When a uranium isotope $^{235}_{92}U$ is bombarded with a neutron, it generates $^{89}_{36}Kr$, three neutrons and :

Q: When a uranium isotope \$^{235}_{92}U\$ is bombarded with a neutron, it generates \$^{89}_{36}Kr\$, three neutrons and :

\$_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy)\$\$\sum\$ Atomic number on \$LHS = 89 + 0 = 89 + 3\$\$\sum\$ Atomic number on \$RHS = Q + 36 + 3 \times 0\$\$\therefore LHS = RHS\$\$Q+36 = 92\$\$Q = 56\$also, \$\sum\$ atomic mass number on \$LHS = 235 + 1 = 236\$\$\sum\$ Atomic mass number on \$RHS=P + 89 + 3 \times 1\$\$\therefore LHS = RHS\$\$P + 92= 236\$\$P = 144\$So, the element is \$_{56}Ba^{144}\$

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Accepted Answer

$_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy)$$\sum$ Atomic number on $LHS = 89 + 0 = 89 + 3$$\sum$ Atomic number on $RHS = Q + 36 + 3 \times 0$$\therefore LHS = RHS$$Q+36 = 92$$Q = 56$also, $\sum$ atomic mass number on $LHS = 235 + 1 = 236$$\sum$ Atomic mass number on $RHS=P + 89 + 3 \times 1$$\therefore LHS = RHS$$P + 92= 236$$P = 144$So, the element is $_{56}Ba^{144}$

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