In the nuclear reaction $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + x^{1}_{0}n$, the value of x is

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3. Conservation of nucleon number requires: 235 + 1 = 141 + 92 + x(1), giving 236 = 233 + x, so x = 3. Conservation of proton number also checks out: 92 + 0 = 56 + 36 + 0. Three neutrons are released in this fission reaction.

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A) 1

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B) 2

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C) 3

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D) 4

In the nuclear reaction $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + x^{1}_{0}n$, the value of x is

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