^{290}{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^{+}\ } Z \xrightarrow{\ e^{-}\ } P \xrightarrow{\ e^{-}\ } Q

In the nuclear emission stated above, the mass number and atomic number of the pro

Question

^{290}{82}X \xrightarrow{\ \alpha\ } Y \xrightarrow{\ e^{+}\ } Z \xrightarrow{\ e^{-}\ } P \xrightarrow{\ e^{-}\ } Q

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

Vishal Gupta · Accepted Answer

C. An α-emission lowers A by 4 and Z by 2: (286, 80). Each β-emission leaves A unchanged; a positron lowers Z by 1 and an electron raises it by 1. Net over e^+, e^-, e^-: Z = 80 - 1 + 1 + 1 = 81. So Q is (286, 81).

Nuclei

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