Chapter 45607P45.37(a)Starting with N=0 radioactive atoms at t=0, the rate of increase is (production – decay)dNdtRN=−λsodNRN dtbg.The variables are separable.dNdtNt−=zz00:−−FHGIKJ=1lnRtsolnRt−FHGIKJandRet−FHGIKJ=−.Therefore,1−=−RNetNRet−1ej.(b)The maximum number of radioactive nuclei would be R.Additional ProblemsP45.38(a)Suppose each 235U fission releases 208 MeV of energy. Then, the number of nuclei thatmust have undergone fission isN==××=×−total releaseenergy per nucleiJ208 MeVJ MeVnuclei510160 1015 10131324af...(b)massnuclei6.02 10 nuclei molg molkg23=××FHGIKJ≈2350 624..P45.39(a)At 6 108×K, the average kinetic energy of a carbon atom is3215 862 106 108 10584kTBeV KKeV×=×−afejNote that 6 108×K is about 6362=times larger than 1 5 107.×K , the core temperature ofthe Sun. This factor corresponds to the higher potential-energy barrier to carbon fusion
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