Identify the missing product in the given reaction.
$_{92}^{235} U +_{0}^{1} n \to ? +_{36}^{92} K r + 3_{0}^{1} n$

_{56}^{141} B a

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_{56}^{139} B a

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_{54}^{139} B a

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_{54}^{141} B a

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Solution

The correct option is A $_{56}^{141} B a$
$92 + 0 = Z + 36 + 0 \Rightarrow Z = 56$
$235 + 1 \to A + 92 + 3$
$∴ A = 141$
Missing nuclide is $_{56}^{141} B a$

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