The half-life of a radioactive substance is $10 \, \text{minutes}$. If $n_1$ and $n_2$ are the number of atoms decayed in $20 \, \text{minutes}$ and $30 \, \text{minutes}$ respectively, then $n_1 : n_2 =$

The ratio of the number of active nuclei in two different samples is $2:3$. Their half-lives are $2 \ hours$ and $3 \ hours$ respectively. After $12 \ hours$,the ratio of their activities will be:

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 decay constant for the nuclide $^{64}Cu$ is $1.6 \times 10^{-5} \, s^{-1}$. Find the activity of a $1 \, mg$ sample of $^{64}Cu$ in $Ci$. (Given: Atomic mass of Copper = $64 \, g/mol$,$1 \, Ci = 3.7 \times 10^{10} \, Bq$)

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 for $\alpha$ and $\beta$ emission of a radioactive material are $16$ years and $48$ years respectively. When the material decays by giving $\alpha$ and $\beta$ emission simultaneously,the time in which $3/4$th of the material decays is ....... years.