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 $\mathrm{dps}$?

Answer 1

Step 1. Given data

Activity reduces from $6000$ to $3000\mathrm{dps}$ i. e. half.

${\mathrm{T}}_{\frac{1}{2}}=140$ days

$280$ days $=2$ half life's

After $280$ days $=6000\mathrm{dps}$ activity

Step 2. Concept to be used

The 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,

$\mathrm{\lambda }=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{A}}_0}{\mathrm{A}}$

Here, ${\mathrm{A}}_0$ is the initial activity and $\mathrm{A}$ is the activity at time $\mathrm{t}$.

Step 3. Calculate the initial activity

Now, we will taking the number of days as time.

Substitute the values in the above expression, we get,

$\mathrm{\lambda }=\frac{2.303}{280}\log \frac{{\mathrm{A}}_0}{6000}$ -------- $\left(1\right)$

and

$\mathrm{\lambda }=\frac{2.303}{420}\log \frac{{\mathrm{A}}_0}{3000}$ ------- $\left(2\right)$

Equate the both equations $\left(1\right)$ and $\left(2\right)$, we get,

$\frac{2.303}{280}\log \frac{{\mathrm{A}}_0}{6000}=\frac{2.303}{420}\log \frac{{\mathrm{A}}_0}{3000}$

$\log \frac{{\mathrm{A}}_0}{6000}=\frac{1}{1.5}\log \frac{{\mathrm{A}}_0}{3000}$

Step 4. Solve by using exponential rule.

Take the exponential on both sides and solving the above equation we get,

$\log \frac{{\mathrm{A}}_0}{6000}=\frac{1}{1.5}\log \frac{{\mathrm{A}}_0}{3000}$

Multiply both sides by $1.5$.

So,

${\log }_{10}\left(\frac{{\mathrm{A}}_0}{6000}\right)\times 1.5=\frac{1}{1.5}{\log }_{10}\left(\frac{{\mathrm{A}}_0}{3000}\right)\times 1.5$

${\log }_{10}\left(\frac{{\mathrm{A}}_0}{6000}\right)\times 1.5={\log }_{10}\left(\frac{{\mathrm{A}}_0}{3000}\right)$

Use log rules ${\log }_{\mathrm{a}}{\mathrm{x}}^{\mathrm{n}}={\mathrm{nlog}}_{\mathrm{a}}\mathrm{x}$, we get,

${\left(0.000166666{\mathrm{A}}_0\right)}^{1.5}=\left(\frac{{\mathrm{A}}_0}{3000}\right)$

multiply both sides by $3000$.

So,

${\left(0.000166666{\mathrm{A}}_0\right)}^{1.5}\times 3000=\left(\frac{{\mathrm{A}}_0}{3000}\times 3000\right)$

${\left(0.000166666{\mathrm{A}}_0\right)}^{1.5}\times 3000={\mathrm{A}}_0$

Take both sides of the equation to the power of $2$, we get,

${\left({\left(0.000166666{\mathrm{A}}_0\right)}^{1.5}\times 3000\right)}^2={\left({\mathrm{A}}_0\right)}^2$

$0.000041666{{\mathrm{A}}_0}^3={\left({\mathrm{A}}_0\right)}^2$

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 principle

If $\mathrm{ab}=0$ 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 $24,000\mathrm{dps}$.