Answer
Verified
396k+ views
Hint: 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,
$\begin{align}
& 1=\dfrac{\ln 2}{{{\lambda }_{1}}} \\
& {{\lambda }_{1}}=\ln 2 \\
\end{align}$
And,
$\begin{align}
& 2=\dfrac{\ln 2}{{{\lambda }_{2}}} \\
& {{\lambda }_{2}}=\dfrac{\ln 2}{2} \\
\end{align}$
To find the ratio of active nuclei at the end of 6 hours is
$\dfrac{{{N}_{1}}}{{{N}_{2}}}$
Substituting,
${{N}_{1}}=2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}$
${{N}_{2}}=3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}$
\[\begin{align}
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}}{3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\
\end{align}\]
Substituting the values of decay constants in the above equation,
\[\begin{align}
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-6\dfrac{\ln 2}{2}}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-3\ln 2}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{\ln \left( {{2}^{-6}} \right)}} \right)}{3\left( 1-{{e}^{\ln \left( {{2}^{-3}} \right)}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}}^{\ln \left( e \right)} \right)}{3\left( 1-{{2}^{-3}}^{\ln \left( e \right)} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}} \right)}{3\left( 1-{{2}^{-3}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-\dfrac{1}{64} \right)}{3\left( 1-\dfrac{1}{8} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( \dfrac{63}{64} \right)}{3\left( \dfrac{7}{8} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\times 63\times 8}{3\times 64\times 7} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{3}{4} \\
\end{align}\]
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.
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,
$\begin{align}
& 1=\dfrac{\ln 2}{{{\lambda }_{1}}} \\
& {{\lambda }_{1}}=\ln 2 \\
\end{align}$
And,
$\begin{align}
& 2=\dfrac{\ln 2}{{{\lambda }_{2}}} \\
& {{\lambda }_{2}}=\dfrac{\ln 2}{2} \\
\end{align}$
To find the ratio of active nuclei at the end of 6 hours is
$\dfrac{{{N}_{1}}}{{{N}_{2}}}$
Substituting,
${{N}_{1}}=2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}$
${{N}_{2}}=3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}$
\[\begin{align}
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}-2{{N}_{o}}{{e}^{-6{{\lambda }_{1}}}}}{3{{N}_{o}}-3{{N}_{o}}{{e}^{-6{{\lambda }_{2}}}}} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3{{N}_{o}}\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6{{\lambda }_{1}}}} \right)}{3\left( 1-{{e}^{-6{{\lambda }_{2}}}} \right)} \\
\end{align}\]
Substituting the values of decay constants in the above equation,
\[\begin{align}
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-6\dfrac{\ln 2}{2}}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{-6\ln 2}} \right)}{3\left( 1-{{e}^{-3\ln 2}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{e}^{\ln \left( {{2}^{-6}} \right)}} \right)}{3\left( 1-{{e}^{\ln \left( {{2}^{-3}} \right)}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}}^{\ln \left( e \right)} \right)}{3\left( 1-{{2}^{-3}}^{\ln \left( e \right)} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-{{2}^{-6}} \right)}{3\left( 1-{{2}^{-3}} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( 1-\dfrac{1}{64} \right)}{3\left( 1-\dfrac{1}{8} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\left( \dfrac{63}{64} \right)}{3\left( \dfrac{7}{8} \right)} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{2\times 63\times 8}{3\times 64\times 7} \\
& \dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{3}{4} \\
\end{align}\]
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE