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Question
A sample of 32P initially shows activity of one Curie. After 303 days, the activity falls to 1.5 × 104 dps. What is the half-life of 32P?
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Solution
Given: 1 Ci = 3.7 × 1010 dps,
`("-dN"_0)/"dt" = 3.7 xx 10^10 "dps" and ("-dN"/"dt") = 1.5 xx 10^4` dps, t = 303 days
To find: t1/2
Formulae:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
- `"t"_(1//2) = 0.693/lambda`
Calculation:
- `lambda = 2.303/"t" log_10 ("N"_0/"N")`
`lambda = 2.303/303 log_10 ((3.7 xx 10^10)/(1.5 xx 10^4)) ....[therefore (- "dN")/"dt" prop "N"]`
= 0.04859 d-1 - `"t"_(1//2) = 0.693/lambda = 0.693/0.0459`
= 14.27 d (by using log table)
Half-life of 32P is 14.27 days
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