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Question

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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