HW5-7b7c

# HW5-7b7c - 5-6 Radioactivity Chap 5(b Tl/2 = In 2>" =...

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Unformatted text preview: 5-6 Radioactivity Chap. 5 (b) Tl/2 = In 2/>" = 1.18 X 106 S = 326.5 h = 13.6 d = 1.94 wk. (c) t = 1/>" = 1.69 X 106 S = 471 h = 19.6 d = 2.80 wk. 8. The isotope 1321 ecays by /3- emissionto 132Xe ith a half-life of 2.3 h. (a) d w How long will it take for 7/8 of the original number of 1321 uclides to decay? n (b) How long will it take for a sample of 1321o lose 95% of its activity? t Solution: (a) After 3 half-lives, the activity is 1/23 = 1/8 of the original activity. Thus the time to obtain 1/8 of the original activity is t = 3 X Tl/2 = 6.9 h. (b) From the radioactive decay law A(t)/A(O) = exp( -At), upon solving for t, the time required to reach a specified value of A(t)/A(O) is t = -X1 In [ A(ij A(t) In this problem>. = ln2/Tl/2 = 0.3014 h-l and A(t)/A(O) = 0.95 Thus from the above equation we have 9. How many grams of 32p are there in a 5 mCi source? Solution: Since the activity A is related to the mass m of a radionuclide source by and, hence, the mass of the radionuclide is m (g) = AA ~. (P5.4) From Ap. A the half-life of 32p is 14.28 d-1 so that>. = In2/T1/2 = 5.618 X 10-7 s-l. Then from Eq. (P5.4) we find the 32p mass is 1.75 X 10-8 g = 11.5 pg. July 24. 2002 ...
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