Answer to Question #164553 in General Chemistry for sabine jeune

Question #164553

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)




1
Expert's answer
2021-02-22T05:48:31-0500

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} \\\\\n{^{249}_{97}Bk} + 1{^0_{+1}\\beta} = {^{249}_{98}Cf}\\\\\n{^{249}_{98}Cf} + 3{^0_0\\gamma} = {^{249}_{98}Cf}"


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


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