Question

The half life of a radioactive material is T. If 80% of the sample decay after time t, then find out the time when 90% of the sample decay.
A. $T+t$
B. $T+2t$
C. $2T+t$
D. $\dfrac{t}{T}$

Answer 1

During a radioactive decay, the amount of the substance left in the sample of the radioactive material at time t is given as $N={{N}_{0}}{{e}^{-\lambda t}}$. Use this formula and the formula for half life of a radioactive material.

Formula used:
$N={{N}_{0}}{{e}^{-\lambda t}}$
Where N is the amount of substance left in the sample at time t, ${{N}_{0}}$ is the initial amount of the sample (i.e. at time $t=0$) and $\lambda$ is a constant called decay constant.
${{T}_{1/2}}=\dfrac{\ln 2}{\lambda }$

Complete step by step answer:
During a radioactive decay, the amount of the substance left in the sample of the radioactive material at time t is given as $N={{N}_{0}}{{e}^{-\lambda t}}$ …. (i),
The half life of a radioactive material is given as ${{T}_{1/2}}=\dfrac{\ln 2}{\lambda }$.
It is given that the half life of the radioactive material is T.
$\Rightarrow {{T}_{1/2}}=T=\dfrac{\ln 2}{\lambda }$.
$\Rightarrow \lambda =\dfrac{\ln 2}{T}$
Substitute the value of $\lambda$ in (i).
$\Rightarrow N={{N}_{0}}{{e}^{-\dfrac{\ln 2}{T}t}}$ …. (ii).
When 90% of the sample decays, 10% of the initial amount of the sample is left. Let the time at which this happens be t’.
Therefore, at time t’ the value of N is $N=\dfrac{10}{100}{{N}_{0}}$.
Substitute this value of N and $t=t'$ in (ii).
$\Rightarrow \dfrac{10}{100}{{N}_{0}}={{N}_{0}}{{e}^{-\dfrac{\ln 2}{T}t}}$
$\Rightarrow \dfrac{1}{10}={{e}^{-\dfrac{\ln 2}{T}t'}}$
$\Rightarrow \ln \left( \dfrac{1}{10} \right)=-\dfrac{\ln 2}{T}t'$
$\Rightarrow \ln 10=\dfrac{\ln 2}{T}t'$
$\Rightarrow \dfrac{t'}{T}=\dfrac{\ln 10}{\ln 2}$ … (iii).
Now, it is said that 80% of the sample decays in time t. That means at time t, 20% of the initial amount of the sample is left.
i.e. $N=\dfrac{20}{100}{{N}_{0}}$.
Substitute this value of N and the value of $\lambda$ in (i).
$\Rightarrow \dfrac{20}{100}{{N}_{0}}={{N}_{0}}{{e}^{-\dfrac{\ln 2}{T}t}}$
$\Rightarrow \dfrac{2}{10}={{e}^{-\dfrac{\ln 2}{T}t}}$
$\Rightarrow \ln \left( \dfrac{2}{10} \right)=-\dfrac{\ln 2}{T}t$
By using properties of log….
$\Rightarrow \ln \left( \dfrac{10}{2} \right)=\dfrac{\ln 2}{T}t$
$\Rightarrow \dfrac{\ln \left( \dfrac{10}{2} \right)}{\ln 2}=\dfrac{t}{T}$
Now, $\ln \left( \dfrac{10}{2} \right)$ can be written as $\ln \left( \dfrac{10}{2} \right)=\ln 10-\ln 2$
Then,
$\Rightarrow \dfrac{t}{T}=\dfrac{\ln 10-\ln 2}{\ln 2}=\dfrac{\ln 10}{\ln 2}-1$
$\Rightarrow \dfrac{\ln 10}{\ln 2}=\dfrac{t}{T}+1$.
Substitute this value in (iii).
$\Rightarrow \dfrac{t'}{T}=\dfrac{t}{T}+1$
$\Rightarrow \dfrac{t'}{T}=\dfrac{t+T}{T}$
$\therefore t'=T+t$.

Hence, the correct option is A.

Note: Half life of a radioactive material is the time at which half of the initial amount of the substance is left in the sample or we can also say that it is the time at which half of the substance is decayed. The value of decay constant varies with the radioactive materials.