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

The natural logarithm of the activity $R$ of a radioactive sample varies with time $t$ as shown. At $t=0$,there are $N_0$ undecayed nuclei. Then,$N_0$ is equal to [Take $e^2=7.5$].

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 .......... $years$.

Half-life of a radioactive substance is $20 \text{ min}$. The time between $20 \%$ and $80 \%$ decay will be: (in $\text{ min}$)

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 radioactive substance $A$ is two times the half-life of another radioactive substance $B$. Initially, the number of nuclei of $A$ and $B$ are $N_A$ and $N_B$ respectively. After three half-lives of $A$, the number of nuclei of both are equal. Then $\frac{N_A}{N_B}$ is