HW6-12 - 5-8 Radioactivity Chap. 5 12. A 6.2 mg sample of...

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Unformatted text preview: 5-8 Radioactivity Chap. 5 12. A 6.2 mg sample of gOSr(half-life 29.12 y) is in secular equilibrium with its daughter gOy(half-life 64.0 h). (a) How many Bq of gOSr re present? (b) How a many Bq of gOy are present? (c) What is the massof gOypresent? (d) What will the activity of gOybe after 100 y? Solution: The decay chain of concernis 90Sr --+ 29.12 y 90y --+ 64.0 h 90Zr (stable). First some preliminary calculations. The relevant decayconstants are >'Sr = 1n2 Tl/2 = In2 = 7.543 X 10-10 S-I, (29.12 y)(3.1557 x 107 sly) and The number of atoms of 90Srinitially present is Nsr= mSrNa Asr = (0.0062g)(6.022 x 1023atoms/mol) = 4.149 X 1019atoms. 90 gfmol (a) The activity of 90Sris Asr = >'srNsr = 3.129 X 1010 Bq. (b) In secular equilibrium ASr = Ay = >'SrNSr = >.yNy. Thus Ay = Asr = 3.129 X 1010 Bq. (c) Because Ny = myNa/Ay = Ay/Ay we find AyAy AyNa (d) After 100 y A(lOO y) = Asr(lOO y) = Asr(O)exp[ -AsrlOO] = 3.129 X 1010 (Bq) exp[-7.543 X 10-10 5-1 X 3.156 X 109 5] = 2.89 X 109 Bq. July 24, 2002 my= ...
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