Question 218: Atoms & Nuclei

Question

The activity of a radioactive material is 2.56 $$	imes$$ 10$$-$$3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 $$	imes$$ 10$$-$$5 Ci ?

Accepted Answer

By Radioactive Decay law,
$$
\begin{aligned}
& \mathrm{R}=\mathrm{R}_{\mathrm{o}} e^{-\lambda t} \\
& \Rightarrow 2 	imes 10^{-5}=2.56 	imes 10^{-3} e^{-\lambda t}
\end{aligned}
$$
[Where, $\mathrm{R} =$ Activity at time $t$
$\lambda =$ Activity constant of Radioactive sample
and $$
\lambda = \frac{\ln 2}{\mathrm{~T}_{1 / 2}}
$$
Taking logarithm on both sides
$$\ln \left(2 	imes 10^{-5}ight)=\ln \left(2.56 	imes 10^{-3}ight)+\ln \left(e^{-\lambda t}ight)$$
$$\Rightarrow\ln \left(2 	imes 10^{-5}ight)-\ln \left(2.56 	imes 10^{-3}ight)=-\lambda t$$
$\Rightarrow \ln \left(\frac{2 	imes 10^{-5}}{2.56 	imes 10^{-3}}ight)=-\lambda t$
$\Rightarrow \ln \left(\frac{1}{128}ight)=-\lambda t$
$\Rightarrow-\ln 128=-\lambda t$
$\Rightarrow \ln 2^7=\frac{\ln 2}{\mathrm{~T}_{1 / 2}} t$
$\Rightarrow 7 \ln 2=\frac{\ln 2}{\mathrm{~T}_{1 / 2}} t$
$\Rightarrow t=7 \mathrm{~T}_{1 / 2}$
$\Rightarrow t=7 	imes 5=35$ days

Answer

30 days

Answer

35 days

Answer

40 days

Answer

25 days

Atoms & Nuclei

The activity of a radioactive material is 2.56 $\times 10-3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 \times 10-$5 Ci ?

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