91Ó°ÊÓ

Half-life of ${}^{40}\mathrm{Ca}$and ${}^{40}Ar$,$T_{1/2}=1.26×{10}^{9}y$

Ratio of Ca to Ar is$N_{ca}/N_k=8.54/1$.

Ratio of K to Ar is $N_{Ar}/N_k=1$.

The concept of radioactive dating is used to know the age of the fossil using the given decay rates. Here, the problem is of a parent nucleus that can decay into two products. Being the products of the decay, the number of atoms of calcium and argon were initially potassium atoms before the decay. Thus, the total number of initial potassium atoms can be given as the sum of all the three atoms present in the rock after the decay at the given time.

Formulae:

The disintegration constant, $\mathrm{λ}=\frac{\ln 2}{T_{1/2}}…....\left(1\right)$

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

The undecayed sample remaining after a given time, $N=N_0{e}^{-\mathrm{λ}t}............\left(2\right)$

We note that every calcium-40 atom and krypton-40 atom found now in the sample was once one of the original numbers of potassium atoms.

Thus, the total number of initial potassium-40 atoms can be given as:

$N_{ko}=N_k+N_{Ar}+N_{ca}$

Thus, substituting equation (1) in (2) with the above data, we can get the age of the sample as follows:

$$\begin{gathered} \ln \left(\frac{N_k}{N_{ko}}\right)=-\left(\frac{\ln 2}{T_{1/2}}\right)t \\ \ln \left(\frac{N_k}{N_k+N_{Ar}+N_{ca}}\right)=-\left(\frac{\ln 2}{1.26×{10}^{9}y}\right)t \\ \ln \left(\frac{1}{1+1+8.54}\right)=-\left(0.55501×{10}^{-9}/y\right)t \\ \ln \frac{1}{10.54}=\left(0.55501×{10}^{-9}/y\right)t \\ t=\frac{2.3552}{0.55501×{10}^{-9}/yt} \\ =4.24×{10}^{9}y \end{gathered}$$

Hence, the age of the sample is $4.24×{10}^{9}y$.