${}_{82}^{290}X\stackrel{\mathrm{\alpha <!--\; \alpha \; -->}}{\to }Y\stackrel{e^{+}}{\to }Z\stackrel{{\mathrm{\beta <!--\; \beta \; -->}}^{-<!--\; -\; -->}}{\to }P\stackrel{e^{-<!--\; -\; -->}}{\to }Q$

In the nuclear emission stated above, the mass number and atomic number of the product Q respectively, are :

We start with the given nuclide written in the usual notation ${}_{82}^{290}$ X, i.e. mass number A = 290 and atomic number Z = 82.

1. **α‑decay**
An α particle carries away 2 protons and 2 neutrons, so A decreases by 4 and Z decreases by 2.
${}_{82}^{290}\to {\alpha }^{}{}_{80}^{286}$
Thus after this step we have Y = ${}_{80}^{286}$.

2. **Positron (e⁺) emission**
A positron emission converts a proton into a neutron, decreasing Z by 1 while A stays the same.
${}_{80}^{286}\to {\mathrm{e⁺}}^{}{}_{79}^{286}$
The product is Z = ${}_{79}^{286}$.

3. **β⁻ (electron) emission**
A β⁻ decay turns a neutron into a proton, raising Z by 1 with no change in A.
${}_{79}^{286}\to {\mathrm{\beta ⁻}}^{}{}_{80}^{286}$
Now we have P = ${}_{80}^{286}$.

4. **Another β⁻ (electron) emission**
A second β⁻ decay again increases Z by 1, leaving A unchanged.
${}_{80}^{286}\to {\mathrm{e⁻}}^{}{}_{81}^{286}$
The final product Q therefore has mass number A = 286 and atomic number Z = 81.

Hence, the mass number and atomic number of Q are **286, 81**.