${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are
$286,80$
$288,82$
$286,81$
$280,81$
When $_{90}T{h^{228}}$ transforms to $_{83}B{i^{212}}$, then the number of the emitted $\alpha$- and $\alpha$- particles is, respectively
In the disintegration series
$_{92}^{238}U\xrightarrow{\alpha }x\xrightarrow{\beta }_Z^AY$
The value of $Z$ and $A$ respectively will be
When $_3Li^7$ nuclei are bombarded by protons, and the resultant nuclei are $_4Be^8$, the emitted particles will be
An atom of mass number $15$ and atomic number $7$ captures an $\alpha - $ particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
A nucleus with $Z = 92$ emits the following in a sequence: $\alpha ,\,{\beta ^ - },\,{\beta ^ - },\,\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },\,{\beta ^ - },\alpha ,\,{\beta ^ + },\,{\beta ^ + },\,\alpha $. The $Z$ of the resulting nucleus is