Question

The isotope of carbon $C^{14}$ has a half life of $5730yrs$. If the ratio of $C^{14}$ to $C^{12}$ is $1.3\times {10}^{-13}$ in a wood sample , then find activity of sample of $1g$ of wood”
$$\begin{array}{rl}& A.2\times {10}^{-12}Bq\\ & B.0.02Bq\\ & C.2\times {10}^{-10}Bq\\ & D.2\times {10}^{10}Bq\end{array}$$

Answer 1

Hint: Radioactive decay is the spontaneous breakdown of an atomic nucleus, with emission of energy and matter. To calculate the activity, we need to find the number of nuclei which are a part of the reaction and the decay constant.

Formula used: $A=\lambda N$,
$T_{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{ln2}{\lambda }}$ and $N={\displaystyle \frac{k}{12}}{N}_A$

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.
We know that half life $T_{{\displaystyle \frac{1}{2}}}$ is given as $T_{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{ln2}{\lambda }}$
Given that $T_{{\displaystyle \frac{1}{2}}}=5730yr$
$⟹\lambda ={\displaystyle \frac{0.693}{5730}}yr{s}^{-1}$
We know that the activity $A$ is given as $A=\lambda N$ where $N$ is the number of nuclei after reaction, and $\lambda$ is the decay constant.
Given that the ratio of $C^{14}$ to $C^{12}$ $k=1.3\times {10}^{-13}$, then we get, $N={\displaystyle \frac{k}{12}}{N}_A$, where $N_A$ is Avogadro number. Here $$12$$ is taken as we compare the amount of the element $C^{14}$ to ${\displaystyle \frac{1}{12}}$ of $C^{12}$. This is a standard way of defining radioactivity.

$⟹N={\displaystyle \frac{1.3\times {10}^{-13}}{12}}\times 6.022\times {10}^{23}$
$⟹N=6.52\times {10}^9$
Then, $A={\displaystyle \frac{0.693}{5730\times 365\times 24\times 3600}}\times 6.52\times {10}^9=0.025Bq$

So, the correct answer is “Option B”.

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 Curie, it is the amount of radiation emitted in equilibrium with one gram of pure radium isotope. Radioactivity is also defined as the amount of the element present with respect to ${\displaystyle \frac{1}{12}}$ of $C^{12}$.