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Numerical

A sample of ^{32}P initially shows activity of one Curie. After 303 days, the activity falls to 1.5 × 10^{4} dps. What is the half-life of ^{32}P?

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#### Solution

**Given:** 1 Ci = 3.7 × 10^{10} dps,

`("-dN"_0)/"dt" = 3.7 xx 10^10 "dps" and ("-dN"/"dt") = 1.5 xx 10^4` dps, t = 303 days

**To find:** t_{1/2}

**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 ^{32}P is 14.27 days

Concept: Radioactive Decays

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