Problem 84 - Chemistry: An Atoms-Focused Approach

Question

For each of the following fission reactions, determine the identity of the unknown nuclide: a. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{147} \mathrm{Pm}+?+2_{0}^{1} \mathrm{n}\) b. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{94} \mathrm{Kr}+?+2_{0}^{1} \mathrm{n}\) c. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{95} \mathrm{Sr}+?+3_{0}^{1} \mathrm{n}\)

Step-by-Step Solution

Answer

+2_{0}^{1} \mathrm{n}\) b. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{94} \mathrm{Kr}+?+2_{0}^{1} \mathrm{n}\) c. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{95} \mathrm{Sr}+?+3_{0}^{1} \mathrm{n}\) Answer: a. The unknown nuclide is \(^{86}\mathrm{Ga}\). b. The unknown nuclide is \(^{139}\mathrm{Ba}\). c. The unknown nuclide is \(^{137}\mathrm{Xe}\).

Step 1: Balance atomic numbers (protons)

On the left side of the equation, we have 92 protons and on the right side, we have 61 protons in Pm. So, we need to find an element with atomic number 92-61=31, which is Gallium (Ga).

Step 2: Balance mass numbers (nucleons)

On the left side, we have 235 nucleons and on the right side, we have 147 in Pm and 2 in the neutrons. So, we need to find an isotope of Ga with a mass number of 235-147-2=86. The unknown nuclide is \(^{86}\mathrm{Ga}\). b. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{94} \mathrm{Kr}+?+2_{0}^{1} \mathrm{n}\)

Step 1: Balance atomic numbers (protons)

On the left side of the equation, we have 92 protons and on the right side, we have 36 protons in Kr. So, we need to find an element with atomic number 92-36=56, which is Barium (Ba).

Step 2: Balance mass numbers (nucleons)

On the left side, we have 235 nucleons and on the right side, we have 94 in Kr and 2 in the neutrons. So, we need to find an isotope of Ba with a mass number of 235-94-2=139. The unknown nuclide is \(^{139}\mathrm{Ba}\). c. \(^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \rightarrow^{95} \mathrm{Sr}+?+3_{0}^{1} \mathrm{n}\)

Step 1: Balance atomic numbers (protons)

On the left side of the equation, we have 92 protons and on the right side, we have 38 protons in Sr. So, we need to find an element with atomic number 92-38=54, which is Xenon (Xe).

Step 2: Balance mass numbers (nucleons)

On the left side, we have 235 nucleons and on the right side, we have 95 in Sr and 3 in the neutrons. So, we need to find an isotope of Xe with a mass number of 235-95-3=137. The unknown nuclide is \(^{137}\mathrm{Xe}\).