$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \, s$ and $10 \, s$ respectively. Initially,the mixture has $40 \, g$ of $A_1$ and $160 \, g$ of $A_2$. The amount of the two in the mixture will become equal after ......... $s$.

There are three lumps of a given radioactive substance. Their activity is in the ratio of $1 : 2 : 3$ now. What will be the ratio of their activities at any further date?

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 $t = 0$?

$A$ radio-isotope has a half-life of $5$ years. The fraction of the atoms of this material that would decay in $15$ years will be

The half-life of $^{131}I$ is $8 \, days$. Given a sample of $^{131}I$ at time $t = 0$,we can assert that

Two radioactive elements $R$ and $S$ disintegrate as:
$R \rightarrow P + \alpha; \lambda_R = 4.5 \times 10^{-3} \, \text{years}^{-1}$
$S \rightarrow P + \beta; \lambda_S = 3 \times 10^{-3} \, \text{years}^{-1}$
Starting with the number of atoms of $R$ and $S$ in the ratio of $2:1$,what will be this ratio after the lapse of three half-lives of $R$?