Áp dụng định luật bảo toàn số khối và bảo toàn điện tích.
a)
\(^{235}_{92}U+^1_0n\rightarrow^{94}_{38}Sr+^{140}_{54}Xe+2^1_0n\left(1\right)\)
\(^{235}_{92}U+^1_0n\rightarrow^{141}_{56}Ba+^{92}_{36}Kr+3^1_0n\left(2\right)\)
b)
\(^{94}_{38}Sr\rightarrow^{94}_{40}Zr+2\beta\left(3\right)\\ ^{140}_{54}Xe\rightarrow^{140}_{58}Ce+4\beta\left(4\right)\)
Cộng từng vế (1), (3), (4) có:
\(^{235}_{92}U\rightarrow^{94}_{40}Zr+^{140}_{58}Ce+6\beta+n\)
\(m_0=m\left(^{235}U\right)=235,0493amu\)
\(m=m\left(^{94}Ze\right)+m\left(^{140}Ce\right)+m\left(n\right)=93,9063+139,9054+1,00862=234,82032\left(amu\right)\)
\(\Rightarrow\Delta E=\left(m_0-m\right).c^2=\left(235,0493-234,82032\right).\left(931,5MeV/c^2\right).c^2=213,29487MeV\)
