InterviewSolution
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InterviewSolution
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(a) On analysis a sample of uranium ore was found to contain `0.277g` of `._(82)Pb^(206)` and `1.667g` of `._(92)U^(238)`. The half life period of `U^(238)` is `4.51xx10^(9)` year. If all the lead were assumed to have come from decay of `._(92)U^(238)`, What is the age of earth? (b) An ore of `._(92)U^(238)` is found to contain `._(92)U^(238)` and `._(82)Pb^(206)` in the weight ratio of `1: 0.1` The half-life period of `._(92)U^(238)` is `4.5xx10^(9)` year. Calculate the age of ore. |
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Answer» (a) given `""""_(92)""^(238)U= 1.667 g = (1.667)/(238) " mole" = 7.004 xx 10""^(-3) " mole" """"_(82)""^(206)pb = 0.227 g = (0.277)/(206) = 1.345 xx 10""^(-3) " mole"` In this case all the lead comes from the decay of uranium `therefore`No. of mole of Pb formed `= (0.277)/(206) = 1.345 xx 10""^(-3) " mole"` No. of mole of U decayed= `1.345 xx 10""^(-3) " mole"` Initial no of mole of U before decay `= 7.004 xx 10""^(-3) + 1.345 xx 10""^(-3)` Moles of U after decay `N = 7.004 xx 10""^(-3) " year"""^(-1)` `lambda = (0.693)/(t""_(0.5)) = (0.693)/(4.5 xx 10""^(9)) = 1.536 xx 10""^(-10) " year"""^(-1)` As we know `t = (2.303)/(lambda) "log" (N""_(0))/(N) = (2.303)/(1.536 xx 10""^(10)) "log" (8.349 xx 10""^(-3))/(7.004 xx 10""^(-3))` `therefore` Age of earth `(t) = 46.85 xx 10""^(-4) " mole"` b) `N""_(0)=(1)/(238)+(0.1)/(206)=46.85xx10""^(-4)" mole"` `N=(1)/(238)=42.0xx10""^(-4)" mole"` `lamda=(0.693)/(4.5xx10""^(9))=1.54xx10""^(-4)year""^(-1)` `t=(2.303)/(1.54xx10""^(10))log.(46.85xx10""^(-4))/(42.0xx10""^(-4))` `t=7.097xx10""^(8)year` Note: Let a radioactive element A decay to daughter element B A to B `lamda""_(A)" and "lamda""_(B)" are decay constant of A and B `" `"Maximum activity time of daughter element can be"` `"calculated as "t""_(max)=(2.303)/((lamda""_(B)-lamda""_(A)))log""_(10)[lamda""_(B)/lamda""_(A)]` |
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