5. Calculate the energy released in the following rare spontaneous fission reaction: 238 U → 95 Sr + Xe+ 3n, 140 '95 The atomic masses are m (238 U) = 238.050784 u, m (5Sr) = 94.919388 u m (140 Xe) = 139.921610 u, and m (n) = 1.008665 u.

College Physics
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Author:Paul Peter Urone, Roger Hinrichs
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Chapter32: Medical Applications Of Nuclear Physics
Section: Chapter Questions
Problem 44PE: (a) Calculate the energy released in the neutroninduced fission reaction n+235U92Kr+142Ba+2n, given...
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5.
Calculate the energy released in the following rare spontaneous fission reaction:
238 U → 95 Sr +140 Xe+ 3n,
'95
The atomic masses are m (238U) = 238.050784 u, m (5Sr) = 94.919388 u,
m (140 Xe) = 139.921610 u, and m (n) = 1.008665 u.
Transcribed Image Text:5. Calculate the energy released in the following rare spontaneous fission reaction: 238 U → 95 Sr +140 Xe+ 3n, '95 The atomic masses are m (238U) = 238.050784 u, m (5Sr) = 94.919388 u, m (140 Xe) = 139.921610 u, and m (n) = 1.008665 u.
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