Been estimated that earth has 9 1 10 11 kg of natural

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beenestimatedthatEarthhas9.1×1011kg of natural uranium that canbe economically mined. Of this total, 0.70%is235U.If all the world’s energy needs (7×1012J/s)were supplied by235U fission, how long wouldthis supply last?Assume that 208 MeV ofenergy is released per fission event and themass of235U is about 3.9×10-25kg.Correct answer: 7.76533×1010s.Explanation:Let :m= 9.1×1011kg,
rguig (tr7356) – Homework 13 – yao – (59020)7mU= 0.007m ,P= 7×1012J/s,R= 208 MeV/event= 2.08×108eV/event,andmU-235= 3.9×10-25kg/atom.Etot=PΔtΔt=EtotP=NRP=mUmU-235·RP=0.007 (9.1×1011kg)3.9×10-25kg/atom×2.08×108eV/event7×1012J/s×1.6×10-19J1 eV=7.76533×1010s.02010.0pointsThe energy released in the fusion reaction is11H +21H-→32He +γThe mass of11H is 1.00782 u,the mass of21His 2.0141 u,and the mass of32He is 3.01603 u,1.2.54 MeV2.1.38 MeV3.5.49 MeVcorrect4.4.29 MeV5.7.65 MeVExplanation:Let :mH-1= 1.00782 u,mH-2= 2.0141 u,mHe-3= 3.01603 u,andc2= 931.5 MeV/u.The energy released in the reaction isQ= (Δm)c= (mH-1+mH-2-mHe-3)c2= (1.00782 u + 2.0141 u + 3.01603 u)×(931.5 MeV/u)=5.49399 MeV.2

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