As the U 238 decays exponentially, the amount of Pb 206 grows correspondingly:
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.
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"))#
The half - life, #sf(t_(1/2))#, of U 238 is #sf(4.47xx10^(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/(4.47xx10^(9))=0.155xx10^(-9)" " "yr"^(-1))#
We know that:
#:.##sf(1+(Pb_(t))/U_(t)=e^(lambda"t"))#
Taking natural logs of both sides we get:
#sf(ln[(Pb_t)/U_(t)+1]=lambdat)#
Putting in the numbers:
#sf(ln[(Pb_t)/U_(t)+1]=0.155xxcancel(10^-9)xx4.47xxcancel(10^9))#
#sf(ln[(Pb_t)/U_(t)+1]=0.69286)#
From which:
#sf((Pb_t)/(U_(t))+1=1.9994)#
#:.##sf((Pb_(t))/(U_(t))=0.9994)#
This ratio can represent the number of moles of #sf(Pb_(t):U_(t))#
It seems a reasonable result since about one half - life has elapsed so we would expect the ratio to be close to 1.
#sf(U_(t)=m/A_(r)=1.305/238=0.005483)#
#:.##sf(Pb_(t)=0.9994xx0.005483=0.0054267)#
#:.##sf(m_(Pb_(t))=Pb_(t)xxA_(r)=0.0054267xx206=1.118color(white)(x)g)#
