InterviewSolution
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InterviewSolution
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The age of a rock containing lead and uranium is equal to `1.5xx10^9` years. The uranium is decaying into lead with half life equal to `4.5xx10^9` years. Find the ratio of lead to uranium present in the rock, assuming that initially no lead was present in the rock (given `2^(1/3)`=1.259) |
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Answer» `N_0` is initial amount of uranium . N is present amount of uranium after `1.5xx10^9` years . The amount of lead present is `N_0`-N `therefore` The required ratio is `(N_0-N)/N=(1-e^(-lambdat))/(e^(-lambdat))` Now , `T_(1//2) ="In2"/lambda rArr lambda="ln2"/T_(1//2)` `therefore(N_0-N)/N=(1-e^(-"ln2"^(t/(T//2))))/e^(-"ln2"^(t/(T//2)))=(1-2^(-t/T_(1//2)))/2^(-t/T_(1//2))=(1-2^(-1/3))/(2^(-1/3))=2^(1//3)-1=0.259` |
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