A) \[_{55}^{138}Cs\]
B) \[_{56}^{140}Ba\]
C) \[_{54}^{144}Xe\]
D) \[_{38}^{90}Sr\]
Correct Answer: A
Solution :
\[_{92}^{235}U+_{0}^{1}n\xrightarrow{{}}_{37}^{95}Rb+{{3}_{0}}{{n}^{1}}+_{55}^{138}Cs\] The other new nuclide formed is cesium\[(_{55}^{138}Cs)\].
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