$A$ radioactive substance emits $100$ beta particles in the first $2 \,s$ and $50$ beta particles in the next $2 \,s$. The mean life of the sample is

  • A
    $4 \,s$
  • B
    $2 \,s$
  • C
    $\frac{2}{0.693} \,s$
  • D
    $2 \times 0.693 \,s$

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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:

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:

An alloy is composed of two radioactive materials $A$ and $B$ having equal weight. The half-lives of $A$ and $B$ are $10 \ yrs$ and $20 \ yrs$ respectively. After time $t$, the alloy was found to consist of $(1/e) \ kg$ of $A$ and $1 \ kg$ of $B$. If the atomic weights of $A$ and $B$ are the same, then the value of $t$ is (Assume $\ln 2 = 0.7$):

The activity $R$ of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:
$t (h)$$0$$1$$2$$3$$4$
$R (MBq)$$100$$35.36$$12.51$$4.42$$1.56$

$(i)$ Plot the graph of $R$ versus $t$ and calculate the half-life from the graph.
$(ii)$ Plot the graph of $\ln \left( \frac{R}{R_0} \right)$ versus $t$ and obtain the value of the half-life from the graph.

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