91Ó°ÊÓ

Beta-minus decay: it is defined as the decay in which an energetic negative electron is emitted and produce a daughter nucleus of one higher atomic number and have the same

The expression of the number of undecided nuclei decrease exponentially is,

$N=N_o{\left(\frac{1}{2}\right)}^{{\frac{t}{t}}_{\left(\frac{t}{2}\right)}}$

Here, N is the radioactive nuclei, t is the time and$t_{\left(\frac{1}{2}\right)}$is the half-life period.

The age of the rock is

The ratio of $\frac{{}^{40}Ar}{{}^{40}K}$is $0.013$

Here, ${}^{40}Ar$is the number of argon atom in the sample and ${}^{40}K$is the number of potassium atom in the sample.

${}^{40}K_0={}^{40}Ar+{}^{40}K$

role="math" $\frac{{}^{40}Ar}{0.013}={}^{40}K$

Substitute $\frac{{}^{40}Ar}{0.013}$for ${}^{40}K$in the above expression of ${}^{40}K$

$$\begin{gathered} {}^{40}K_0={}^{40}Ar+\frac{{}^{40}Ar}{0.013} \\ =\left(\frac{1.013}{0.013}\right){}^{40}Ar \end{gathered}$$

Substitute $\left(\frac{1.013}{0.013}\right){}^{40}Ar$for $N_0$in the above expression of half time period.

$$\begin{gathered} {}^{40}Ar=\left(\frac{1.013}{0.013}\right){}^{40}Ar{\left(\frac{1}{2}\right)}^{{\frac{t}{t}}_{\frac{1}{2}}} \\ 1=\left(\frac{1.013}{0.013}\right){\left(\frac{1}{2}\right)}^{{\frac{t}{t}}_{\frac{1}{2}}} \\ \frac{1.013}{0.013}={\left(\frac{1}{2}\right)}^{{\frac{t}{t}}_{\frac{1}{2}}} \end{gathered}$$

Take natural log both side,

$$\begin{gathered} \ln \left(\frac{0.013}{1.013}\right)=\ln \left({\left(\frac{1}{2}\right)}^{{\frac{t}{t}}_{\frac{1}{2}}}\right) \\ -4.356={\frac{t}{t}}_{\frac{1}{2}}\ln \left(\frac{1}{2}\right) \\ -4.356={\frac{t}{t}}_{\frac{1}{2}}(-0.693) \\ t=\frac{4.356t_{{\displaystyle \frac{1}{2}}}}{0.693} \end{gathered}$$

Substitute $1.28$billion for $t_{\frac{1}{2}}$

$$\begin{gathered} t=\left(\frac{4.356}{0.693}\right)(1.28billion) \\ =\left(\frac{4.356}{0.693}\right)(1.28billionyear) \\ =8.04billionyear \end{gathered}$$

Therefore, The age of the rock is$t=8.04$billion years