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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
$^{91}_{40}Zr$
B
$^{101}_{36}Kr$
C
$^{103}_{36}Kr$
D
$^{144}_{56}Ba$
Detailed Solution
$^{235}_{92}U + {}^1_0n \rightarrow {}^{89}_{36}Kr + 3\,{}^1_0n + {}^A_ZX$
Atomic number: 92 + 0 = 36 + Z ⇒ Z = 56
Mass number: 235 + 1 = 89 + 3 + A ⇒ A = 144
So $^{144}_{56}Ba$ is generated.