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The specific activity of a preperation consisting of radioactive C0^(58) and nonradioactive Co^(59) is equal to 2.2.10^(12) dis(s.g). The half-life of Co^(58) is 71.3 days. Find the ratio of the mass of radioactive cobalt in that preperation to the total mass of the preparation (in per cent). |
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Answer» Solution :We see that Specific activity of the SAMPLE `(1)/(M+M)` { Activity of `M gm` of `Co^(58)` in the sample} Here `M` and `M'` are the masses of `Co^(58)` and `Co^(59)` in the sample. Now activity of `M` gm of `Co^(58)` `=(M)/(58)xx6.023xx10^(23)xx(In 2)/(71.3xx86400) dis//sec` `= 1.168xx10^(15)M` Thus from the PROBLEM `1.168xx10^(15)(M)/(M+M')= 2.2xx10^(12)` or `(M)/(M+M)= 1.88xx10^(-3) i.e., 0.188%` |
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