Two samples $A$ and $B$ contain equal amount of radioactive substances. If $(1/8)^{th}$ of sample $A$ and $(1/128)^{th}$ of sample $B$ remain after $9 \text{ hours}$, then the ratio of half-life period of $B$ to that of $A$ is

At time $t=0$, a material is composed of two radioactive atoms $A$ and $B$, where $N_{A}(0)=2 N_{B}(0)$. The decay constant of both kinds of radioactive atoms is $\lambda$. However, $A$ disintegrates to $B$ and $B$ disintegrates to $C$. Which of the following figures represents the evolution of $N_{B}(t) / N_{B}(0)$ with respect to time $t$?
$N_{A}(0) = \text{Number of } A \text{ atoms at } t=0$
$N_{B}(0) = \text{Number of } B \text{ atoms at } t=0$

The activity of a radioactive sample is $1.6 \, curie$ and its half-life is $2.5 \, days$. Its activity after $10 \, days$ will be .......... $curie$.

Two radioactive samples $A$ and $B$ have half-lives $T_1$ and $T_2$ $(T_1 > T_2)$ respectively. At $t=0$,the activity of $B$ was twice the activity of $A$. Their activity will become equal after a time:

The half-life period of a radioactive element $x$ is the same as the mean life time of another radioactive element $y$. Initially,they have the same number of atoms. Then:

$A$ $16\, g$ sample of a radioactive element is taken from Bombay to Delhi in $2\, hours$ and it is found that $1\, g$ of the element remains (undisintegrated). The half-life of the element is: