24.135 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Uranium-238  undergoes a slow decay step $(t_{t/2}=4.5\times {10}^{9}yr)$ followed by a series of fast steps to form the stable isotope ²⁰⁶pb. Thus, on a time scale of billions of years, ²³⁸U effectively decays “directly” to ²⁰⁶pb, and the relative amounts of these isotopes are used to find the age of some rocks .Two students derive equations relating number of half lives (n) since the rock formed to the amounts of the isotopes: 

Student 1: ${\left(\frac{1}{2}\right)}^n=\frac{{}_{92}{}^{238}U}{{}_{82}{}^{206}Pb}$

Student 2: ${\left(\frac{1}{2}\right)}^n=\frac{{}_{92}{}^{238}U}{{}_{92}{}^{238}U{+}_{82}^{206}Pb}$

 

(a) Which equation is correct, and why? 

(b) If a rock contains exactly twice as much ²³⁸U as ²⁰⁶pb, what is its age in years?

Step-by-Step Solution

Answer

a. Student 2 is correct.

b. The age is $2.6323\times {10}^9$ years old.

Step 1: Definition of Nuclear Reactions

A nuclear reaction occurs when two atom nuclei or nuclei meet with a subatomic particle from outside the atom. Such reactions can occur spontaneously or artificially to provide “nuclear energy” on demand in a regulated manner.

Step 2: Find the correct equation

a. Student 2 is entirely accurate. The remaining fraction of ²³⁰U is equal to ^2sU divided by its starting value.

Therefore, student 2 is correct.

Step 3: Find its age in years

b. ${}^{238}U=2(Pb-206)$

 

Let ${}^{238}U=2x$and $2(Pb-206)=x$;

 

Fraction of ²³⁸U:

$\frac{{}_{92}{}^{238}U}{{}_{92}{}^{238}U{+}_{82}^{206}Pb}=\frac{2x}{2x+x}=\frac{2}{3}$

 

We must use the half-life formula to find the constant (k) in this situation.

$$\begin{gathered} t_{\frac{1}{2}}=\frac{\ln 2}{k}\to k=\frac{\ln 2}{t_{\frac{1}{2}}} \\ k=\frac{\ln 2}{t_{\frac{1}{2}}}=\frac{\ln 2}{4.5x{10}^{9}yr} \\ =\frac{1.540327\times {10}^{-10}}{yr} \end{gathered}$$

 

 

We can solve for the value of t using the expression for determining the number of nuclei remaining.

$\ln \frac{N_0}{N_t}=kt$

$$\begin{gathered} \ln \frac{1}{\frac{2}{3}}=\frac{1.540327\times {10}^{-10}}{yr}\left(t\right) \\ t=2.6323\times {10}^9 \end{gathered}$$

 

Therefore, the age is $2.6323\times {10}^9$years old.