Answer to Question #164553 in General Chemistry for sabine jeune

Question

After emitting 2 alpha particles, 1 positron, and 3 gamma particles, the resulting daughter nuclide was ²⁴¹₉₃Np. What was the initial parent nuclide before any particles were emitted? (Incorrect formatting will result in your answer being incorrect)

Answer 1

The decay equations usually start from a parent and produce a daughter. Depending on the type of reaction together with a daughter, other particles emit. In the current case, we know a daughter(${^{241}_{93}Np}$ ) and number and type of emitted particles (${2^4_2He}, {1^0_{+1}\beta}, {3^0_0 \gamma}$ ). To determine a parent nuclide, we will go to the opposite direction by summation of the daughter nuclide and all emmited particles:

${^{241}_{93}Np} + 2{^4_2He} = {^{249}_{97}Bk} \\ {^{249}_{97}Bk} + 1{^0_{+1}\beta} = {^{249}_{98}Cf}\\ {^{249}_{98}Cf} + 3{^0_0\gamma} = {^{249}_{98}Cf}$

The initial parrent nuclide is ${^{249}_{98}Cf}$