A sample of rock is known to contain the isotopes U 238 and Pb 206 in the mass ratio of 2:1. What is the age of the sample assuming all the Pb 206 has originated from the decay of U 238 ?The half - life of U 238 is #4.468xx10^9# years.

1 Answer

Answer:

#sf(2.94xx10^9color(white)(x)"yr"#

Explanation:

The ratio of the amounts of U 238 and Pb 206 in a rock sample enables the age of the rock to be estimated using the technique of radiometric dating.

U 238 forms a decay chain in which it undergoes a sequence of 8 #alpha# and 6 #beta# decays:

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It moves back in the periodic table until the isotope falls in the band of stability at Pb 206.

Each step has its own individual half - life but the first decay to Th 234 is about 20,000 times slower than the other decay steps.

Those of you who are familiar with chemical kinetics will know that it is the slowest step in a mechanism which determines the overall rate of reaction, the so - called "rate determining step".

This is the case here in the conversion of U 238 to Pb 206.

As the U 238 decays exponentially, the amount of Pb 206 grows correspondingly:

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The half - life of U 238 is about 4.5 billion years. As time passes, the ratio of Pb 206 to U 238 will increase and it is this which enables the age of the rock to be estimated.

We need to do some "math" to show this:

Radioactive decay is a first order process:

#sf(U_(t)=U_(0)e^(-lambda"t"))#

#sf(U_(0))# is the number of undecayed U 238 atoms initially.

#sf(U_t)# is the number of undecayed U 238 after time #sf(t)#.

#sf(lambda)# is the decay constant.

Since the decay of 1 U 238 atom will result in the formation of 1 atom of Pb 206 we can say that:

#sf(U_(0)=U_(t)+Pb_(t))#

Where #sf(Pb_t)# is the number of Pb 206 atoms formed after time #sf(t)#.

The decay equation can therefore be written:

#sf(U_(t)=(U_(t)+Pb_(t))e^(-lambda"t"))#

#:.##sf((U_(t))/((U_(t)+Pb_(t)))=e^(-lambda"t"))#

#:.##sf((U_(t)+Pb_(t))/(U_(t))=e^(lambdat)#

#:.##sf(1+(Pb_(t))/U_(t)=e^(lambda"t"))#

#:.##sf((Pb_(t))/U_(t)=e^(lambda"t")-1)#

The half - life, #sf(t_(1/2))#, of U 238 is #sf(4.468xx10^(9))# years.

We can get the value of the decay constant from the expression:

#sf(lambda=0.693/(t_(1/2))#

#sf(lambda=0.693/(4.468xx10^(9))=0.1551xx10^(-9)" " "yr"^(-1))#

We are given the ratio of Pb206:U238 by mass. If "n" refers to the number of moles we can write:

#sf(nPb206=(mPb206)/(A_(r)Pb206))#

#sf(nU238=(mU238)/(A_(r)U238))#

Dividing gives#rArr#

#sf((nPb206)/(nU238)=(mPb206)/(A_(r)Pb206)xx(A_(r)U238)/(mU238))#

This becomes:

#sf((nPb206)/(nU238)=0.5xx(238.050784)/(205.974449)=e^(lambda"t")-1)#

#:.##sf(e^(lambda"t")-1=0.577864)#

#sf(e^(lambda"t")=1.577864)#

Taking natural logs of both sides gives:

#sf(lambda"t"=ln(1.577864)=0.4560725)#

#sf(t=0.4560725/(0.1551xx10^(-9))color(white)(x)"yr")#

#sf(t=2.94xx10^(9)color(white)(x)"yr")#