If a radioactive material remains $25 \%$ after $16$ days, then its half life will be ......... days

A radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to $(a)$ $3.125\% $ $(b)$ $1\% $ of its original value?

In a radioactive decay process, the activity is defined as $A=-\frac{\mathrm{d} N}{\mathrm{~d} t}$, where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t=0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their $3^{\text {rd }}$ and $7^{\text {th }}$ half-lives, respectively, the ratio $A_1 / A_2$ is. . . . . . .

The half-life of a radioactive element $A$ is the same as the mean-life of another radioactive element $B.$ Initially both substances have the same number of atoms, then

Give the different units of radioactivity and define them. 

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