1238_Physics ProblemsTechnical Physics

1238_Physics ProblemsTechnical Physics - Chapter 44 *P44.32...

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Chapter 44 579 *P44.32 (a) dN dt 2 = rate of change of N 2 = rate of production of N 2 – rate of decay of N 2 = rate of decay of N 1 – rate of decay of N 2 =− λλ 11 22 NN (b) From the trial solution Nt N ee tt 2 10 1 12 21 a f ej = −− λ ∴= −+ dN dt N 0 1 (1) ∴+ = −++− = = −−−− dN dt N N eeee N e N tttt t 2 10 1 2122 10 1 21 21 1 λλλλ bg So dN dt 2 as required. (c) The functions to be plotted are e e e t 1 02236 2 00259 1000 11308 1 af = L N M O Q P . .. . min min min From the graph: t m 10 9 . m in 1 200 1 000 800 600 400 200 0 0 10 20 30 40 Po Pb time (min) Decay of and Po 218 Pb 214 FIG. P44.32(c) (d) From (1), dN dt 2 0 = if = . e t 1 2 . Thus, m == ln . With 1 1 = . m i n , 2 1 = . min , this formula gives t m = 10 9 . min , in agreement with the result of part (c).
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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 .

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