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The decay constant (lambda) of radium is 0.0356 per year (1.13 xx 10^(-9) per sec). How much time is taken to reduce a. to (1//10)^(th) of the original value? b. by (1//10)^(th) of the original value? |
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Answer» Solution :Data supplied, a. `N = (N_0)/(N) , "" lambda = 0.0356` PER year `t = 1/(lambda) XX 2.303 LOG ((N_0)/(N)) = (1 xx 2.303 xx log 10)/(0.0356) = 64.69 ` years b. `N = (9N_0)/(10) "" :. (N_0)/N = 10/9` `t = 1/lambda xx 2.303 log ((N_0)/(N)) = (2.303 xx log (10/9))/(0.0356)` `= (2.303)/(0.0356) xx (log 10 - log 9) = (2.303)/(0.0356) xx 0.0458 = 2.96` years . |
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