei Therefore Energy released from = 1.0 kg deuterium (1.505 ×10 )(4.03) 26 6 26 = 6.07 × 10 MeV 33
- 34. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.2.4 Nuclear fusion in the sun The sun is a small star which generates energy on its own by means of nuclear f i in it i t i f l fusion i its interior. The fuel of fusion reaction comes from the protons available in the sun. The protons undergo a set of fusion reactions, producing isotopes of hydrogen and also isotopes of helium. However, the helium nuclei th h li l i themselves undergo nuclear reactions which l d l ti hi h produce protons again. This means that the protons go through a cycle which is then repeated. Because of this proton-proton cycle, nuclear fusion in the sun can be self sustaining. The set of fusion reactions in the proton-proton cycle can be illustrated by Figure 9 7 9.7. 34
- 35. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION positron (beta plus) 1 1 2 0 1 H +1 H→ 1 H + 1 e + v + Q 2 3 H +1 H→ 2 He + 1 1 neutrino γ +Q 3 3 4 1 1 2 He+ 2 He→ 2 He+1 H +1 H +Q Figure 9.7 Fi 97 The amount of energy released per cycle is about 25 MeV. Nuclear f i occurs i the i N l fusion in h interior of the sun b i f h because the h temperature of the sun is very high (approximately 1.5 × 107 K). 35
- 36. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION 9.2.5 Comparison between fission and fusion Table 9.2 shows the differences between fission and fusion reaction. reaction Fission Fusion Splitting a heavy nucleus into two C bi Combines t two small nuclei t f ll l i to form small nuclei. a larger nucleus. It occurs at very high temperature It occurs at temperature can be (108 K). controlled. Difficult to controlled and a Easier to controlled and sustained controlled reaction has sustained. sustained not yet been achieved. Table 9.2 The similarity between the fission and fusion reactions is both reactions produces energy energy. Graph of binding energy per nucleon against the mass number in Figure 9.8 is used to explain the occurrence of fission and 36 fusion reactions.
- 37. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Binding energy per n nucleon (MeV/nu ucleon) Greatest stability Fission The falling part of the binding energy curve shows that very heavy elements such as uranium can produce energy by fission of their nuclei to nuclei of smaller mass number number. Fusion The rising part of the binding energy curve shows that elements with low mass number can produce energy by fusion fusion. Figure 9.8 Mass number A 37
- 38. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION Exercise 9 2 : E i 9.2 Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u, 1. 1 Complete the following nuclear reaction equations: 3 ]→2 He+ 4 He 3 2 b. 58 Ni+ 2 H → [ ]+1H 28 1 1 a. 6 Li + [ 1 1 c. 235 U + 0 n →138 Xe +50 n + 92 54 [ ] ( ) _____(n, p )16 N 7 d. 9 Be α , ____ 12 C 4 6 e. 2. Calculate the energy released in joule for the following fusion 1 reaction: 2 H + 2 H→ 4 He+ 0 n 1 1 2 (Given the mass of deuterium =3.345×10−27 kg, mass of tritium =5 008×10−27 kg mass of He = 6 647×10−27 kg and =5.008×10 27 kg, 6.647×10 27 mass of neutron =1.675×10−27 kg) 38 ANS. : 2.8×10−12 J 2.8×
- 39. DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION All The Best ll h t Final Exam – Oct 2012 39

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