Q.Complete the following equations describing nuclear decays.

[!TLDR] (a) $_{88}^{226}$Ra $\rightarrow$ $_{86}^{222}$Rn + $\alpha$; (b) $_8^{19}$O $\rightarrow$ $_9^{19}$F + $e^-$ + antineutrino; (c) $_{90}^{228}$Th $\rightarrow$ $_{88}^{224}$Ra + $\alpha$; (d) $_7^{12}$N $\rightarrow$ $_6^{12}$C + $e^+$ + neutrino …

Each equation is completed using conservation of mass number A and atomic number Z, together with the type of decay each format implies. (a) Radium-226 (Z=88, A=226) undergoing alpha decay (emitting $_2^4He$) must produce a daughter with Z=88-2=86 and A=226-4=222 -- this is radon-222, so $_{88}^{226}Ra\rightarrow\,_{86}^{222}Rn+\,_2^4He$; this is in fact the very first step of the natural uranium-238 decay chain. (b) Oxygen-19 (Z=8, A=19) emitting an electron ($\beta^-$ decay: a neutron converts to a proton) must produce a daughter with Z=8+1=9 and A unchanged at 19 -- this is fluorine-19, so $_8^{19}O\rightarrow\,_9^{19}F+e^-+\bar\nu$ (a real, known decay of the radioactive oxygen isotope $^{19}$O, half-life about 26 s, to stable fluorine-19). (c) Thorium-228 (Z=90, A=228) undergoing alpha decay must produce a daughter with Z=90-2=88 and A=228-4=224 -- this is radium-224, so $_{90}^{228}Th\rightarrow\,_{88}^{224}Ra+\,_2^4He$. (d) Nitrogen-12 (Z=7, A=12) conv …

Balance mass number A and atomic number Z on both sides of each nuclear equation, using the fact that alpha decay reduces Z by 2 and A by 4, $\beta^-$ decay increases …

Balancing mass number A but forgetting to also balance atomic (charge) number Z, or vice versa -- both quantities must b …