A rock recovered from far underground is found to contain 0.86 mg of ${}^{\mathbf{238}}\mathit{U}$, 0.15 mg of${}^{\mathit{206}}\mathit{P}\mathit{b}$ , and 1.6 mg of${}^{\mathbf{40}}\mathit{A}\mathit{r}$ . How much${}^{\mathbf{40}}\mathit{K}$ will it likely contain? Assume that${}^{\mathbf{40}}\mathit{K}$ decays to only${}^{\mathbf{40}}\mathit{A}\mathit{r}$ with a half-life of$\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathit{y}$ . Also assume that${}^{\mathbf{238}}\mathit{U}$ has a half-life of$\mathbf{4}\mathbf{.}\mathbf{47}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathit{y}$ .

a) Mass of ${}^{238}U$ ,$m_U=0.86mg$

b) Mass of${}^{206}Pb$ ,$m_{Pb}=0.15mg$

c) Mass of ${}^{40}Ar$, $m_{Ar}=0.86mg$

d) Hal-life of${}^{40}Ar$ , $T_{1/2Ar}=1.25\times {10}^{9}y$

e) Half-life of${}^{238}U$ ,$T_{1/2,U}=4.47\times {10}^{9}y$

The rock at the time of formation was having only uranium and thus with every decay of one uranium atom, we get one lead atom. Thus, the change in the number of atoms of uranium atoms is also the production of lead atoms through the decay of uranium. Similarly, the decay of potassium only to argon through the beta decay process also marks that one potassium atom decay gives rise to an argon atom. Thus, using this concept in the mass of potassium, we can get the required value.

Formulae:

The disintegration constant,$\lambda =\frac{\ln 2}{T_{1/2}}$ (1)

where,$T_{1/2}$is the half-life of the substance.

The undecayed sample remaining after a given time,$N=N_0{e}^{-\lambda t}$ (2)

The number of undecayed atoms in a given mass of a substance,

$N=\frac{m}{A}{N}_A,where{N}_A=6.022\times {10}^{23}atom/mol$ (3)

where A is the molar mass of the substance.

At first, we can get the age of the rock using masses of${}^{238}U$and${}^{206}Pb$ . As the initial mass of the uranium can be given as:$m_{U0}=m_u+\frac{238}{206}{m}_{Pb}$

Thus, using equation (1) in equation (2) and the above value, we can get the age of the rock as follows:

$$\begin{gathered} t=\frac{T_{1/2}}{\ln 2}\ln \left(\frac{m_u+{\displaystyle \frac{238}{206}}{m}_{Pb}}{m_u}\right)\left(fromequation\left(3\right),N\propto m\right) \\ =\frac{4.47\times {10}^{9}y}{\ln 2}\ln \left[1+\left(\frac{238}{206}\right)\left(\frac{0.15mg}{0.86mg}\right)\right] \\ =1.18\times {10}^{9}y \end{gathered}$$

Now, for the $\beta -$decay of ${}^{40}K$, the initial mass of ${}^{40}K$(using the same concept of uranium) is given by, (as given in the problem that${}^{40}K$ decays only to${}^{40}Ar$ )

$$\begin{gathered} m_{K0}=m_K+\frac{40}{40}{m}_{Ar} \\ =m_K+m_{Ar} \end{gathered}$$

So using equations (1) and (2), the mass of${}^{40}K$in the rock can be given as:

$$\begin{gathered} m_K=m_{K0}{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}\left(fromequation\left(3\right)N\propto m\right) \\ {m}_K=\left(m_K+m_{Ar}\right)e^{-\left(\frac{\ln 2}{T_{1/2}}\right)t} \\ {m}_K=\left(m_K{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}+m_{Ar}{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}\right) \\ {m}_K\left(1-e^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}\right)=m_{Ar}{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t} \\ {m}_K=\frac{m_{Ar}{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}}{\left(1-e^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}\right)} \\ {m}_K=\frac{m_{Ar}}{\left(1-e^{-\left(\frac{\ln 2}{T_{1/2}}\right)t}-1\right)} \\ {m}_K=\frac{1.6mg}{e^{\left({\displaystyle \frac{\ln 2}{1.25\times {10}^{9}y}}\right)1.18\times {10}^{9}Y}-1} \\ {m}_K=1.7mg \end{gathered}$$

Hence, the amount of is ${}^{40}K$in 1.7 mg the rock.