Draw a graph showing the variation of decay rate with number of active nuclei.
According to formula,
$\mathrm{I}=-\lambda \mathrm{N}$
$\therefore \mathrm{I}=(-\lambda) \mathrm{N}+0$
Form of above equation is like equation of a straight line, $y=m x+c .$ Hence graph of $\mathrm{I} \rightarrow \mathrm{N}$ is a straight line with slope $(-\lambda)$.
Here graph is obtained in $4^{\text {th }}$ quadrant because $N$ is positive and $I$ is negative.
There are two radionuclei $A$ and $B.$ $A$ is an alpha emitter and $B$ is a beta emitter. Their distintegration constants are in the ratio of $1 : 2.$ What should be the ratio of number of atoms of two at time $t = 0$ so that probabilities of getting $\alpha$ and $\beta$ particles are same at time $t = 0.$
Half lives of two radioactive nuclei $A$ and $B$ are $10\, minutes$ and $20\, minutes$, respectively. If, initially a sample has equal number of nuclei, then after $60$ $minutes$ , the ratio of decayed numbers of nuclei $A$ and $B$ will be
What is the half-life (in years) period of a radioactive material if its activity drops to $1 / 16^{\text {th }}$ of its initial value of $30$ years?
A fraction $f_1$ of a radioactive sample decays in one mean life, and a fraction $f_2$ decays in one half-life.
A sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$, which in turn disintegrates to some other stable nucleus $C.$ Plot of a graph showing the variation of number of atoms of nucleus $B$ vesus time is :
(Assume that at ${t}=0$, there are no ${B}$ atoms in the sample)