Modern Physics for Engineers

Problem Set 8

Problem Set - Due in class, Wednesday, Nov. 17, 1999.

    Problems:
  1. SM&M 13-46

    Part A) Energy is not conserved. The mass of the products is greater than the mass of the of the reactants.

    Part B) Now the binding energy is sufficient such that the mass of the products is less than the mass of the reactants.

    Part C) 

    	delta E = delta m c^2 = (m_N - m_C - 2 m_e) c^2
	        =(13.005739 u - 13.003355 u - 2*.000549 u) * 931.5 MeV/c^2/u * c^2
	        = 1.2 MeV
  2. SM&M 13-50 
    U235->Th231
        |
       \_/
      Pa231->Ac227->Fr223->At219->Bi215
               |      |      |      |
              \_/    \_/    \_/    \_/
             Th227->Ra223->Rn219->Po215->Pb211
                                           |
                                          \_/
                                         Bi211->Tl207
                                           |      |
                                          \_/    \_/
                                         Po211->Pb207
  3. SM&M 14-2 
    Part A)	KE = 3/2 k T = 1.5 8.6e-5 eV/oK 6e8 oK = 7.7e4 eV = 77 keV
Part B)	Ne20 formation
		Delta E = (2 m_C - m_Ne - m_He) c^2
		        = (2 * 12 u - 19.992440 u - 4.002603 u)*931.5 MeV/u
		        = 4.62 MeV
		Mg24 formation
		Delta E = (2 m_C - m_Mg) c^2
		        = (2 * 12 u - 23.985042 u)*931.5 MeV/u
		        = 13.9 MeV
Part C)
	Number of Carbon atoms = 2e3 gm * 6.02e23 atoms/12 gm = 1e26 atoms
	Energy released = 1e26 atoms * 4.62e6 eV/2 atoms * 1.6e-19 joules/eV
	                = 3.7e13 joules = 3.7e13 watt sec 
	                = 3.7e10 kwatt sec *1 min/60 sec *1 hr/60 min
	                = 1e7 kwatt hrs = 10 Gwatt-hrs
  4. SM&M 14-34 
    part A)
	Number of Pu atoms = 1e3 g * 6.02e23 atoms/239.05 g = 2.5e24 atoms
	Energy = 2.5e24 * 2e8 eV * 4.44e-26 kwatt-hr/eV = 2.2e7 kW-hr
Part B)
	Delta M = m_H^3 + m_H^2 - m_He^4 - m_n
	        = (3.016049 + 2.014102 - 4.002603 - 1.008665 )u
	        = .018883 u
	Delta E = Delta M c^2
	        = .018883 u * 931.4943 MeV/c^2/u c^2
	        = 17.59 MeV
Part C)
	Number of deutrium = 1e3 g 6.022e23 atoms/2.014102 g = 2.99e26 atoms
	Energy released = 2.99e26 atoms * 17.59e6 eV/atom * 4.44e-26 kwatt-hr/eV
	                = 2.3e8 kW-hrs
Part D)
	Number of C atoms = 1e3 g 6.022e23 atoms/12 g = 5e25 atoms
	Energy released = 5e25 atoms *4.2 eV/atom 4.44e-26 kwatt-hrs/eV
	                = 9.4 kwatt-hrs
part E)
	Various.
  5. SM&M 14-49
    The technician takes 8 X-rays/day * 5 days/week * 52 weeks/year = 2080 X-rays/year. This implies a dose of 5 rem/year/2080 X-Rays/year = .0024 rem/X-ray. This is approximately 1/100 of thebackground radiation/X-Ray. However, the total yearly exposure is about 10 times background.

    Please note that the person receiving the X-Ray receives a much higher does (about 100 times higher) and can receive the equivalent of the yearly dose of background radiation in one X-ray.