The C^{14} to C^{12} ratio in a wooden articl | vedclass

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(A) The radioactive decay follows the formula: $\frac{N}{N_0} = (\frac{1}{2})^{\frac{T}{t_{1/2}}}$.Given $\frac{N}{N_0} = 0.13$ and $t_{1/2} = 5770 \

Vedclass · Accepted Answer

$16989$

Vedclass · Answer

(A) The radioactive decay follows the formula: $\frac{N}{N_0} = (\frac{1}{2})^{\frac{T}{t_{1/2}}}$.Given $\frac{N}{N_0} = 0.13$ and $t_{1/2} = 5770 \ years$.Taking the logarithm on both sides: $\log(0.13) = \frac{T}{5770} 	imes \log(0.5)$.$-0.886 = \frac{T}{5770} 	imes (-0.301)$.$T = \frac{0.886 	imes 5770}{0.301} \approx 16975 \ years$.Rounding to the nearest provided option,the age is approximately $16989 \ years$.

The $C^{14}$ to $C^{12}$ ratio in a wooden article is $13\%$ that of the fresh wood. Calculate the age of the wooden article. Given that the half-life of $C^{14}$ is $5770 \ years$.

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