Question

Ratio of initial nuclei in two different samples is $2:3$. Their half-lives are 1 hour and 2 hours respectively. Ratio of active nuclei at the end of 6 hours will be?
$\begin{aligned} & \text{A}\text{. }1:1 \\\ & \text{B}\text{. }1:12 \\\ & \text{C}\text{. 4}:3 \\\ & \text{D}\text{. 3}:4 \\\ \end{aligned}$

Answer 1

The process by which an unstable atomic nucleus loses energy by radiation is called radioactive decay. The radioactive decays per unit time are directly proportional to the number of nuclei of radioactive compounds in the sample. We will use the expression for radioactive decay relating the number of decayed nuclei to the initial number of nuclei and the decay constant.

Formula used:
Radioactive decay law, $\dfrac{dN}{dt}=-\lambda N$
Number of nuclei decayed in particular time:
$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$

Complete step by step answer:
Radioactive decay is described as the process by which an unstable atomic nucleus loses energy by radiation. A sample material containing radioactive nuclei is considered as radioactive. The decay of radioactive elements occurs at a fixed constant rate.
According to the radioactive decay law, the radioactive decays per unit time are directly proportional to the number of nuclei of radioactive compounds in the sample. Let the number of nuclei in a sample is $N$ and the number of radioactive decays per unit time $dt$ is $dN$, then,
$\dfrac{dN}{dt}=-\lambda N$
Where,
$\lambda$ is the decay constant
Decay constant is defined as the proportionality between the size of population of radioactive atoms and the rate at which this population decreases as a result of radioactive decay. Population of radioactive atoms refers to the number of radioactive nuclei.
The decay of radioactive elements occurs at a fixed constant rate. The half-life of a radioisotope is the time required for one half of the concentration of unstable substance to degrade into a more stable material. We can say that half-life is the time required for a radioactive sample to reduce to half of its initial value. The half-life of a radioactive sample is represented by ${{T}_{\dfrac{1}{2}}}$.
Integrating both sides,
$\int\limits_{{{N}_{o}}}^{N}{\dfrac{dN}{N}}=-\lambda \int\limits_{{{t}_{o}}}^{T}{dt}$
Taking initial time${{t}_{o}}=0$, we get,
$\ln \dfrac{N}{{{N}_{o}}}=-\lambda T$
By definition of half-life, we have,
${{N}_{t}}=\dfrac{{{N}_{o}}}{2}$
Therefore,
$\ln 2=\lambda T={{\log }_{e}}2$
Relation between Decay constant and Half-life:
${{T}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\lambda }$
Number of nuclei decayed in particular time being is given as,
$N={{N}_{o}}-{{N}_{o}}{{e}^{-\lambda t}}$
Where,
$N$ is the number of decayed nuclei
${{N}_{o}}$ is the number of initial nuclei
$\lambda$ is the decay constant
$t$ is the time
Now, as given ratio of initial nuclei in two different samples is $2:3$
Let the initial nuclei be in two different samples be $2{{N}_{o}}$ and $3{{N}_{o}}$
The decay constant of two different samples be ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$
And, nuclei at time $t$ of two different samples be ${{N}_{1}}$ and ${{N}_{2}}$
So, number of nuclei decayed in particular time being is given as,
${{N}_{1}}=2{{N}_{o}}-2{{N}_{o}}{{e}^{-{{\lambda }_{1}}t}}$
${{N}_{2}}=3{{N}_{o}}-3{{N}_{o}}{{e}^{-{{\lambda }_{2}}t}}$
Now, the half-life time is:
${{T}_{\dfrac{1}{2}}}=\dfrac{\ln 2}{\lambda }$
Half lives of two different samples are 1 hour and 2 hours respectively,
So,
$\begin{aligned} & 1=\dfrac{\ln 2}{{{\lambda }_{1}}} \\\ & {{\lambda }_{1}}=\ln 2 \\\ \end{aligned}$
And,
$\begin{aligned} & 2=\dfrac{\ln 2}{{{\lambda }_{2}}} \\\ & {{\lambda }_{2}}=\dfrac{\ln 2}{2} \\\ \end{aligned}$
To find the ratio of active nuclei at the end of 6 hours is
$\dfrac{{{N}_{1}}}{{{N}_{2}}}$
Substituting,
${{N}_{1}}=2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}$
${{N}_{2}}=3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}$

& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}}{3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\\ \end{aligned}$$ Substituting the values of decay constants in the above equation, $$\begin{aligned} & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-6\dfrac{\ln 2}{2}}} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-3\ln 2}} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{\ln \left( {{2}^{-6}} \right)}} \right)}{3\left( 1-{{e}^{\ln \left( {{2}^{-3}} \right)}} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}}^{\ln \left( e \right)} \right)}{3\left( 1-{{2}^{-3}}^{\ln \left( e \right)} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}} \right)}{3\left( 1-{{2}^{-3}} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-\dfrac{1}{64} \right)}{3\left( 1-\dfrac{1}{8} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( \dfrac{63}{64} \right)}{3\left( \dfrac{7}{8} \right)} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\times 63\times 8}{3\times 64\times 7} \\\ & \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{3}{4} \\\ \end{aligned}$$ The ratio of active nuclei in the samples is $\dfrac{3}{4}$ **Hence, the correct option is D.** **Note:** Half-life means how much time a radioactive sample takes to become half of its original concentration. The half-life of a radioactive sample depends on the elimination rate of the sample and the initial concentration of the sample. The calculation of the number of active nuclei at the moment is done by subtracting the number of decayed nuclei till the time from the number of initial nuclei.