91Ó°ÊÓ

When a nucleus undergoes fission, it breaks into smaller fragments. For each reaction, we can write a general equation as follows: \(^{A_1} \mathrm{X} + _0^1 \mathrm{n} \rightarrow ^{A_2} \mathrm{Y} + ^{A_3} \mathrm{Z} + n_1\cdot _0^1 \mathrm{n}\) Here, X is the initial nucleus, Y and Z are the fragments produced after fission, and \(n_1\) is the number of neutrons released during the process.

For each reaction, we can use the conservation of atomic mass and atomic number to find the missing fragment: a. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{137} \mathrm{I}+?+2_{0}^{1} \mathrm{n}\) Using conservation of atomic mass and atomic number, we can determine the unknown nuclide: Atomic Number: 92 + 0 = 53 + Z Z = 92 - 53 = 39 Atomic Mass: 235 + 1 = 137 + A + 2(1) A = 235 + 1 - 137 - 2 = 96 So, the unknown nuclide in reaction (a) is \(_{39}^{96} \mathrm{Y}\) (Yttrium-96). b. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{137} \mathrm{Cs}+?+3_{0}^{1} \mathrm{n}\) Using conservation of atomic mass and atomic number, we can determine the unknown nuclide: Atomic Number: 92 + 0 = 55 + Z Z = 92 - 55 = 37 Atomic Mass: 235 + 1 = 137 + A + 3(1) A = 235 + 1 - 137 - 3 = 95 So, the unknown nuclide in reaction (b) is \(_{37}^{95} \mathrm{Rb}\) (Rubidium-95). c. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{141} \mathrm{Ce}+?+2_{0}^{1} \mathrm{n}\) Using conservation of atomic mass and atomic number, we can determine the unknown nuclide: Atomic Number: 92 + 0 = 58 + Z Z = 92 - 58 = 34 Atomic Mass: 235 + 1 = 141 + A + 2(1) A = 235 + 1 - 141 - 2 = 92 So, the unknown nuclide in reaction (c) is \(_{34}^{92} \mathrm{Se}\) (Selenium-92).