Question

What is the age in years of a bone in which the $\frac{{}^{14}C}{{}^{12}C}$ratio is measured to be$1.65\times {10}^{-13}$?

Short answer

The age of bone is$17,058,783$Years

Step-by-step solution

Given Calculations:

Ratio of ${}^{14}C$to ${}^{12}C$ in living organisms is,

localid="1649227741643" $\frac{N_0\left({}^{14}C\right)}{N_0\left({}^{12}C\right)}=1.3\times {10}^{-12}$

Ratio of ${}^{14}C$to ${}^{12}C$ in bone is,

localid="1649224990460" $\frac{N\left({}^{14}C\right)}{N\left({}^{12}C\right)}=1.65\times {10}^{-13}$

As, localid="1649226770511" $N_0\left({}^{12}C\right)=N_0\left({}^{12}C\right)$

localid="1649227753217" $\frac{N\left({}^{14}C\right)}{N_0\left({}^{14}C\right)}=\frac{1.65\times {10}^{-13}}{1.3\times {10}^{-12}}$

localid="1649226377005" $=1.27\times {10}^{-1}$

$=0.127$

Therefore, $N{(}^{14}C)=0.127N_0{(}^{14}C)$

Given Calculations:

Decay equation in terms of half-life is,

$N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{t_1}}$

${\left(\frac{1}{2}\right)}^{\frac{t}{t_1}}=\frac{N}{N_0}$

Here, $N$is the present number of ${}^{14}C$atoms, $N_0$represent the initial number of atoms of ${}^{14}C$atoms and localid="1649226999766" $t_{\frac{1}{2}}$ represents the half-life of${}^{14}C$ .
Substitute 5730 years for $t_{\frac{1}{2}}$and $0.127$for $\frac{N}{N_0}$,

localid="1649227250557" ${\left(\frac{1}{2}\right)}^{\frac{t}{5730Years}}=0.127$

localid="1649227227925" ${\left(0.5\right)}^{\frac{t}{5730Years}}=0.127$

Take natural log on both sides,

localid="1649227331822" $\frac{t}{5730Years}\ln \left(0.5\right)=\ln \left(0.127\right)$

$t=\frac{\ln \left(0.127\right)}{\ln \left(0.5\right)}\left(5730Years\right)$

$t=(2.9771)(5730years)$

$t=17,058,783$

Hence, the age of bone is localid="1649227458973" $17,058,783$Years.