$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then:

The modern treatment method $P.E.T.$ is based on:

$A$ 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:

Radioactivity is a/an ...... process.

$A$ hypothetical radioactive nucleus decays according to the following series:
$A \xrightarrow{\alpha} A_1 \xrightarrow{\beta^-} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4$
If the mass number and atomic number of $A$ are respectively $180$ and $72$,then the atomic number and mass number of $A_4$ will be respectively:

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 $\ln 10 = 2.3$]