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

\({V}=\frac{{{3.7}\times{10}^{{{4}}}}}{{{373}/{60}}}={5951.7}{c}{m}^{{{3}}}={5.951}{l}{i}{t}{r}{e}\)

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

\({V}=\frac{{{3.7}\times{10}^{{{4}}}}}{{{373}/{60}}}={5951.7}{c}{m}^{{{3}}}={5.951}{l}{i}{t}{r}{e}\)

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