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

A
$$^{144}_{56}Ba$$
B
$$^{91}_{40}Zr$$
C
$$^{101}_{36}Kr$$
D
$$^{103}_{36}Kr$$
Solution
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#### 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 $$_{56}Ba^{144}$$

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