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hw11_solutions

# hw11_solutions - PHY 251 Fall 2009 homework problem set 11...

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PHY 251 Fall 2009: homework problem set 11, due in the PHY 251 drop box in room A-129 by noon on Friday, Dec. 11. 1. Serway 14.3 Answer: The reaction is 9 4 Be( α ,n) 12 6 C or 9 4 Be + 4 2 He 12 6 C + 1 0 n The mass change is Δ m = 9 . 012 182 + 4 . 002 603 12 . 000 000 + 1 . 008 665 = 0 . 006 120 u so Q = Δ m c 2 = (0 . 006 120 u ) · (931 . 49 MeV /c 2 ) = 5 . 70 MeV. 2. Serway 14.7 Answer: The value of Q for 4 2 He + 14 7 N 17 8 O + 1 1 H is Q = (4 . 002 603 + 14 . 003 074 16 . 999 131 1 . 007 825) 931 . 494 MeV /c 2 amu c 2 = 1 . 19 MeV . Therefore we need the incoming α particle to have a kinetic energy of at least 1.19 MeV to drive the reaction. Next, we want to find the Q of 1 1 H + 7 3 Li 4 2 He + 4 2 He, which is Q = (1 . 007 825 + 7 . 016 004 4 . 002 603 4 . 002 603) 931 . 494 MeV /c 2 amu c 2 = 17 . 35 MeV so the alpha particles that emerge are very energetic. 3. Serway 14.8 Answer: The reaction 4 2 He + 9 4 Be 12 6 C + 1 0 n has Q = (4 . 002 603 + 9 . 012 182 12 . 000 000 1 . 008 665 u ) · 931 . 49 MeV /c 2 u = 5 . 70 MeV as calculated in problem 14.3. The reaction 2 1 H + 2 1 H 3 2 He + 1 0 n has Q = (2 . 014 102 + 2 . 014 102 3 . 016 029 1 . 008 665 u ) · 931 . 49 MeV /c 2 u = 3 . 27 MeV which is positive (meaning mass is lost) so it is exothermic (releases heat). 1

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4. Serway 14.22 Answer: In example 14.4 on page 513 of Serway, we find that one 235 U fission event releases 208 MeV. Therefore the number x of fission events needed per day to generate 1000 MW is x = ( 1 decay 208 × 10 6 eV ) · ( eV 1 . 602 × 10 - 19 Joules ) · ( 10 9 Joules sec ) · ( 24 · 60 · 60 sec 1 day ) or x = 2 . 59 × 10 24 decays per day. The mass m of the number of 235 U nuclei which must undergo fission in a day is then m = (2 . 59 × 10 24 atoms ) · (235 . 043 amu atom ) · (1 . 661 × 10 - 27 kg amu ) = 1 . 01 kg . Because density ρ is mass per volume, we have ρ = m (4 / 3) πr 3 r = parenleftBigg 3 m 4 πρ parenrightBigg 1 / 3 = parenleftBigg 3 (1010 g ) 4 π (19 . 0 g/cm 3 ) parenrightBigg 1 / 3 = 2 . 33 cm or a sphere with a diameter of less than two inches. 5. Serway 14.24 Answer: The estimate for the total mass m of recoverable 235 U is m = (10 9 tons U ) · ( 2000 pounds ton ) · ( 1 kg 2 . 204 pounds ) · ( 0 . 007 235 U U ) = 6 . 35 × 10 9 kg The total amount of energy E that can be released from this is E = (6 . 35 × 10 9 kg ) · ( 1 amu 1 . 661 × 10 - 27 kg ) · ( 1 atom 235 . 043 amu ) · ( 208
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hw11_solutions - PHY 251 Fall 2009 homework problem set 11...

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