When a nucleus with atomic number $Z$ and mass number $A$ undergoes a radioactive decay process:

Assertion : The ionising power of $\beta$-particle is less compared to $\alpha$-particles but their penetrating power is more.
Reason : The mass of $\beta$-particle is less than the mass of $\alpha$-particles.

The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are 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).

$_6^{12}C$ absorbs an energetic neutron and emits a beta particle. The resulting nucleus is

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