In gamma ray emission from a nucleus,

When $_{90}^{228}Th$ transforms to $_{83}^{212}Bi$,the number of emitted $\alpha$- and $\beta$- particles is,respectively:

$A$ 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 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

$A$ nucleus with $Z = 92$ emits the following in a sequence: $\alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$. The $Z$ of the resulting nucleus is

An element $A$ decays into element $C$ by a two-step process:
$A \to B + {\;_2}He^4$
$B \to C + 2e^-$
Then: