Part a
plutonium−239 ⟹ α−decay94239Pa→92235U+24αplutonium-239 \implies \alpha - decay\\\\ _{94}^{239} Pa \to _{92}^{235} U + _{2}^{4} \alphaplutonium−239⟹α−decay94239Pa→92235U+24α
Part b
Indium−120 ⟹ α−decay49120In→50120Sn+−10eIndium-120 \implies \alpha - decay\\\\ _{49}^{120} In \to _{50}^{120} Sn + _{-1}^{0} eIndium−120⟹α−decay49120In→50120Sn+−10e
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments