A radioactive nucleus is being produced at a constant rate $\alpha$ per second. Its decay constant is $\alpha$. If $N_{0}$ is the number of nuclei at the $t=0$, then maximum number of nuclei possible are:
\[\begin{align}
& A.{{N}_{0}}+\dfrac{\alpha }{\lambda } \\
& B.{{N}_{0}} \\
& C.\dfrac{\lambda }{\alpha }+{{N}_{0}} \\
& D.\dfrac{\alpha }{\lambda } \\
\end{align}\]
Answer
645.9k+ views
Hint: Radioactive decay is the spontaneous breakdown of an atomic nucleus, with emission of energy and matter. To calculate the time, we need to use the rate of radioactive decay, given by $N=N_{0}e^{-\lambda t}$. Since the number of initial nuclei and the rate of decay is given, can calculate the time taken for the decay.
Formula used:
$N=N_{0}e^{-\lambda t}$
Complete step-by-step answer:
We know that radioactive decay is the spontaneous breakdown of an atomic nucleus, with emission of energy and matter. The elements which undergo radioactive decay are generally unstable. The common radioactive decays are $\alpha,\beta$ and$\gamma$ decay. These are generally found in nuclear reactions. Also from the first law of radioactive decay, we know that $N=N_{0}e^{- \lambda t}$, where $N$ is the number of nuclei after reaction, $N_{0}$ is the number of nuclei before the reaction, $\lambda$ is the decay constant, and $t$ is the time taken.
Given that, the decay constant is $\alpha$ and the rate of the decay is $\dfrac{dN}{dt}=\alpha$
But, the rate of formation of the nuclei is given as $\dfrac{dN}{dt}=N_{0}\lambda$ where, $N_{0}$ is the number of nuclei before the reaction.
But, the maximum number $N$ of nuclei during decay is the rate of formation of the nuclei
Equating, we get, $N\lambda=\alpha$
$\implies N=\dfrac{\alpha}{\lambda}$
Hence the correct answer is\[D.\dfrac{\alpha }{\lambda }\]
So, the correct answer is “Option D”.
Note: The SI unit of radioactive decay is becquerel (Bq) named after Henri Becquerel. One Bq is one nucleus decay of one seconds. Or in terms of curie (Cu) named after Cuire, it is the amount of radiation emitted in equilibrium with one gram of pure radium isotope. Be careful, when calculating and comparing the powers of $e$.
Formula used:
$N=N_{0}e^{-\lambda t}$
Complete step-by-step answer:
We know that radioactive decay is the spontaneous breakdown of an atomic nucleus, with emission of energy and matter. The elements which undergo radioactive decay are generally unstable. The common radioactive decays are $\alpha,\beta$ and$\gamma$ decay. These are generally found in nuclear reactions. Also from the first law of radioactive decay, we know that $N=N_{0}e^{- \lambda t}$, where $N$ is the number of nuclei after reaction, $N_{0}$ is the number of nuclei before the reaction, $\lambda$ is the decay constant, and $t$ is the time taken.
Given that, the decay constant is $\alpha$ and the rate of the decay is $\dfrac{dN}{dt}=\alpha$
But, the rate of formation of the nuclei is given as $\dfrac{dN}{dt}=N_{0}\lambda$ where, $N_{0}$ is the number of nuclei before the reaction.
But, the maximum number $N$ of nuclei during decay is the rate of formation of the nuclei
Equating, we get, $N\lambda=\alpha$
$\implies N=\dfrac{\alpha}{\lambda}$
Hence the correct answer is\[D.\dfrac{\alpha }{\lambda }\]
So, the correct answer is “Option D”.
Note: The SI unit of radioactive decay is becquerel (Bq) named after Henri Becquerel. One Bq is one nucleus decay of one seconds. Or in terms of curie (Cu) named after Cuire, it is the amount of radiation emitted in equilibrium with one gram of pure radium isotope. Be careful, when calculating and comparing the powers of $e$.
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