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:

Consider a radioactive nuclide which follows the decay rate given by $A(t) = A_0 2^{-(t/t_0)}$, where $A(t)$ is the fraction of radioactive material remaining after time $t$ from the initial $A_0$ at zero time. Let $A_1$ be the fraction of original activity which remains after $120 \ h$. Likewise, $A_2$ is the fraction of original activity remaining after $200 \ h$. If $A_1/A_2 = 16$, then the half-life $(t_0)$ will be: (in $h$)

$A$ radioactive nucleus $A$ has a single decay mode with half-life $\tau_A$. Another radioactive nucleus $B$ has two decay modes $1$ and $2$. If decay mode $2$ were absent,the half-life of $B$ would have been $\tau_A / 2$. If decay mode $1$ were absent,the half-life of $B$ would have been $3 \tau_A$. If the actual half-life of $B$ is $\tau_B$,then the ratio $\tau_B / \tau_A$ is

Half-lives of two radioactive elements $A$ and $B$ are $30 \text{ minute}$ and $60 \text{ minute}$ respectively. Initially,the samples have an equal number of nuclei. After $120 \text{ minute}$,the ratio of the number of decayed nuclei of $B$ to that of $A$ will be:

If the decay or disintegration constant of a radioactive substance is $\lambda$,then its half-life and mean life are respectively $(log_e 2 = \ln 2)$.

$A$ mixture consists of two radioactive materials $A_1$ and $A_2$ with half-lives of $20 \, s$ and $10 \, s$ respectively. Initially,the mixture has $40 \, g$ of $A_1$ and $160 \, g$ of $A_2$. The amount of the two in the mixture will become equal after ......... $s$.