Question

After $280$ days, the activity of a radioactive sample is $6000\text{dps}$. The activity reduces to $3000\text{dps}$ after another $140$ days. The initial activity of the sample (in $\text{dps}$) is,

• A. $6000$
• B. $9000$
• C. $3000$
• D. $24000$

Answer 1

The correct option is D $24000$

$t_1=280$ days, Activity $A_1=6000\text{dps}$
$t_2=140$ days, Activity $A_2=3000\text{dps}$

Let, initial activity is $A_0$.

Now, for the interval $(0\to t_1)$
$A_1=A_0{e}^{-\lambda t_1}$

$6000=A_0{e}^{-280\lambda }-\left(i\right)$

And, for the interval $t_1\to t_2$
$A_2=A_1{e}^{-\lambda t_2}$

$3000=6000e^{-140\lambda }-\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$, we get,

$\frac{6000}{3000}=\frac{A_0{e}^{-280\lambda }}{6000e^{-140\lambda }}$

$2=\frac{A_0{e}^{-140\lambda }}{6000}-\left(iii\right)$

From $\left(ii\right)$, we can write,
$e^{-140\lambda }=\frac{3000}{6000}=\frac{1}{2}$

Putting this value in $\left(iii\right)$, we get,

$2=\frac{A_0}{6000}\times \frac{1}{2}$

$\therefore A_0=24,000\text{dps}$

Hence, $\left(D\right)$ is the correct answer.