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The `beta-` activity of a sample of `CO_(2)` prepared form a contemporary wood gave a count rate of `25.5` counts per minute `(cp m)`. The same of `CO_(2)` form an ancient wooden statue gave a count rate of `20.5 cp m`, in the same counter condition. Calculate its age to the nearest 50 year taking `t_(1//2)` for `.^(14)C` as 5770 year. What would be the expected count rate of an identical mass of `CO_(2)` form a sample which is 4000 year old? |
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Answer» Correct Answer - `a. t=821 years`,`b.` Coutn rate `=15.5 cpm` The above magnitudes are same which means that the stability of all three atoms is same. Initial activity `(N_(0) =25.5cpm)` Activity after some time `(N_(1)` or `N)=20.5 cpm` `a.` `t=(2.303)/(K)log``(N_(0))/(N_(t))` `=(2.303xx5770year)/(0.69)log``(25.5)/(20.5)` `=19258.4xxlog1.2439` `=19258.4xx0.0947=1821years` `b.` Number of counts decreases in 1821 years `=25.5-20.5=5 cpm` Number of counts of decrease in 4000 years `(5)/(1821)xx4000=10 counts ~~10 counts ` Count rate of a sample which is 4000 years old `=25.5-10=15.5cpm` |
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