Chapter 44593P44.73Kmv=122,sovKm==××=×−−2200400160 10167 10277 1019273....eVJ eVkgmsbgej.The time for the trip is txv××=100 103614..m2.77 10 m ss3.The number of neutrons finishing the trip is given by NNet=−0λ.The fraction decaying is 1110 004 000 400%02236112−=−=−==−−NNeetTlnln.afs624s.P44.74(a)If we assume all the 87Sr came from 87Rb,thenet=−0yieldstNNTNN=−FHGIKJ=FHGIKJ1200lnlnlnwhere=Rb-87andN0=+Sr-87Rb-87t=××+××FHGIKJ475 1021 82 101 07 10391 10101099.lnln.yr1.82 10yr10.(b)It could be no older . The rock could be younger if some 87Sr were originally present.P44.75RRt=−0explets us writelnlnRRt0which is the equation of a straight line withslope=.The logarithmic plot shown in Figure P44.75 is fitted byln..Rt844 0262 .If tis measured in minutes, then decay constant
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .