Consider a radioactive nuclear decay series ABC.... with decay constants AA, AB, AC..., and assume at time t = 0 only nuclei of type A are present. Now integrate your expression to find the number of nuclei of type B at time t, i.e. NB(t). [ Hint: multiply by et and use integration by parts. Don't forget the integration constant!] . In the case of the actinide series (e.g. half-life of 238 U is 4.47 x 10⁹ years, very much greater than the daughter nuclei!) show that the decay chain is in secular equilibrium, i.e. all the daughter nuclei have essentially equal activities.

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Chapter1: Chemical Foundations
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Consider a radioactive nuclear decay series A → B → C → with decay constants
AA, AB, AC..., and assume at time t = 0 only nuclei of type A are present.
Now integrate your expression to find the number of nuclei of type B at time t,
i.e. NB(t).
[ Hint: multiply by et and use integration by parts. Don't forget the integration
constant!]
. In the case of the actinide series (e.g. half-life of 238 U is 4.47 x 10⁹ years, very
much greater than the daughter nuclei!) show that the decay chain is in secular
equilibrium, i.e. all the daughter nuclei have essentially equal activities.
Transcribed Image Text:Consider a radioactive nuclear decay series A → B → C → with decay constants AA, AB, AC..., and assume at time t = 0 only nuclei of type A are present. Now integrate your expression to find the number of nuclei of type B at time t, i.e. NB(t). [ Hint: multiply by et and use integration by parts. Don't forget the integration constant!] . In the case of the actinide series (e.g. half-life of 238 U is 4.47 x 10⁹ years, very much greater than the daughter nuclei!) show that the decay chain is in secular equilibrium, i.e. all the daughter nuclei have essentially equal activities.
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