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NEET PhysicsEasy

When a uranium isotope 92235U^{235}_{92}\text{U} is bombarded with a neutron, it generates 3689Kr^{89}_{36}\text{Kr}, three neutrons and :

A

56144Ba^{144}_{56}\text{Ba}

B

4091Zr^{91}_{40}\text{Zr}

C

36101Kr^{101}_{36}\text{Kr}

D

36103Kr^{103}_{36}\text{Kr}

Step-by-Step Solution

The reaction is 92235U+01n→3689Kr+301n+ZAX^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{89}_{36}\text{Kr} + 3^{1}_{0}\text{n} + ^{A}_{Z}\text{X}. Balancing mass number: 235+1=89+3+A⇒A=144235 + 1 = 89 + 3 + A \Rightarrow A = 144. Balancing atomic number: 92+0=36+Z⇒Z=5692 + 0 = 36 + Z \Rightarrow Z = 56. Thus, 56144Ba^{144}_{56}\text{Ba} is generated.

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