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

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

The correct answer is 286, 81.

Let us analyze the nuclear decay chain step-by-step. The starting nucleus is represented as:
${}_{82}^{290}X$
Here, the mass number (A) of X is 290, and the atomic number (Z) of X is 82.

Step 1: Alpha Decay ($\alpha$-decay) from X to Y
During an alpha decay, the nucleus emits an alpha particle (which is a helium nucleus, ${}_2^4\mathrm{He}$). This reduces the mass number by 4 and the atomic number by 2.
For nucleus Y:
Mass number, $A_Y=290-4=286$
Atomic number, $Z_Y=82-2=80$
Thus, the nucleus Y is:
${}_{80}^{286}Y$

Step 2: Positron Emission ($e^{+}$ decay) from Y to Z
During positron emission, a proton inside the nucleus converts into a neutron, emitting a positron ( ${}_{+1}^{0}e$) and a neutrino. This keeps the mass number unchanged but decreases the atomic number by 1.
For nucleus Z:
Mass number, $A_Z=286$
Atomic number, $Z_Z=80-1=79$
Thus, the nucleus Z is:
${}_{79}^{286}Z$

Step 3: Beta-minus Decay (${\beta }^{-}$ decay) from Z to P
In a beta-minus decay, a neutron converts into a proton, emitting an electron ( ${}_{-1}^0\beta$) and an antineutrino. This keeps the mass number unchanged and increases the atomic number by 1.
For nucleus P:
Mass number, $A_P=286$
Atomic number, $Z_P=79+1=80$
Thus, the nucleus P is:
${}_{80}^{286}P$

Step 4: Electron Emission ($e^{-}$ decay) from P to Q
An electron emission ($e^{-}$) is identical in its effect to a beta-minus decay. The mass number remains constant, and the atomic number increases by 1.
For product nucleus Q:
Mass number, $A_Q=286$
Atomic number, $Z_Q=80+1=81$
Thus, the final product Q is:
${}_{81}^{286}Q$

Therefore, the mass number and atomic number of the product Q are 286 and 81, respectively.