
For a radioactive material, its activity A and rate of charge of its activity R are defined as $A=-\dfrac{dN}{dt},R=-\dfrac{dA}{dt}$ where N(t) is the number of nuclei at time t. Two radioactive source P (mean life $\tau $ ) and Q(mean life $2\tau $ ) have the same activity at $t=2\tau {{R}_{p}},{{R}_{q}}$ respectively, if $\dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{n}{e}$
Answer
557.1k+ views
Hint: Find the number of nuclei present in the radioactive material at any time. As the activity of the two radioactive materials depend on the number of nuclei at any instant, the mean life of the radioactive substance is inversely proportional to the wavelength.
Formula used:
$\begin{align}
& N={{N}_{0}}{{e}^{-\lambda t}} \\
& \tau =\dfrac{1}{\lambda } \\
\end{align}$
Complete step by step answer:
We know, the activity is given and the rate of change of activity is given, the mean life of the two radioactive materials is loads given.
Let us find the activity of the radioactive material
$\begin{align}
& N={{N}_{0}}{{e}^{-\lambda t}} \\
& A=-\dfrac{dN}{dt} \\
& \Rightarrow A=-\lambda {{N}_{0}}{{e}^{-\lambda t}} \\
\end{align}$
The rate of change of activity of the radioactive material will be,
$\begin{align}
& R=-\dfrac{dA}{dt} \\
& \Rightarrow R=-\dfrac{d}{dt}(-\lambda {{N}_{0}}{{e}^{-\lambda t}}) \\
& \Rightarrow R={{N}_{0}}{{\lambda }^{2}}{{e}^{-\lambda t}} \\
& \Rightarrow R=\lambda {{A}_{0}}{{e}^{-\lambda t}} \\
\end{align}$
Now, the rate of activity for individual radioactive materials p,q is as following,
$\begin{align}
& {{R}_{p}}={{\lambda }_{p}}{{A}_{0}}{{e}^{-{{\lambda }_{p}}t}} \\
& {{R}_{q}}={{\lambda }_{q}}{{A}_{0}}{{e}^{-{{\lambda }_{q}}t}} \\
\end{align}$
Ratio between them will be,
$\begin{align}
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{{{\lambda }_{p}}{{A}_{0}}{{e}^{-{{\lambda }_{p}}t}}}{{{\lambda }_{q}}{{A}_{0}}{{e}^{-{{\lambda }_{q}}t}}} \\
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{{{\tau }_{q}}}{{{\tau }_{p}}}\times \dfrac{e}{{{e}^{2}}} \\
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{2}{e} \\
\end{align}$
Therefore, comparing it with the avoided equation given in question, we get n=2.
Additional information:
The radioactive decay law states that “The probability per unit time that a nucleus will decay is a constant, independent of time”. Radioactivity is the phenomenon exhibited by the nuclei of an atom as a result of nuclear instability. In the year 1896, Henry Becquerel discovered this phenomenon. Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation. In a drawer which was having photographic plates, a small amount of Uranium compound was wrapped in a black paper and was kept in it. Examination of these plates later resulted that there has been an exposure. This phenomenon came to be known as Radioactive Decay.
Note:The mean time life of the radioactive material is inversely proportional to the radioactive decay constant. In the above question, the radioactive decay constant is substituted with the inverse of mean life because of this reason. Also, the probability of the materials per unit time is constant and is time independent.
Formula used:
$\begin{align}
& N={{N}_{0}}{{e}^{-\lambda t}} \\
& \tau =\dfrac{1}{\lambda } \\
\end{align}$
Complete step by step answer:
We know, the activity is given and the rate of change of activity is given, the mean life of the two radioactive materials is loads given.
Let us find the activity of the radioactive material
$\begin{align}
& N={{N}_{0}}{{e}^{-\lambda t}} \\
& A=-\dfrac{dN}{dt} \\
& \Rightarrow A=-\lambda {{N}_{0}}{{e}^{-\lambda t}} \\
\end{align}$
The rate of change of activity of the radioactive material will be,
$\begin{align}
& R=-\dfrac{dA}{dt} \\
& \Rightarrow R=-\dfrac{d}{dt}(-\lambda {{N}_{0}}{{e}^{-\lambda t}}) \\
& \Rightarrow R={{N}_{0}}{{\lambda }^{2}}{{e}^{-\lambda t}} \\
& \Rightarrow R=\lambda {{A}_{0}}{{e}^{-\lambda t}} \\
\end{align}$
Now, the rate of activity for individual radioactive materials p,q is as following,
$\begin{align}
& {{R}_{p}}={{\lambda }_{p}}{{A}_{0}}{{e}^{-{{\lambda }_{p}}t}} \\
& {{R}_{q}}={{\lambda }_{q}}{{A}_{0}}{{e}^{-{{\lambda }_{q}}t}} \\
\end{align}$
Ratio between them will be,
$\begin{align}
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{{{\lambda }_{p}}{{A}_{0}}{{e}^{-{{\lambda }_{p}}t}}}{{{\lambda }_{q}}{{A}_{0}}{{e}^{-{{\lambda }_{q}}t}}} \\
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{{{\tau }_{q}}}{{{\tau }_{p}}}\times \dfrac{e}{{{e}^{2}}} \\
& \dfrac{{{R}_{p}}}{{{R}_{q}}}=\dfrac{2}{e} \\
\end{align}$
Therefore, comparing it with the avoided equation given in question, we get n=2.
Additional information:
The radioactive decay law states that “The probability per unit time that a nucleus will decay is a constant, independent of time”. Radioactivity is the phenomenon exhibited by the nuclei of an atom as a result of nuclear instability. In the year 1896, Henry Becquerel discovered this phenomenon. Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation. In a drawer which was having photographic plates, a small amount of Uranium compound was wrapped in a black paper and was kept in it. Examination of these plates later resulted that there has been an exposure. This phenomenon came to be known as Radioactive Decay.
Note:The mean time life of the radioactive material is inversely proportional to the radioactive decay constant. In the above question, the radioactive decay constant is substituted with the inverse of mean life because of this reason. Also, the probability of the materials per unit time is constant and is time independent.
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