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

Hint: 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.