$A$ nucleus $_n^mX$ emits one $\alpha$ particle and two $\beta^-$ particles. The resulting nucleus is

$A$ radioactive decay forms an isotope of the original nucleus with the emission of which of the following particles?

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

$A$ 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: $\alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}$ and $\alpha$. The atomic number of the resulting nucleus is

$A$ certain radioactive nuclide of mass number $m_x$ disintegrates,with the emission of an electron and $\gamma$ radiation only,to give a second nuclide of mass number $m_y$. Which one of the following equations correctly relates $m_x$ and $m_y$?