When a neutron decays,it produces . . . . . . .

Three $\alpha$-particles and one $\beta$-particle decay take place in series from an isotope $_{88}Ra^{238}$. Finally,the isotope obtained will be:

$A$ free neutron decays spontaneously into:

In the disintegration series
$_{92}^{238}U \xrightarrow{\alpha} X \xrightarrow{\beta} _{Z}^{A}Y$
The values of $Z$ and $A$ respectively will be:

$A$ 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).

Some radioactive nucleus may emit