$A$ 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

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

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.

After the decay of a single $\beta$ particle, the parent and daughter nuclei are

$A$ radioactive element $_{90}X^{238}$ decays into $_{83}Y^{222}$. The number of $\beta$-particles emitted is:

$A$ sample of a radioactive nucleus $A$ disintegrates to another radioactive nucleus $B$,which in turn disintegrates to some other stable nucleus $C$. Plot a graph showing the variation of the number of atoms of nucleus $B$ versus time:
(Assume that at $t=0$,there are no $B$ atoms in the sample)