The T_{1/2} of C^{14} isotope is 5770 years. | vedclass

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Question

(A) The decay constant $K$ is given by $K = \frac{0.693}{T_{1/2}} = \frac{0.693}{5770} 	ext{ year}^{-1}$.Using the first-order kinetics equation: $t

Vedclass · Accepted Answer

$2740$

Vedclass · Answer

(A) The decay constant $K$ is given by $K = \frac{0.693}{T_{1/2}} = \frac{0.693}{5770} 	ext{ year}^{-1}$.Using the first-order kinetics equation: $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]_t}$.Substituting the values: $t = \frac{2.303 	imes 5770}{0.693} \log \frac{100}{72}$.$t = 19175.05 	imes (\log 100 - \log 72)$.$t = 19175.05 	imes (2 - 1.8573) = 19175.05 	imes 0.1427 \approx 2736.3 	ext{ years}$.The closest option is $2740$ years.

The $T_{1/2}$ of $C^{14}$ isotope is $5770$ years. The time after which $72\%$ of the isotope is left is .......... years.

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