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Question
What is the amount of \[\ce{_27^60Co}\] necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
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Solution
Data: Activity= 10.0 mCi = 10.0 x 10-3 Ci = (10.0 x 10-3)(3.7 x 1010) dis/s = 3.7 x 108 dis/s
T1/2 = 5.3 years = (5.3)(3.156 × 107)s = 1.673 × 108 s
Decay constant, `lambda = 0.693/("T"_(1//2)) = 0.693/(1.673 xx 10^8) "s"^-1`
= 4.142 x 10-9 s-1
∴ N = `"activity"/lambda = (3.7 xx 10^8)/(4.142 xx 10^-9)` atoms
= 8.933 × 1016 atoms
= 60 grams of \[\ce{_27^60Co}\] contain 6.02 x 1023 atoms
∴ Mass of 8.933 x 1016 atoms of \[\ce{_27^60Co}\]
`= (8.933 xx 10^16)/(6.02 xx 10^23) xx 60 "g"`
= 8.903 x 10-6 g = 8.903 µg
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