Solution a MW 10 57 2 350 MW 900 350 350 3 E N N E P P P P b fissions 1004 8 n

# Solution a mw 10 57 2 350 mw 900 350 350 3 e n n e p

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Solution (a) MW 10 57 . 2 350 . 0 MW 900 ) 350 . 0 ( ) 350 . 0 ( 3 E N N E P P P P (b) fission/s 10 04 . 8 n MeV/fissio 200 J 10 1.60 MeV 1 W 10 571 . 2 fission 19 13 9 N E P n (c)  kg 991 g 10 91 . 9 10 6.02 mol 1 g/mol 04 . 235 s/y 10 16 . 3 /s 10 8.0357 5 23 7 19 m
OpenStax College Physics Chapters 32-34 3. (33-19) (a) What is the uncertainty in the energy released in the decay of a 0 due to its short lifetime? (b) What fraction of the decay energy is this, noting that the decay mode is 0 (so that all the 0 mass is destroyed)? Solution (a) Using 4 h t E we can calculate the uncertainty in the energy, given the lifetime of the 0 from Table 33.2 : eV 9 . 3 J 10 60 . 1 eV 1 J 10 28 . 6 ) s 10 4 . 8 ( 4 s J 10 63 . 6 4 19 19 17 34 t h E (b) The fraction of the decay energy is determined by dividing this uncertainty in the energy by the rest mass energy of the 0 found in Table 33.2 : 8 2 2 6 2 10 9 . 2 eV 10 0 . 135 eV 9256 . 3 0 c c c m E 4. (33-37) (a) How much energy would be released if the proton did decay via the conjectured reaction e p 0 ? (b) Given that the 0 decays to two s and that the e will find an electron to annihilate, what total energy is ultimately produced in proton decay? (c) Why is this energy greater than the proton’s total mass (converted to energy)? Solution (a) The energy released from the reaction is determined by the change in the rest mass energies: 2 f 2 i 2 0 c m m m mc mc E e p Using Table 33.2 , we can then determine this difference in rest mass energies: MeV 803 MeV 8 802 MeV 511 0 MeV 0 135

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• Fall '18
• Afrifa
• Energy, Mass, Mass–energy equivalence, OpenStax College Physics

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