The cloth shroud from around a mummy is found to have a {14}C activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for {14}C decay, 5715 yr, calculate the age of the shroud.

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veiga34 · Accepted Answer

The half life of&amp;nbsp;&amp;nbsp;{14}C&amp;nbsp;is 5715 years.The connection between the half life and the decay constant is shown by the equation:k=0.693t12k=0.6935715yrk=1.21×10−4yr−1The information provided lets us know that&amp;nbsp;N0=16.3&amp;nbsp;and&amp;nbsp;Nt=9.7Use the equation&amp;nbsp;ln⁡NtNo=−ktRearrange to determine the shroud&#039;s age:t=−1kln⁡NtNok=t=−11.21×10−4yr−1ln⁡9.716.3t−4.28×103yr

Paul Mitchell · Accepted Answer

The half life for&amp;nbsp;&amp;nbsp;{40}K&amp;nbsp;decay to&amp;nbsp;&amp;nbsp;{40}Ar&amp;nbsp;is&amp;nbsp;t12=1.27⋅109yrKThe mass ratio of the rock is&amp;nbsp;&amp;nbsp;{40}Arto&amp;nbsp;&amp;nbsp;{40}K=421Hence,&amp;nbsp;No=4.2+1=5.2And,&amp;nbsp;Nt=1Let us calculate the age of the rock.Let&#039;s first determine the decay constant.k=0.6931.27⋅109yr=5.46⋅10−10yr{−1}We can now determine the rock&#039;s age.ln⁡(NtNo)=−ktln⁡(152)=−5.46⋅10−10yr−1⋅t−1.649=−5.46⋅10−10yr−1⋅t=−1.649−5.46⋅10−10yr−1=3.02⋅109yr

Vasquez · Accepted Answer

We know:The half time for decay,&amp;nbsp;t12=5715yrThe disintegration per minute per gram of carbon, N=9.7The disintegration per minute per gram,&amp;nbsp;No=16.3The rate constant&#039;s expression,k=0.693t12=0.693(5715yr(365days1yr)(24hr1day)(60min1hr))=2.31×10−10min−1the expression &quot;radioactive decay,&quot;ln⁡NNo=−ktln⁡9.717.3=−(2.31×10−10)tt=2.5×109minResult: The age of the shroud is&amp;nbsp;2.5×109min

The cloth shroud from around a mummy is found to have a ^{14}C activity of

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