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50 Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a sta- ble, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) 235U – 232Th + ³He, (b) 235U – 23!Th + He, (c) 25U – 230TH + He. The needed atomic masses are 232TH 232.0381 u He 3.0160 u 231'Th 231.0363u 230TH 230.0331 u 235U 235.0429 u "He 4.0026 u SHe 5.0122 u

50 Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a sta- ble, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) 235U – 232Th + ³He, (b) 235U – 23!Th + He, (c) 25U – 230TH + He. The needed atomic masses are 232TH 232.0381 u He 3.0160 u 231'Th 231.0363u 230TH 230.0331 u 235U 235.0429 u "He 4.0026 u SHe 5.0122 u

Modern Physics
Modern Physics
3rd Edition
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Chapter13: Nuclear Structure
Section: Chapter Questions
Problem 46P
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50 Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a sta- ble, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) 235U – 232Th + ³He, (b) 235U – 23!Th + He, (c) 25U – 230TH + He. The needed atomic masses are 232TH 232.0381 u He 3.0160 u 231'Th 231.0363u 230TH 230.0331 u 235U 235.0429 u "He 4.0026 u SHe 5.0122 u
Transcribed Image Text:50 Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a sta- ble, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) 235U – 232Th + ³He, (b) 235U – 23!Th + He, (c) 25U – 230TH + He. The needed atomic masses are 232TH 232.0381 u He 3.0160 u 231'Th 231.0363u 230TH 230.0331 u 235U 235.0429 u "He 4.0026 u SHe 5.0122 u
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