The half-life of a radioactive element is $10 \ h$. The fraction of initial radioactivity of the element that will remain after $40 \ h$ is

In a radioactive decay process,the activity is defined as $A = -\frac{dN}{dt}$,where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources,$S_1$ and $S_2$,have the 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 radioactivity of a sample at time $T_1$ is $R_1$ and at time $T_2$ is $R_2$. If the mean life of the sample is $T$,then the number of nuclei disintegrated in the time interval $(T_2 - T_1)$ is:

After $40 \, days$,the $1/16$th part of a radioactive element remains undecayed. What is its half-life (in $, days$)?

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 $......... \times 10^{24}$)

The activity of a sample of radioactive material is $A_1$ at time $t_1$ and $A_2$ at time $t_2$ $(t_2 > t_1)$. If its mean life is $T$,then which of the following is true?