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Radioactivity is a first-order process. Radioactive carbon in wood sample decays with a half-life of 5770 years. What is the rate constant (in "year"^(-1)) for the decay ? What fraction would remain after 11540 years? |
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Answer» Solution :We have `LAMDA= (0.6932)/(t_((1)/(2)))` `=(0.6932)/(5770)=1.2 xx 10^(-4)` per year Further, as the duration of 11540 years is just the DOUBLE of `t_((1)/(2))` (i.e., 5770 years) we can FIND out the fraction remained after 11540 years without using equation (24) as follows: After 5770 years, half of the SUBSTANCE would remain and after another 5770 years (i.e., a total of 11540 years) half of the half would remain, i.e., one fourth would remain after 11540 years. |
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