In the fission reaction $235 \mathrm { U } + { } ^ { 1 } \mathrm { n } \rightarrow { } ^ { 141 } \mathrm { Ba } + 92 \mathrm { Kr } + x ^ { 1 } \mathrm { n }$, what is the number $x$ of neutrons produced?

Question

Quizplus · Accepted Answer

To find the number of neutrons produced, $x$, we need to conserve the total number of nucleons (protons + neutrons) in the reaction. Uranium-235 has 235 nucleons, and the neutron that initiates the fission adds 1, making a total of 236 nucleons. Barium-141 and Krypton-92 have a combined total of 233 nucleons. To balance the nucleons on both sides of the equation, 236 - 233 = 3 neutrons must be produced.

In the Fission Reaction 235U+1n→141Ba+92Kr+x1n235 \mathrm { U } + { } ^ { 1 } \mathrm { n } \rightarrow { } ^ { 141 } \mathrm { Ba } + 92 \mathrm { Kr } + x ^ { 1 } \mathrm { n }235U+1n→141Ba+92Kr+x1n

In the Fission Reaction $235 \mathrm { U } + { } ^ { 1 } \mathrm { n } \rightarrow { } ^ { 141 } \mathrm { Ba } + 92 \mathrm { Kr } + x ^ { 1 } \mathrm { n }$