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$^{290}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$
In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are:
A
288, 82
B
286, 81
C
280, 81
D
286, 80
Detailed Solution
$^{290}_{82}X \xrightarrow{\alpha} {}^{286}_{80}Y \xrightarrow{e^+} {}^{286}_{79}Z \xrightarrow{\beta^-} {}^{286}_{80}P \xrightarrow{e^-} {}^{286}_{81}Q$
$\alpha$-emission: A decreases by 4, Z decreases by 2
$e^+$ (positron) emission: A unchanged, Z decreases by 1
$\beta^-$ / $e^-$ emission: A unchanged, Z increases by 1 each
So Q has mass number 286 and atomic number 81.