1272_Physics ProblemsTechnical Physics

# 1272_Physics ProblemsTechnical Physics - Chapter 45 For a...

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Chapter 45 613 For a more accurate calculation of the kinetic energy, we should use relativistic expressions. Conservation of momentum gives γγ ααα nnn mm vv += 0 10087 1 40026 1 22 .. v vc v n n = α yielding v c v cv n n 2 2 2 15 746 14 746 = . Then αα nn mc −+ −= 11 1 7 6 bg . M eV and n = 0171 . , implying that γ 4 0 2 eV . P45.54 From Table A.3, the half-life of 32 P is 14.26 d. Thus, the decay constant is λ == = = × −− ln ln . 14 26 0 048 6 5 63 10 12 7 T d d s . N R 0 0 6 7 12 522 10 563 10 928 10 × × . . . decay s s nuclei 1 At t = 10.0 days, the number remaining is NN e e t × = × 0 12 00486 100 12 9 28 10 5 71 10 nuclei nuclei d 1 ej so the number of decays has been 0 12 357 10 −= × . and the energy released is E × = 700 160 10 0400 12 16 ... a f keV J keV J. If this energy is absorbed by 100 g of tissue, the absorbed dose is Dose J .100 kg rad Jkg rad = F H G I K J F H G I K J = 0 1 10 400 2 . . P45.55 (a) The number of Pu nuclei in 602 10 239 05 1000 23 . . . kg nuclei mol
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## This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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