Carbon $ - 14$ decays with half-life of about $5,800\, years$. In a sample of bone, the ratio of carbon $ - 14$ to carbon $ - 12$ is found to be $\frac{1}{4}$ of what it is in free air. This bone may belong to a period about $x$ centuries ago, where $x$ is nearest to

Two radioactive substances $A$ and $B$ have decay constants $5\lambda $ and $\lambda $ respectively. At $t = 0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(\frac {1}{e})^2$ will be

The rate of disintegration was observed to be ${10^{17}}$ disintegrations per sec when its half life period is $1445$ years. The original number of particles are

At a given instant, say $t = 0,$ two radioactive substances $A$ and $B$ have equal activates. The ratio $\frac{{{R_B}}}{{{R_A}}}$ of their activities. The ratio $\frac{{{R_B}}}{{{R_A}}}$ of their activates after time $t$ itself decays with time $t$ as $e^{-3t}.$ If the half-life of $A$ is $ln2,$ the half-life of $B$ is

At any instant, two elements $X _1$ and $X _2$ have same number of radioactive atoms. If the decay constant of $X _1$ and $X _2$ are $10 \lambda$ and $\lambda$ respectively. then the time when the ratio of their atoms becomes $\frac{1}{e}$ respectively will be

Two radioactive nuclei $P$ and $Q,$ in a given sample decay into a stable nucleus $R.$ At time $t = 0,$ number of $P$ species are $4\,\, N_0$ and that of $Q$ are $N_0$. Half-life of $P$ (for conversion to $R$) is $1$ minute where as that of $Q$ is $2$ minutes. Initially there are no nuclei of $R$ present in the sample. When number of nuclei of $P$ and $Q$ are equal, the number of nuclei of $R$ present in the sample would be