Question

The half life for radioactive decay of C – 14 is 5730 years an archaeological artifact containing wood had only 80% of the C-14 found in a living tree. The age of the sample is

• A. 1485 years
• B. 1845 years
• C. 530 years
• D. 4767 years

Answer 1

Answer: (B)

1845 years

Answer 2

Radioactive decay follows first order kinetics therefore,

Decay constant $(\lambda) = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730}$

Given, $R_{0} = 100\therefore R = 80$

And $t = \frac{2.303}{\lambda}\log\frac{\lbrack R\rbrack_{0}}{\lbrack R\rbrack} = \frac{2.303}{\left( \frac{0.693}{5730} \right)}\log\frac{100}{80}$

$= \frac{2.303 \times 5730}{0.693} \times 0.0969 = 1845years$