$^{290}_{82}X\xrightarrow{\alpha}Y\xrightarrow{e^+}Z\xrightarrow{\beta^-}P\xrightarrow{e^-}Q$ In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are:

Question

Dr. Abubacker · Accepted Answer

[Solution by Claude - not from source] $\alpha$: (286, 80); $e^+$: (286, 79); $\beta^-$: (286, 80); $e^-$ ($\beta^-$ emission): (286, 81). Answer (2).

Q. $^{290}_{82}X\xrightarrow{\alpha}Y\xrightarrow{e^+}Z\xrightarrow{\beta^-}P\xrightarrow{e^-}Q$ In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are:

Explanation

Reviewed by Dr. Abubacker

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