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## QuestionPhysicsClass 12

A small quantity of solution containing $${N}{e}^{{{24}}}$$ radio nucliode (half life $$={15}{h}{o}{u}{r}$$) of activity $${1.0}$$ microcurlar is injected into the blood of a person A sample of the blood of volume $${1}{c}{m}^{{{3}}}$$ taken a after $${5}$$ hour shown an activity of the blood in the body of the person . Assume that redicative solution mixed uniformly in the blood of the person $${\left({1}{c}{u}{r}{i}{e}={3.7}\times{10}^{{{10}}}\right.}$$ disntegrations per sound)

$${t}_{{{1}/{2}}}={15}{h}{o}{u}{r}$$
Actually $${A}_{{{0}}}={10}^{{-{6}}}$$ curie (in small quentyty of solution of $$^{\left({24}\right)}{N}{a}{)}={3.7}\times{10}^{{{4}}}{d}{p}{s}$$
Observation of blood of volume $${1}{c}{m}^{{{3}}}$$
After $${5}{h}{o}{u}{r}.,{A}={296}{d}{p}{s}$$
The initial activity can be found by the formula
$${t}=\frac{{{2.303}}}{{\lambda}}\frac{{{\log}_{{{10}}}{\left({A}_{{{0}}}\right)}}}{{{A}}}\Rightarrow{5}=\frac{{{2.303}}}{{{0.693}/{15}}}\times\frac{{{\log}_{{{10}}}{\left({A}_{{{0}}}\right)}}}{{{296}}}$$
$$\Rightarrow\frac{{{\log}_{{{10}}}{\left({A}_{{{0}}}\right)}}}{{{296}}}=\frac{{{5}\times{0.693}}}{{{2.303}\times{15}}}=\frac{{{0.33010}}}{{{3}}}={0.10033}$$
$$\Rightarrow\frac{{{A}_{{{0}}}}}{{{296}}}={1.26}\Rightarrow{A}_{{{0}}}={373}{d}\pm=\frac{{{373}}}{{{60}}}{d}{p}{s}$$
This is the activity level in $${1}{c}{m}^{{{3}}}$$ Comparing it with the initial activity level of $${3.7}\times{10}^{{{4}}}{d}{p}{s}$$we find the volume of blood
$${V}=\frac{{{3.7}\times{10}^{{{4}}}}}{{{373}/{60}}}={5951.7}{c}{m}^{{{3}}}={5.951}{l}{i}{t}{r}{e}$$