$A$ radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3$ $\alpha$-particles and $2$ positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be:

$A$ nucleus decays by ${\beta ^ + }$ emission followed by a gamma emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$ respectively, the corresponding numbers for the daughter nucleus are respectively.

$A$ free neutron decays spontaneously into :

$A$ nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$,then the minimal condition that the $\beta^{-}$ decay $X_Z^A \rightarrow Y_{Z+1}^A + \beta^{-} + \bar{\nu}_e$ will occur is ($m_e$ denotes the mass of the $\beta^{-}$ particle and the neutrino mass $m_{\nu}$ can be neglected).

$A$ 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

$A$ radioactive element $_{90}X^{238}$ decays into $_{83}Y^{222}$. The number of $\beta$-particles emitted is: