Alpha particles are

When ${}_{92}U^{238}$ changes into ${}_{82}Pb^{206}$,the number of $\alpha$ and $\beta^-$ particles emitted are:

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

For the $\beta^{+}$ (positron) emission from a nucleus,there is another competing process known as electron capture (an electron from an inner orbit,say,the $K$-shell,is captured by the nucleus and a neutrino is emitted).
$_{z}^{A} X + e^{-} \rightarrow _{z-1}^{A} Y + \nu$
Show that if $\beta^{+}$ emission is energetically allowed,electron capture is necessarily allowed,but not vice-versa.

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

An element $A$ decays into element $C$ by a two-step process:
$A \to B + {\;_2}He^4$
$B \to C + 2e^-$
Then: