Use the following table of atomic masses to determine the energy released in the alpha decay of 243, 95Am. Atomic number Z Mass Atomic mass (1) number A (Neutron) 1 1.008 665 1 H 1 1.007 825 2 2.014 102 3 3.016 049 Не 3 3.016 029 4 4.002 602 90 Th 230 230.033 134 232 232.038 056 91 Pa 231 231.035 884 92 233 233.039 636 234 234.040 952 235 235.043 930 236 236.045 562 2.

Use the following table of atomic masses to determine the energy released in the alpha decay of 243, 95Am. Atomic number Z Mass Atomic mass (1) number A (Neutron) 1 1.008 665 1 H 1 1.007 825 2 2.014 102 3 3.016 049 Не 3 3.016 029 4 4.002 602 90 Th 230 230.033 134 232 232.038 056 91 Pa 231 231.035 884 92 233 233.039 636 234 234.040 952 235 235.043 930 236 236.045 562 2.

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter30: Nuclear Physics

Section: Chapter Questions

Problem 18P

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Use the following table of atomic masses to determine the energy released in the alpha decay of 243,
95Am.
Atomic
number Z
Mass
Atomic mass (1)
number A
(Neutron)
1
1.008 665
1
H
1
1.007 825
2
2.014 102
3
3.016 049
Не
3
3.016 029
4
4.002 602
90
Th
230
230.033 134
232
232.038 056
91
Pa
231
231.035 884
92
233
233.039 636
234
234.040 952
235
235.043 930
236
236.045 562
238
238.050 784
93
Np
237
237.048 168
239
239.052 939
94
Pu
238
238.049 560
239
239.052 164
240
240.053 Sos
241.056 852
241
242
242.058 737
244
244.064 205
95
Am
241
241.056 823
243
243.061 375
O 4.98 MeV
5.43 MeV
5.26 MeV
O 5.32 MeV
2.

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