Exercise J.3.1. Answers 1. m = 800kg v = 70kmh -1 = 70 103 = 60 60 19.4ms-1 KK. EE = 1 2 mmvv2 = 800 19.42 2 = 150544JJ Using E= mc 2 and the kinetic energy of the car we obtain mm = EE cc 2 = 151235 (3 10 8 ) 2 = 1.68 10 12 kkkk 2. (a) EE = mmcc 2 = 1kkkk (3 10 8 ) 2 = 9 10 16 JJ (b) The type of matter is irrelevant since the energy is directly proportional to mass only (c) Energy released by a 50g mars bar; EE = mmcc 2 = 0.05kkkk (3 10 8 ) 2 = 4.5 10 15 JJ The power rating of the light bulb is 100W. Using power = energy / time we obtain tt = EE PP = 4.5 1015 JJ 100WW = 4.5 1013 ss = 1.43 mmmmmmmmmmmmmm yyyyyyyyyy 3. Nuclear fission occurs when the Coulombic force pushing protons away from each other and the strong nuclear force holding the nucleons together becomes unbalanced in large nuclei causing them to become unstable. In natural fission the nucleus randomly splits into two new and more stable nuclei, releasing energy and up to three s in the process. Induced fission is when a is fired at an unstable nuclei, initiating the fission event. Thus fission; involves splitting apart occurs in large unstable nuclei can be random or induced
Fusion is the fusing together of lighter nuclei. With sufficient energy coulombic repulsion can be overcome and two light nuclei can be brought so close together that the strong nuclear force is able to hold them together to form a new heavier nucleus, releasing energy in the process. Fusion is the fuelling process for our own sun. Thus fusion involves putting together only occurs for lighter nuclei requires very high thermal energies to initiate 4. fission fragment X U-235 fission fragment Y 5. The mass defect is given by; m = 9.111822 10-31 kg 9.11 10-31 kg = 1.822 10-34 kg The energy released in the form of the kinetic energy of the electron is then EE = mmcc 2 = 1.822 10 34 (3 10 8 ) 2 = 1.6398 10 17 JJ Rearranging KE = 1/2mv 2 to make v the subject and using the mass of the electron and the energy calculated above we obtain vv = 2KKKK mm = 2 1.6398 10 17 9.11 10 31 = 6 10 6 mmss 1
Exercise J.3.2. Answers 1. Induced fission occurs randomly in unstable nuclei when they split to form two smaller and more stable nuclei. Induced fission occurs when a of the correct energy is fired at unstable nuclei and absorbed, causing the nuclei to split. 2. 1024 3. In a nuclear reactor the moderator (usually water) acts to increase the rate of reactions by slowing high-energy s via collisions to speeds where they can cause new fission reactions. The control rods reduce the rate of reaction by absorbing s before they can initiate a fission event. The control rods can be raised and lowered s desired. The lower the control rods the more s they absorb and the slower the rate of reaction. 4. (a). Either boron, cadmium, silver or indium. (b). The U-235 isotope is used as a fuel for a nuclear reactor because during the fission process the U-235 nucleus splits and produces enough new s that a chain reaction may be sustained. The U-238 isotope produces to few s when it splits, and therefore for U-238 a chain reaction is not sustainable. 5. You can simply multiply by 2 each time until the number of nuclei in the 10g sample is reached (74 th generation). Alternatively if you understand logarithms, since the number of free s initiating an event raised to the power of the number of generations gives the number of events; therefore 2 nn = 2.5 10 22 n log(2) = log (2.5 10 22 ) n = log(2.5 1022 ) log(2) The sample is used up during the 74 th generation. = 74.4 Since there is 2 10-6 s between generations, this occurs in 2 10-6 s 74 = 0.15 milliseconds.
Exercise J.3.3. Answers 1. Note that in the following, a HH 1 1 nucleus is simply a proton Stage 1: Stage 2: Stage 3: 1 1 2 1HH + 1HH 1 HH + 1ee + 3 0 1 2 1HH + 1HH 2HHHH 3 3 4 2HHHH + 2HHHH 2HHHH + 1HH + 1 1 1 HH 2. Using E= mc 2, mm = EE cc 2 = 4.11 10 12 JJ (3 10 8 ) 2 = 4.57 10 29 kkkk 3. The proton-electron-proton or PEP-chain 4. Two protons are brought together at high energy forcing one of the protons to turn into a, forming a hydrogen isotope (deuterium) and releasing a positron. Another proton is then added forming a helium isotope, consisting of 2 protons and 1. These two stages must then repeat to form a second helium isotope, before the two may be fused together to form a helium nucleus (2p, 2n) and releasing 2 protons. 5. Given that the Sun has a mass of 2 10 30 kg and converts mass into 3.8 10 26 J of energy every second, calculate how many years it will take for the mass of the sun to reduce by ¼. A loss of ¼ of the total mass of the sun represents a loss of 2 10 30 0.25 = 5 10 29 kg. From E = mc 2, every second the sun looses mm = EE cc 2 = 3.8 1026 JJ (3 10 8 ) 2 = 4. 2 10 9 kkkk then 5 10 29 4. 2 10 = 1.19 9 1020 ss = 3.8 10 12 yyyyyyyyyy