Question #a57d5

1 Answer
Apr 20, 2017

Answer:

#sf(1.118color(white)(x)g)#

Explanation:

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

scifun.chem.wisc.edu

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)#