Question

A bone fragment found in a cave contains 0.2 times as much ${}_6^{14}$C as an equal amount of carbon in the air when the organism containing bone died.The approximate age of fragment is (t$\frac{1}{2}$ of ${}_6^{14}$C = 5730 years )

• A. 1.15 x10${}^4$ y
• B. 1.3 x10${}^4$ y
• C. 1.24 x10${}^4$ y
• D. 1.4 x10${}^4$ y

Answer 1

Answer: (B)

1.3 x10${}^4$ y

$t_{1/2}=\frac{0.693}{\lambda }$ ; where, $\lambda =$ decay constant.

$\lambda =\frac{0.693}{5730}{y}^{-1}$

$N=N_0{e}^{-\lambda t}$

$ln\left(\frac{N}{N_0}\right)=-\lambda t$

$t=\frac{1}{\lambda }ln\left(\frac{N_0}{N}\right)=\frac{5730}{0.693}ln\left(\frac{N_0}{0.2N_0}\right)=1.3\times {10}^4$ y