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Q.

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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a

$_{45} {Rh}^{117}$

b

$_{45} {Rh}^{116}$

c

$_{46} {Pd}^{117}$

d

$_{46} {Pd}^{116}$

answer is A.

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Detailed Solution

$_{92} U^{234} \to_{46} {Pd}^{116} +_{46} {Pd}^{116} + 2 {{}_{0}n}^{1}$ .

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