A piece of wood from the ruins of an ancient building was found to have a $^{14}C$ activity of $12$ disintegrations per minute per gram of its carbon content. The $^{14}C$ activity of the living wood is $16$ disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of $^{14}C$ is $5760$ years.
According to exponential law,
$\quad \mathrm{I}=\mathrm{I}_{0} e^{-\lambda t}$
$\therefore 12=16 e^{-\lambda t}$
$\therefore \frac{12}{16}=e^{-\lambda t}$
$\therefore e^{\lambda t}=\frac{16}{12}=1.333$
$\therefore \lambda t \ln e=\ln (1.333)$
$\therefore \lambda t(1)=2.303 \log (1.333)$
$\therefore t=\frac{(2.303)(0.1249)\left(\tau_{1 / 2}\right)}{0.693}$
$\therefore t=\frac{(2.303)(0.1249)(5760)}{0.693}$
$\therefore t \approx 2389 \text { years }$
Define $SI$ unit of radioactivity ?
Two radioactive samples $A$ and $B$ have half lives $T_1$ and $T_2\left(T_1 > T_2\right)$ respectively At $t=0$, the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time
Consider an initially pure $M$ gm sample of$_ A{X}$, an isotope that has a half life of $T$ hour, what is it’s initial decay rate ($N_A$ = Avogrado No.)
The half life of radium is $1620$ years and its atomic weight is $226\, k\,gm$ per kilomol. The number of atoms that will decay from its $1\, gm$ sample per second will be
(Avogadro's number $N = 6.02 \times {10^{26}}$atom/kilomol)
A radioactive sample decays by two modes by $\alpha $ decay and by $\beta -decay$. $66.6 \%$ of times it decays by $\alpha -decay$ and $33.3 \%$ of times, it decays by $\beta -decay$. If half life of sample is $60$ years then what will be half life of sample, if it decays only by $\alpha - decay$. ............ $years$