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The isotope of U^(238) and U^(235) occur...

The isotope of `U^(238) and U^(235)` occur in nature in the ratio `140:1`. Assuming that at the time of earth's formation, they were present in equal ratio, make an estimate of the age of earth. The half lives of `U^(238)` and `U^(235)` are `4.5xx10^9` years and `7.13xx10^8` years respectively. Given `log_(10)140=2.1461 and log_(10)^2=0.3010.`

Text Solution

Verified by Experts

As is known, `N_1=N_0e^(-lambda_1t) , N_2=N_0e^(-lambda_2t)`
`(N_1)/(N_2)=e^((lambda_2-lambda_1)t)`
`t=(log_e(N_1//N_2))/(lambda_2-lambda_1)`
Now,`lambda=0.693/T=(log_e^2)/T`
`:. t=(log_e(N_1//N_2))/(log_e^2(1/(T_2)-1/(T_1)))=(log_(10)(N_1//N_2))/(log_(10)^2((T_1-T_2)/(T_1T_2)))`
As `N_1//N_2=140`.
Therefore, `t=2.1461/0.3010((4.5xx10^(9)xx7.13xx10^8)/(4.5xx10^(9)-7.13xx10^(8)))`
`t=(2.1461xx4.5xx7.13xx10^(17))/(0.3010xx37.87xx10^(8))`
`t=6.04xx10^9 years`
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