If $N_t = N_o e^{-\lambda t}$,then the number of nuclei decaying between time $t_1$ and $t_2$ $(t_2 > t_1)$ is given by:

At any instant,two elements $X_1$ and $X_2$ have the same number of radioactive atoms. If the decay constants of $X_1$ and $X_2$ are $10\lambda$ and $\lambda$ respectively,then the time when the ratio of their atoms becomes $\frac{1}{e}$ will be:

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 $e^2=7.5$].

In a mean life of a radioactive sample,

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 $t$?
$N_{A}(0) = \text{Number of } A \text{ atoms at } t=0$
$N_{B}(0) = \text{Number of } B \text{ atoms at } t=0$

After five half-lives, the percentage of original radioactive atoms left is ...... $ \%$