Question

ODE Decay Question :
In the rock being examined there is 100 mg of uranium and 14 mg of lead. Uranium decays with a half-life of 4.5 *10^9 years and after full decay of 238g of uranium 206 g of lead is left. Determine the rock age. Assume that when the rock formed it contained no lead, and neglect all intermediate decay-chain elements( since their half-life is much smaller than uranium )

Answer 1

So I think I found the decay equation U(t)= quantity of Uranium \[U(t)=U _{0}e ^{\frac{ \ln119/238}{4.5*10^9}t}\] is this somewhat right? I then put 0.01 and t=4.5*10^9 to find the original amount of Uranium but I can't seem to get it right.

Answer 2

I got the previous from solving U'(t)=KU(t) and then used the info I had to solve for the constants.

Answer 3

so I got a slightly better equation
\[U_{t}=U_{0}exp(\frac{ln(1/2)}{4.5∗109}t)\]

Answer 4

so I tried finding the initial amount of uranium in the rock which gave me 0.2 g , now , is this right? and I can't seem to know how to get the age. I got the ration 0.014g of lead/0.2g of uranium = 0.07 but I don't know what to do then

Answer 5

The Uranium decays according to \[U=U _{0}\exp \left( -\frac{ t }{4.5x10^{9} } \right)\]and as the Uranium decays the lead is building up according to \[Pb=Pb _{0} \left( 1- \exp \left( -\frac{ t }{4.5x10^{9} } \right) \right)\] for, for every U decay you will eventually get one Pb. Now You are give the ratios U/Pb and Uo/Pbo. Can you take it from here?