During a negative beta decay,

$A$ 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).

Originally, radioactive beta decay was thought to be a decay of a nucleus with the emission of electrons only (Case $I$). However, in addition to the electron, another (nearly) massless and electrically neutral particle is also emitted (Case $II$). Based on the figure below, which of the following is correct?

$A$ nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

In the following nuclear reaction,$D \xrightarrow{\alpha} D_{1} \xrightarrow{\beta^-} D_{2} \xrightarrow{\alpha} D_{3} \xrightarrow{\gamma} D_{4}$. The mass number of $D$ is $182$ and the atomic number is $74$. The mass number and atomic number of $D_{4}$ respectively will be:

The $\beta$-particles of a radioactive metal originate from