At a certain moment,the amounts of two radioactive substances are in the ratio $2:1$. Their half-lives are $12 \text{ hours}$ and $16 \text{ hours}$ respectively. What will be the ratio of their remaining amounts after $2 \text{ days}$?

If $f$ denotes the ratio of the number of nuclei decayed $(N_{d})$ to the number of nuclei at $t=0$ $(N_{0})$,then for a collection of radioactive nuclei,the rate of change of $f$ with respect to time is given as: [$\lambda$ is the radioactive decay constant]

Two radioactive materials $A$ and $B$ having decay constants $7 \lambda$ and $\lambda$ respectively,initially have the same number of nuclei. The time taken for the ratio of the number of nuclei of material $B$ to that of $A$ to be $e$ is:

An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half-lives of $A$ and $B$ are $10 \ yrs$ and $20 \ yrs$ respectively. After time $t$, the alloy was found to consist of $(1/e) \ kg$ of $A$ and $1 \ kg$ of $B$. If the atomic weights of $A$ and $B$ are the same, then the value of $t$ is (Assume $\ln 2 = 0.7$):

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 $e^2=7.5$].

Substance $A$ has an atomic mass number of $16$ and a half-life of $1$ day. Another substance $B$ has an atomic mass number of $32$ and a half-life of $0.5$ day. If both $A$ and $B$ start undergoing radioactivity simultaneously with an initial mass of $320 \, g$ each,how many total atoms of $A$ and $B$ combined would be left after $2$ days? (Answer in $......... \times 10^{24}$)