The radioisotope $_{15} P^{32}$ is used in biochemical studies. A sample containing this isotope has an activity 1000 times the detectable limit. The experiment could be run with the sample for $x$ days before the radioactivity could not be detected, then $x$ is
$t_{1 / 2}$ of $P^{32}$ is $14.2$ days.
(write the value to the nearest integer)

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Solution

Rate of decay = $1000$ $\times$ Rate of detectable limit
or $\frac{R a t e o f d e c a y}{R a t e o f d e t e c t a b l e l i m i t} = 1000$
$∴ t = \frac{2.303}{K} l o g \frac{N_{0}}{N}$
$= \frac{2.303}{K} l o g \frac{R a t e o f d e c a y}{R a t e o f d e t e c t a b l e l i m i t}, (∵ R a t e \propto N_{0})$
$= \frac{2.303 \times 14.2}{0.693}$ $l o g 1000$
$= 141.57$ days

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