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 y e a r s

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1845 y e a r s

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530 y e a r s

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4767 y e a r s

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Solution

The correct option is B $1845 y e a r s$
Radioactive decay follows first order kinetics.
Therefore,
Decay constant $(λ) = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5730}$
Given, $[R_{0}] = 100 ∴ [R] = 80$
and $t = \frac{2.303}{λ} log \frac{[R]_{0}}{[R]} = \frac{2.303}{(\frac{0.693}{5730})} log \frac{100}{80}$
$= \frac{2.303 \times 5730}{0.693} \times 0.0969 = 1845 y e a r s$ .

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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 ?

The correct option is B 1845 yearsRadioactive decay follows first order kinetics.Therefore,Decay constant (λ)=0.693t1/2=0.6935730Given, [R0]=100∴[R]=80and t=2.303λlog[R]0[R]=2.303(0.6935730)log10080=2.303×57300.693×0.0969=1845 years.

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 ?