In a radioactive decay chain, ${ }_{90}^{232} Th$ nucleus decays to ${ }_{82}^{212} Pb$ nucleus. Let $N _\alpha$ and $N _\beta$ be the number of $\alpha$ and $\beta^{-}$particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

$(A)$ $N _\alpha=5$  $(B)$ $N _\alpha=6$  $(C)$ $N _\beta=2$  $(D)$ $N _\beta=4$

At time $t=0$, a material is composed of two radioactive atoms ${A}$ and ${B}$, where ${N}_{{A}}(0)=2 {N}_{{B}}(0)$ The decay constant of both kind of radioactive atoms is $\lambda$. However, A disintegrates to ${B}$ and ${B}$ disintegrates to ${C}$. Which of the following figures represents the evolution of ${N}_{{B}}({t}) / {N}_{{B}}(0)$ with respect to time $t$ ?

${N}_{{A}}(0)={No} . \text { of } {A} \text { atoms at } {t}=0$

${N}_{{B}}(0)={No} . \text { of } {B} \text { atoms at } {t}=0$

Two radioactive substances $A$ and $B$ have decay constants $5\lambda $ and $\lambda $ respectively. At $t = 0$, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(\frac {1}{e})^2$ will be

A sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$, which in turn disintegrates to some other stable nucleus $C.$ Plot of a graph showing the variation of number of atoms of nucleus $B$ vesus time is :

(Assume that at ${t}=0$, there are no ${B}$ atoms in the sample)

In Fig. $X$ represents time and $Y$ represents activity of a radioactive sample. Then the activity of sample, varies with time according to the curve

A radioactive material decays by simultaneous emissions of two particles with half lives of $1400\, years$ and $700\, years$ respectively. What will be the time after which one third of the material remains? (Take In $3=1.1$ ) (In $years$)