When a uranium isotope $$^{235}_{92}U$$ is bombarded with a neutron, it generates $$^{89}_{36}Kr$$, three neutrons and :
Correct option is D. $$^{144}_{56}Ba$$
$$_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy)$$
$$\sum$$ 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 $$_{56}Ba^{144}$$