Question

The half-life of ${}_6{C}^{14}$ is 5730 year. What fraction of it's original $C^{14}$ would left year 22920 year of storage?

• A. 0.5
• B. 0.25
• C. 0.125
• D. 0.0625

Answer 1

The correct option is D 0.0625

$t_{1/2}=\frac{0.693}{k}$

$k=1.21\times {10}^{-4}{yrs}^{-1}$

$k=\frac{2.303}{t}log\frac{1}{1-x}$

$1-x=\frac{1}{16}=0.0625$