A radioactive material decays by simultaneous emission of two particles with respective half lives $1620$ and $810$ years. The time (in years) after which one- fourth of the material remains is
$1080$
$2430$
$3240$
$4860$
If $t_{1/2}$ is the half life of a substance then $t_{3/4}$ is the time in which substance
Number of nuclei of a radioactive substance at time $t = 0$ are $1000$ and $900$ at time $t = 2$ $s$. Then number of nuclei at time $t = 4$ $s$ will be
The graph which represents the correct variation of logarithm of activity $(log\, A)$ versus time, in figure is
The half-life of a sample of a radioactive substance is $1$ hour. If $8 \times {10^{10}}$ atoms are present at $t = 0$, then the number of atoms decayed in the duration $t = 2$ hour to $t = 4$ hour will be
Two radioactive elements $R$ and $S$ disintegrate as
$R \rightarrow P + \alpha; \lambda_R = 4.5 × 10^{-3} \,\, years^{-1}$
$S \rightarrow P + \beta; \lambda_S = 3 × 10^{-3} \,\, years^{-1}$
Starting with number of atoms of $R$ and $S$ in the ratio of $2 : 1,$ this ratio after the lapse of three half lives of $R$ will be :