The nucleus _{92}U^{234} splits exactly in ha | vedclass

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Question

(A) The fission reaction is given by: $_{92}U^{234} \rightarrow 2(_{Z}X^{A}) + 2(_{0}n^{1})$.Applying the law of conservation of mass number: $234 = 2

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$_{46}Pd^{116}$

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(A) The fission reaction is given by: $_{92}U^{234} \rightarrow 2(_{Z}X^{A}) + 2(_{0}n^{1})$.Applying the law of conservation of mass number: $234 = 2A + 2(1) \Rightarrow 234 = 2A + 2 \Rightarrow 2A = 232 \Rightarrow A = 116$.Applying the law of conservation of atomic number: $92 = 2Z + 2(0) \Rightarrow 92 = 2Z \Rightarrow Z = 46$.The element with atomic number $Z = 46$ is Palladium $(Pd)$.Thus,the resultant nuclei are $_{46}Pd^{116}$.

The nucleus $_{92}U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are

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