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 :-

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

Tardigrade Admin · Accepted Answer

Radioactive decay follow first order kinetics. Therefore,

= \frac{0.693}{t _{1/2}} = \frac{0.693}{5730}

Given,

R_{o} = 100 ∵ Δ R = 80

and

t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ]} = \frac{2.303}{[ \frac{0.693}{5730} ]} lo g \frac{100}{80}

t = \frac{2.303 \times 5730}{0.693} \times 0.0969 = 1845

years

Q. 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 :-

1485 years

1845 years

530 years

4767 years

Radioactive decay follow first order kinetics. Therefore, =t1/2​0.693​=57300.693​ Given, Ro​=100∵ΔR=80 and t=k2.303​log[R][R]0​​=[57300.693​]2.303​log80100​ t=0.6932.303×5730​×0.0969=1845 years

Q. 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 :-