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)
