Give the equation form of the exponential law of radioactive decay.

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 initial activity of a certain radioactive isotope was measured as $16000 \ counts \ min^{-1}$. Given that the only activity measured was due to this isotope and that its activity after $12 \ h$ was $2000 \ counts \ min^{-1}$,its half-life,in hours,is nearest to

The activity of a radioactive sample decreases to $(1/3)$ of its original value in $3\, \text{days}$. Then, in $9\, \text{days}$, its activity will become:

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

Half-lives of two radioactive nuclei $A$ and $B$ are $10 \, minutes$ and $20 \, minutes$,respectively. If,initially,a sample has an equal number of nuclei,then after $60 \, minutes$,the ratio of the number of decayed nuclei of $A$ and $B$ will be: