An ancient discovery found a sample where $75 \%$ of the original carbon $(C^{14})$ remains. The age of the sample is: $\left(T_{1/2}(C^{14}) = 5730 \text{ years}, \ln 0.5 = -0.7, \ln 0.75 = -0.3\right)$ (in $\text{ years}$)

The activity of a radioactive sample is $I_0$ counts/minute at $t = 0$ and $I_0/e$ counts/minute at $t = 5$ minutes. At what time (in minutes) will its activity decrease to half of its initial value?

The normal activity of living carbon-containing matter is found to be about $15$ decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive $_{6}^{14}C$ present with the stable carbon isotope $_{6}^{12}C$. When the organism is dead,its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ($5730$ years) of $_{6}^{14}C$,and the measured activity,the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14}C$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of $9$ decays per minute per gram of carbon. Estimate the approximate age (in years) of the Indus-Valley civilisation.

Half-life of a substance is $10$ years. In what time,it becomes $\frac{1}{4}$ of the initial amount?

There are two radionuclei $A$ and $B.$ $A$ is an alpha emitter and $B$ is a beta emitter. Their disintegration constants are in the ratio of $1 : 2.$ What should be the ratio of the number of atoms of the two at time $t = 0$ so that the probabilities of getting $\alpha$ and $\beta$ particles are the same at time $t = 0$?

The half-life of a radioactive element is $10 \ h$. The fraction of initial radioactivity of the element that will remain after $40 \ h$ is