2.25 C^{14} atoms dissociate from 1 g of car | vedclass

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(B) The radioactive decay follows first-order kinetics. The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ 	ext{yea

Vedclass · Accepted Answer

$15800 \ 	ext{years}$

Vedclass · Answer

(B) The radioactive decay follows first-order kinetics. The decay constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ 	ext{year}^{-1}$.Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \left( \frac{N_0}{N_t} ight)$.Here,$N_0 = 15.3$ and $N_t = 2.25$.$t = \frac{2.303 	imes 5730}{0.693} \log \left( \frac{15.3}{2.25} ight)$.$t = 19042.7 	imes \log(6.8)$.$t = 19042.7 	imes 0.8325 \approx 15853 \ 	ext{years}$.Rounding to the nearest significant value,the age is approximately $1.58 	imes 10^{4} \ 	ext{years}$.

$2.25$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon in an old statue. $15.3$ $C^{14}$ atoms dissociate from carbon in a living statue. How many years old is the statue? (Half-life of $C^{14} = 5730 \ \text{years}$).

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