6. Radioactive decay chain: A rock sample contains traces of 238U, 235U, 232Th,

Question

6. Radioactive decay chain: A rock sample contains traces of 238U, 235U, 232Th, 208Pb, 207Pb, and 206Pb. Careful analysis shows that the ratio of the amount of 238 U to 206Pb is 1.208. (a) Assume that the rock originally contained no lead, determine the age of the rock. (b) What should be the ratios of 235U to 207Pb and of 232Th to 208Pb so that they would yield the same age for the rock? Ignore the minute amounts of the intermediate decay products in the decay chains. Note that this form of multiple dating gives reliable geological dates. (Use the information on radioactive decay chains, including their parent nuclei, half-lives, and the stable end pfoducts, from this week's lecture slides.)]

Explanation / Answer

a)

Let NU be the number of U-238 atoms initial in the rock

half life of U-238 T1/2 = 4.468E+9 yrs

number of U-238 atoms present at any instant t is given by

N(t) = NU exp(-0.693t/T1/2 )

Number of Pb-206 atoms NP = NU - N(t) = NU (1- exp(-0.693t/T1/2 ))

ratio of U-238 to Pb-206 = N(t)/Np =exp(-0.693t/T1/2 )/ (1-exp(-0.693t/T1/2 ) ) = 1.208

exp(0.693t/T1/2 ) = 2.208/1.208

t = 0.603* 4.468E+9/0.693 = 3.89E+9 yrs , age of the sample

b) half life of U-235 T1/2 = 7.04E8 yrs

end product   - Pb-207 , age of the rock t = 1.218E+9 Yrs , same as with U-238 dating

ratio = exp(-0.693t/T1/2 ) / 1- exp(-0.693t/T1/2 ) =

exp(-0.693*4.468E+9 /7.04E+8) / (1-exp(-0.693*4.468E+9 /7.04E+8)

           = 0.0125

Th-232 decay:

half life of Th-232 = 1.405E+10 yrs

end product Pb-208

ratio = exp(-0.693t/T1/2 )/ 1- exp(-0.693t/T1/2 ) =

exp(-0.693*1.218E+9/1.405E+10)/ ( 1- exp(-0.693*1.218E+9/1.405E+10) )

                                                 = 16.15

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