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1252_Physics ProblemsTechnical Physics

# 1252_Physics ProblemsTechnical Physics - Chapter 44 P44.73...

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Chapter 44 593 P44.73 Km v = 1 2 2 , so v K m == × × 2 200400 160 10 167 10 277 10 19 27 3 .. . . eV J eV kg ms bg ej . The time for the trip is t x v × × = 100 10 361 4 . . m 2.77 10 m s s 3 . The number of neutrons finishing the trip is given by NN e t = 0 λ . The fraction decaying is 1 1 1 0 004 00 0 400% 0 2 236 1 12 −= = = = N N ee tT ln ln . af s624 s . P44.74 (a) If we assume all the 87 Sr came from 87 Rb, then e t = 0 yields t N N T N N = F H G I K J = F H G I K J 1 2 0 0 ln ln ln where = Rb-87 and N 0 =+ Sr-87 Rb-87 t = × ×+× × F H G I K J 475 10 2 1 82 10 1 07 10 391 10 10 10 9 9 . ln ln . yr 1.82 10 yr 10 . (b) It could be no older . The rock could be younger if some 87 Sr were originally present. P44.75 RR t =− 0 exp lets us write ln ln RRt 0 which is the equation of a straight line with slope = . The logarithmic plot shown in Figure P44.75 is fitted by ln . . Rt 844 0262 . If t is measured in minutes, then decay constant
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