In the fission reaction $\mathrm{n}+\mathrm{U}_{92}^{235} \rightarrow \mathrm{Ba}_{56}^{141}+$ ? $+4 \mathrm{n}$, what are the $\mathrm{Z}$ and $\mathrm{A}$ for the unknown fission product?

Question

Quizplus · Accepted Answer

The unknown fission product can be determined by balancing the atomic numbers (Z) and mass numbers (A) on both sides of the equation. The sum of atomic numbers on the left is 92 + 0 = 92, and on the right, it's 56 + Z (unknown) + 0 (for neutrons), so Z = 36. The sum of mass numbers on the left is 235 + 1 = 236, and on the right, it's 141 + A (unknown) + 4(1) (for neutrons), so A = 236 - 141 - 4 = 91. Therefore, the unknown fission product has Z = 36 and A = 91, which corresponds to choice A.

In the Fission Reaction n+U92235→Ba56141+\mathrm{n}+\mathrm{U}_{92}^{235} \rightarrow \mathrm{Ba}_{56}^{141}+n+U92235​→Ba56141​+ +4n+4 \mathrm{n}+4n , What Are The

In the Fission Reaction $\mathrm{n}+\mathrm{U}_{92}^{235} \rightarrow \mathrm{Ba}_{56}^{141}+$ $+4 \mathrm{n}$ , What Are The