Question

# After $280$ days, the activity of a radioactive sample is $=6000\mathrm{dps}$. The activity reduces to$3000\mathrm{dps}$ after another $140$ days. What Is The initial activity of the sample in $\mathrm{dps}$?

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## Step 1. Given dataActivity reduces from $6000$ to $3000\mathrm{dps}$ i. e. half. ${\mathrm{T}}_{\frac{1}{2}}=140$ days$280$ days $=2$ half life'sAfter $280$ days $=6000\mathrm{dps}$ activityStep 2. Concept to be usedThe decays constant $\left(\mathrm{\lambda }\right)$ is the inverse of the mean lifetime (average lifetime of a radioactive particle before decay).So, the rate equation in first order reaction is,$\mathrm{\lambda }=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{{\mathrm{A}}_{0}}{\mathrm{A}}$Here, ${\mathrm{A}}_{0}$ is the initial activity and $\mathrm{A}$ is the activity at time $\mathrm{t}$.Step 3. Calculate the initial activityNow, we will taking the number of days as time.Substitute the values in the above expression, we get, $\mathrm{\lambda }=\frac{2.303}{280}\mathrm{log}\frac{{\mathrm{A}}_{0}}{6000}$ -------- $\left(1\right)$and $\mathrm{\lambda }=\frac{2.303}{420}\mathrm{log}\frac{{\mathrm{A}}_{0}}{3000}$ ------- $\left(2\right)$Equate the both equations $\left(1\right)$ and $\left(2\right)$, we get, $\frac{2.303}{280}\mathrm{log}\frac{{\mathrm{A}}_{0}}{6000}=\frac{2.303}{420}\mathrm{log}\frac{{\mathrm{A}}_{0}}{3000}$ $\mathrm{log}\frac{{\mathrm{A}}_{0}}{6000}=\frac{1}{1.5}\mathrm{log}\frac{{\mathrm{A}}_{0}}{3000}$Step 4. Solve by using exponential rule.Take the exponential on both sides and solving the above equation we get, $\mathrm{log}\frac{{\mathrm{A}}_{0}}{6000}=\frac{1}{1.5}\mathrm{log}\frac{{\mathrm{A}}_{0}}{3000}$Multiply both sides by $1.5$.So, ${\mathrm{log}}_{10}\left(\frac{{\mathrm{A}}_{0}}{6000}\right)×1.5=\frac{1}{1.5}{\mathrm{log}}_{10}\left(\frac{{\mathrm{A}}_{0}}{3000}\right)×1.5$ ${\mathrm{log}}_{10}\left(\frac{{\mathrm{A}}_{0}}{6000}\right)×1.5={\mathrm{log}}_{10}\left(\frac{{\mathrm{A}}_{0}}{3000}\right)$Use log rules ${\mathrm{log}}_{\mathrm{a}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{nlog}}_{\mathrm{a}}\mathrm{x}$, we get, ${\left(0.000166666{\mathrm{A}}_{0}\right)}^{1.5}=\left(\frac{{\mathrm{A}}_{0}}{3000}\right)$multiply both sides by $3000$. So, ${\left(0.000166666{\mathrm{A}}_{0}\right)}^{1.5}×3000=\left(\frac{{\mathrm{A}}_{0}}{3000}×3000\right)$ ${\left(0.000166666{\mathrm{A}}_{0}\right)}^{1.5}×3000={\mathrm{A}}_{0}$Take both sides of the equation to the power of $2$, we get,${\left({\left(0.000166666{\mathrm{A}}_{0}\right)}^{1.5}×3000\right)}^{2}={\left({\mathrm{A}}_{0}\right)}^{2}$ $0.000041666{{\mathrm{A}}_{0}}^{3}={\left({\mathrm{A}}_{0}\right)}^{2}$Substract ${{\mathrm{A}}_{0}}^{2}$ from both sides, we get,$0.000041666{{\mathrm{A}}_{0}}^{3}-{{\mathrm{A}}_{0}}^{2}{={\mathrm{A}}_{0}}^{2}-{{\mathrm{A}}_{0}}^{2}$$0.000041666{{\mathrm{A}}_{0}}^{3}-{{\mathrm{A}}_{0}}^{2}={{\mathrm{A}}_{0}}^{2}-{{\mathrm{A}}_{0}}^{2}$ $0.000041666{\mathrm{A}}_{0}-1=0$Step 5. Use zero factor principleIf $\mathrm{ab}=0$ then $\mathrm{a}=0$ or $\mathrm{b}=0$.So,$0.000041666{\mathrm{A}}_{0}-1=0$ ${\mathrm{A}}_{0}=\frac{1}{0.000041666}$ ${\mathrm{A}}_{0}=24000\mathrm{dps}$Hence, the initial activity is $24,000\mathrm{dps}$.

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