$1$ curie represents

$A$ certain radioactive material $_ZX^A$ starts emitting $\alpha$ and $\beta$ particles successively such that the end product is $_{Z-3}Y^{A-8}$. The number of $\beta$ and $\alpha$ particles emitted are

In a radioactive decay,an element $^A_Z X$ emits four $\alpha$-particles,three $\beta$-particles and eight $\gamma$-photons. The atomic number and mass number of the resulting final nucleus are

In a radioactive sample,${ }_{19}^{40} K$ nuclei decay into stable ${ }_{20}^{40} Ca$ nuclei with a decay constant of $4.5 \times 10^{-10} \text{ per year}$ or into stable ${ }_{18}^{40} Ar$ nuclei with a decay constant of $0.5 \times 10^{-10} \text{ per year}$. Given that in this sample,all the stable ${ }_{20}^{40} Ca$ and ${ }_{18}^{40} Ar$ nuclei are produced by the ${ }_{19}^{40} K$ nuclei only. In time $t \times 10^9 \text{ years}$,if the ratio of the sum of stable ${ }_{20}^{40} Ca$ and ${ }_{18}^{40} Ar$ nuclei to the radioactive ${ }_{19}^{40} K$ nuclei is $99$,the value of $t$ will be: [Given $\ln 10 = 2.3$]

$A$ radioactive nuclide of element $X$ decays to form element $Y$. The graph representing the rate of production of $Y$ in a sample of $X$ is:

Assertion: ${}_{Z}{X^{A}}$ undergoes $2\alpha$-decays,$2\beta$-decays,and $2\gamma$-decays,and the daughter product is ${}_{Z-2}{X^{A-8}}$.
Reason: In $\alpha$-decays,the mass number decreases by $4$ and the atomic number decreases by $2$. In $\beta$-decays,the mass number remains unchanged,but the atomic number increases by $1$.