Problem 82 - Chemistry: An Atoms-Focused Approach

Question

For each of the following fission reactions, determine the identity of the unknown nuclide: a. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{137} \mathrm{I}+?+2_{0}^{1} \mathrm{n}\) b. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{137} \mathrm{Cs}+?+3_{0}^{1} \mathrm{n}\) c. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{141} \mathrm{Ce}+?+2_{0}^{1} \mathrm{n}\)

Step-by-Step Solution

Answer

+ 2 Neutrons (\(_{0}^{1}\mathrm{n}\)) The unknown nuclide is __________. b. Uranium-235 (\(^{235}\mathrm{U}\)) + Neutron (\(_{0}^{1}\mathrm{n}\)) → Cesium-137 (\(^{137}\mathrm{Cs}\)) + ? + 3 Neutrons (\(_{0}^{1}\mathrm{n}\)) The unknown nuclide is __________. c. Uranium-235 (\(^{235}\mathrm{U}\)) + Neutron (\(_{0}^{1}\mathrm{n}\)) → Cerium-141 (\(^{141}\mathrm{Ce}\)) + ? + 2 Neutrons (\(_{0}^{1}\mathrm{n}\)) The unknown nuclide is __________.

Step 1: Determine the missing mass and atomic numbers for each reaction

For all reactions, we have the same initial reactants: uranium-235 (\(^{235}\mathrm{U}\)) and a neutron (\(_{0}^{1}\mathrm{n}\)). Analyze each product and find the difference in mass and atomic numbers. This will give the missing mass and atomic numbers for the unknown nuclide. a. Uranium-235 (\(^{235}\mathrm{U}\)) + Neutron (\(_{0}^{1}\mathrm{n}\)) → Iodine-137 (\(^{137}\mathrm{I}\)) + ? + 2 Neutrons (\(_{0}^{1}\mathrm{n}\)) b. Uranium-235 (\(^{235}\mathrm{U}\)) + Neutron (\(_{0}^{1}\mathrm{n}\)) → Cesium-137 (\(^{137}\mathrm{Cs}\)) + ? + 3 Neutrons (\(_{0}^{1}\mathrm{n}\)) c. Uranium-235 (\(^{235}\mathrm{U}\)) + Neutron (\(_{0}^{1}\mathrm{n}\)) → Cerium-141 (\(^{141}\mathrm{Ce}\)) + ? + 2 Neutrons (\(_{0}^{1}\mathrm{n}\))

Step 2: Apply the conservation laws to determine the unknown nuclide

Use the conservation laws of mass number (A) and atomic number (Z) to find the unknown nuclide for each reaction. a. Mass number: \(235 + 1 = 137 + A + 2(1)\) --> \(A = 97\) Atomic number: \(92 + 0 = 53 + Z + 2(0)\) --> \(Z = 39\) The unknown nuclide is Yttrium-97 (\(^{97}\mathrm{Y}\)). b. Mass number: \(235 + 1 = 137 + A + 3(1)\) --> \(A = 96\) Atomic number: \(92 + 0 = 55 + Z + 3(0)\) --> \(Z = 37\) The unknown nuclide is Rubidium-96 (\(^{96}\mathrm{Rb}\)). c. Mass number: \(235 + 1 = 141 + A + 2(1)\) --> \(A = 93\) Atomic number: \(92 + 0 = 58 + Z + 2(0)\) --> \(Z = 34\) The unknown nuclide is Selenium-93 (\(^{93}\mathrm{Se}\)).