Question

A rock is found containing $Uranium - 238$ and also $lead - 206$ . Scientists analyze the rock for these two elements and find that the total mass of uranium in the rock is $2.40g$ , while the amount of lead is $1.11g$ . How old is this rock?
$Uranium - 238$ $\;(\lambda \; = \;4.5 \times {10^9}yr)$ decays through several steps until it finally decays into $lead - 206$ .

Answer 1

Radioactive decay is known as the process in which an unstable nucleus loses energy to form a stable nuclei. It is a first order reaction. In this case, we know that the decay of one atom of $U - 238$ will result in the formation of one atom of $Pb - 206$ . So, we will apply the first order radioactive decay equation to find the relation between the ratio of these two nuclei and time. Then, we will find out the decay constant $(\lambda )$ using the half-life of the nuclei. Using this information, we will find the time.

Complete Step By Step Answer:
We will denote $Uranium - 238$ as $U - 238$ and $lead - 206$ as $Pb - 206.$
As $U - 238$ decays exponentially, the amount of $Pb - 206.$ increases accordingly:
$U - 238$ has a half-life of approximately $4.5$ billion. Over time, the ratio of $Pb - 206$ to $U - 238$ will increase, and it is this ratio that makes it possible to estimate the age of the rock.
Radioactive decay is a first order process:
${U_t} = {\text{ }}{U_0}{e^{ - \lambda t}}$
${U_0}$ is the number of $U - 238$ atoms that did not initially decay.
${U_t}$ is the number of $U - 238$ that has not decayed after time t.
$\lambda$ is known as the decay constant.
Since the decay of one atom of $U - 238$ will result in the formation of one atom of $Pb - 206$ , we can say:
${U_0} = {U_t} + P{b_t}$
where $P{b_t}$ is the number of $Pb - 206$ atoms formed after time t.
Therefore, the decay equation can be written as:
${U_t} = \left( {{U_t} + P{b_t}} \right){e^{ - \lambda t}}$
$\Rightarrow \dfrac{{{U_t}}}{{\left( {{U_t} + P{b_t}} \right)}} = {e^{ - \lambda t}}$
On further simplifying this, we get
$\dfrac{{{U_t} + P{b_t}}}{{{U_t}}} = {e^{\lambda t}}$
$1 + \dfrac{{P{b_t}}}{{{U_t}}} = {e^{\lambda t}}$
$\Rightarrow \dfrac{{P{b_t}}}{{{U_t}}} = {e^{\lambda t}} - 1$
The half-life of $U - 238$ is $4.5 \times {10^9}yr$ . We can obtain the value of the decay constant from the expression:
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
$\lambda = \dfrac{{0.693}}{{4.5 \times {{10}^9}}}$
$\lambda = 0.154 \times {10^{ - 9}}y{r^{ - 1}}$
We can obtain n by dividing the number of moles of each isotope by one mole by dividing the given mass by atomic mass Ar:
${n_{P{b_t}}} = \dfrac{{1.11}}{{206}}$
${n_{P{b_t}}} = 0.005388$
${n_{{U_t}}} = \dfrac{{2.40}}{{\,238}}$
${n_{{U_t}}} = 0.01008$
There is no need to multiply these by the atomic number constant, because we are interested in the ratio of $Pb - 206$ and $\;U - 238$ , so it will be cancelled anyway.
$\dfrac{{0.005388}}{{0.01008}} = {\text{ }}{e^{\lambda t}} - 1$
${e^{\lambda t}} - 1 = 0.53462$
On further solving this equation, we get.
$\Rightarrow {e^{\lambda t}} = 1.53462$
Take the natural logarithm of both sides:
$\lambda t = \ln (1.53462)$
$\Rightarrow \lambda t = 0.428282$
Now, we will find the value of t from the above equation as follows:
$t = \dfrac{{0.428282}}{{0.1551 \times {{10}^{ - 9}}}}yr$
$\Rightarrow t = 2.76 \times {10^9}yr$
Therefore, rock is $2.76 \times {10^9}yr$ old.

Note:
We should remember that the radioactive elements are unstable and emit radiation to achieve greater stability. The parent nuclei emit $\alpha$ , $\beta$ , and $\gamma$ particles during their disintegration into the daughter nuclei. These daughter nuclei further disintegrate into stable nuclei. These types of reactions are always first order reactions.