Download Modern Physics Homework: Conservation of Energy and Momentum in Nuclear Reactions and more Assignments Physics in PDF only on Docsity! Homework 13 Physics 2130 Modern Physics Due: Wednesday December 11, 2002 1. T & Z, #13.50. (a)M(U) = mass of uranium consumed per year. Mass of 235U consumed per year =M(235U) = (0:0072)M(U) because 235U constitutes only 0:72% of all uranium. Energy produced by ssion = (30%) (number of ssions ) (energy per ssion) = 0:3 0:0072M(U)6:02 1023 particles/mole 235 g/mole ! 200 MeV/ ssion(1:6 10 13 J/MeV) = 1:77 108 J/gM(U): Energy per year, power plant = 109 W(3:15 107 s/y) = 3:15 1016 J/y: 1:77 108 J/gM(U) = 3:15 1016 J/y M(U) = 2 108 g = 2 105 kg = 200 tonnes (metric tons) (b) Mass of coal = M(coal). Energy produced by coal = (0:40)M(coal)(33 106 J/kg) = (1:31 107 J/kg)M(coal) = 3:15 1016 J M(coal) = 2:4 106 tonnes (c) M(coal) = 24,000 box car loads 240 trains. (d) M(U) = 2 box car loads < 1 train. 2. T & Z, #13.51. m neutron Initial M p p’ m nucleus P M Final Figure 1: Conservation of momentum P = p p0 (1) Conservation of kinetic energy P 2 2M = p2 2m + p02 2m ! P 2 = p2 p02 (2) Note that kinetic energy is conserved because this is an elastic collision. Dividing equation (2) by (1) and adding the result to equation (1) yields: (1 + )P = p! P = 2p 1 + (3) The kinetic energy lost by the neutron, K, will be just equal to the kinetic energy of the recoiling nucleus: K = P 2 2M (3) = 4p2 2M(1 + )2 K K = 4p2 2M(1 + )2 2m p2 = 4m M 1 (1 + )2 = 4 (1 + )2 = f() (4) where we have noted that the relative kinetic energy loss of the incident photon is a function of mu and call that function f(). (b) To maximize equation (4) with respect to we take t he derivative and set it equal to 0: df() d = 4(1 ) (1 + )3 = 0! = 1; (5) that is, energy loss is maximum when the two masses are equal. (c) Deuterium: = 1=2 ! K=K = 0:89. Carbon: = 1=12 ! K=K = 0:28. Note that if K=K = 1, the neutron is brought to a complete halt during the collision. 3. T & Z, #13.55. 1 1H + 12C proton capture! 13N + (6) 13N +-decay! 13C + e+ + (7) 1 1H + 13C proton capture! 14N + (8) 1 1H + 14N proton capture! 15O + (9) 15O +-decay! 15N + e+ + (10) 1 1H + 15N -decay! 12C + 4He (11) (b) 4P ! 3 + 2e+ + 2 + 4He. (c) In the CNO cycle, the highest Coulombic barrier is in the last step, reaction (11): UCNO = 7ke2 r1 : In the p p cycle, the highest barrier is for 3He+ 3He! 4He+ 2p: Upp = 4ke2 r2 : If r1 = r2, UCNO=Upp = 7=4. Since the average kinetic energy is proportional to temperature: TCNO Tpp 7 4 : 4. T & Z, #13.56. (3 pts) T & Z #13.56. a = 2ke2 p 2m=h = 2(1:44 MeV fm q 2(4:003)(931:5)=197 MeV fm = 3:97 (MeV)1=2 b = 8 p mke2=h = 8 q 4:003(931:5 MeV)(1:44 MeV fm)=197 MeV fm = 2:98 (fm) 1=2
