$\ \ _{82}^{290}X\overset{\phantom{\alpha\ }}{\rightarrow}Y\overset{\phantom{e^{+}\ }}{\rightarrow}Z

Question

$\ \ _{82}^{290}X\overset{\phantom{\alpha\ }}{\rightarrow}Y\overset{\phantom{e^{+}\ }}{\rightarrow}Z\overset{\phantom{\beta^{-}\ }}{\rightarrow}P\overset{\phantom{e^{-}\ }}{\rightarrow}Q$

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

Accepted Answer

Sol. [Key Concept: Nuclear decay — identifying the emitted particles] $${\ _{82}^{290}X\overset{\phantom{\alpha\ }}{\rightarrow}\ _{80}^{286}Y\overset{\phantom{e^{+}\ }}{\rightarrow}\ _{79}^{286}Z\overset{\phantom{\beta^{-}\ }}{\rightarrow}\ _{80}^{286}P\overset{\phantom{e^{-}\ }}{\rightarrow}\ _{81}^{286}Q }{A \rightarrow 286 }{Z = 81 }$$

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