An archaeologist analyses the wood in a prehistoric structure and finds that the ratio of $C^{14}$ (half-life $= 5700 \, years$) to $C^{12}$ is only one-fourth of that found in the cells of buried plants. The age of the wood is about .......... $years$.

$A$ radioactive element of mass $1 \,kg$ after $N$ years is left with only $125 \,g$. If the half-life of the element is $12.5 \,y$, then the value of $N$ is:

Two radioactive elements $R$ and $S$ disintegrate as:
$R \rightarrow P + \alpha; \lambda_R = 4.5 \times 10^{-3} \, \text{years}^{-1}$
$S \rightarrow P + \beta; \lambda_S = 3 \times 10^{-3} \, \text{years}^{-1}$
Starting with the number of atoms of $R$ and $S$ in the ratio of $2:1$,what will be this ratio after the lapse of three half-lives of $R$?

What is the relationship between the decay constant $\lambda$ and the half-life $\tau_{1/2}$?

Number of nuclei of a radioactive substance at time $t = 0$ are $1000$ and $900$ at time $t = 2 \, s$. Then number of nuclei at time $t = 4 \, s$ will be

One mole of radium has an activity of $\frac{1}{3.7} \text{ kilo curie}$. Its decay constant is (Avogadro number $= 6 \times 10^{23} \text{ mol}^{-1}$)