Radioactive decay of Uranium 238

[Portuga]

Homework Statement

Analyzing a rock sample, it is found that it contains 1.58 mg of ²³⁸U and 0.342 mg of ²⁰⁶Pb, which is the stable final product of the disintegration of ²³⁸U. Assume that all ²⁰⁶Pb found comes from the disintegration of ²³⁸U originally contained in sample. How old is this rock?

Homework Equations

Radioactive decay:
$$N = N_0 e^{- \lambda t},$$ where $\lambda$ stands for the decay constant, and $N_0$ and $N(t)$ stands for the initial and the number of particles in some instant of time. To convert the mass of the Uranium and Plumbum of the sample, it will be necessary to use the number of Avogadro $N_A = 6.02 \times 10^23$, and their respective molar mass, 238.02891u and 207.2 u.

The Attempt at a Solution

My attempt consisted in convert the mass of Uranium and Plubum to $N_U$ and $N_{Pb}$ using their molar mass, which resulted in $4.00\times 10^{21}$ and $9.94\times 10^{20}$ respectively.
In my first try, I used that $N_0 = N_U + N_Pb$ in the Radioactive decay, and got $9.99\times 10^8$ years.
The author claims that this time is $1.45\times 10^9$ years.
In my second try, I observed the ²³⁸U series here, https://en.wikipedia.org/wiki/Decay_chain#Uranium_series, and verified that there are 8 alpha decays before one Plubum atom appears. With this in mind I used that $N_0 = N_U + 32*N_{Pb}$ (I considered that an alpha decay takes 4u of mass from the sample) and got $9.86\times 10^9$ years.
Please, I really need some help, I just can't figure out any better assumptions to solve this. I must be using some very incorrect reasoning.

 

[phyzguy]

Your first try is correct. I suspect you are confusing half-life and λ. Show us your calculations using your first attempt with N0 = NU + NPb.

 

[mjc123]

Just a couple of pedantic points:
The English name of Pb is Lead.
The molar mass of ²⁰⁶Pb is 206 (205.974, but 206 is close enough for the present purpose). 207.2 is the molar mass of naturally occurring lead, with its mixture of isotopes.

 

[Portuga]

Just a couple of pedantic points:
The English name of Pb is Lead.
The molar mass of ²⁰⁶Pb is 206 (205.974, but 206 is close enough for the present purpose). 207.2 is the molar mass of naturally occurring lead, with its mixture of isotopes.

Just a couple of pedantic points:
The English name of Pb is Lead.
The molar mass of ²⁰⁶Pb is 206 (205.974, but 206 is close enough for the present purpose). 207.2 is the molar mass of naturally occurring lead, with its mixture of isotopes.

Ok, thanks all for the replies!
My calculus:
$$<br /> \begin{aligned} & N=N_{0}e^{-\lambda t}\\<br /> \Rightarrow & -\lambda t=\ln\left(\frac{N}{N_{0}}\right)\\<br /> \Rightarrow & t=\frac{\ln\left(\frac{N_{0}}{N}\right)}{\lambda}\\<br /> & =\frac{\ln\left(\frac{N_{U}+N_{Pb}}{N_{U}}\right)}{\frac{1}{T_{\frac{1}{2}}}}\\<br /> & =\frac{\ln\left(1+\frac{N_{Pb}}{N_{U}}\right)}{\frac{1}{T_{\frac{1}{2}}}}\\<br /> & =\frac{\ln\left[1+\frac{m_{Pb}\left(\frac{m_{mol{}_{Pb}}}{N_{A}}\right)}{m_{U}\left(\frac{m_{mol{}_{U}}}{N_{A}}\right)}\right]}{\frac{1}{T_{\frac{1}{2}}}}\\<br /> & =\frac{\ln\left[1+\frac{\left(m_{Pb}\right)\left(m_{mol{}_{Pb}}\right)}{\left(m_{U}\right)\left(m_{mol{}_{U}}\right)}\right]}{\frac{1}{T_{\frac{1}{2}}}}\\<br /> & =\frac{\ln\left[1+\frac{\left(0.342 \text{mg}\right)\left(206\text{u}\right)}{\left(1.58\text{mg}\right)\left(238.02891\text{u}\right)}\right]}{\frac{1}{4.5\times10^{9}\text{y}}}\\<br /> & \approxeq7.38\times10^{8}\text{y}.<br /> \end{aligned}$$

 

[phyzguy]

As I suspected. λ is not 1 / Half-life. λ is ln(2) / Halflife. If you take your original answer of 1.0E9 years and divide by ln(2) you will have the answer.

 

[Portuga]

As I suspected. λ is not 1 / Half-life. λ is ln(2) / Halflife. If you take your original answer of 1.0E9 years and divide by ln(2) you will have the answer.

Thank you very much!

 

[haruspex]

Thank you very much!

Do you understand why λ=ln(2)/half life?

 

[Portuga]

Do you understand why λ=ln(2)/half life?

Yes, now I got the point!
$$\begin{aligned} & N=N_{0}e^{-\lambda t}\\<br /> \Rightarrow & \frac{N_{0}}{2}=N_{0}e^{-\lambda T_{\frac{1}{2}}}\\<br /> \Rightarrow & \frac{1}{2}=e^{-\lambda T_{\frac{1}{2}}}\\<br /> \Rightarrow & \ln\left(\frac{1}{2}\right)=-\lambda T_{\frac{1}{2}}\\<br /> \Rightarrow & \ln2=\lambda T_{\frac{1}{2}}\\<br /> \Rightarrow & \lambda=\frac{\ln2}{T_{\frac{1}{2}}}<br /> \end{aligned}$$