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$200 \mathrm{mg}$ of ${ }^{14} \mathrm{C}$ will become $25 \mathrm{mg}$ in how many years if the half-life is 5760 years?
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17280 years
Half life for radio active ${ }^{19} \mathrm{C}$ is 5760 years.
It is a first order reaction.
$\begin{aligned} & t \frac{1}{2}=\frac{0.693}{k} \\ & \begin{aligned} & k=\frac{0.693}{5760} \\ &=1.203 \times 10^{-4} \text { years }^{-1} \\ & k=\frac{2.303}{t} \log \frac{a}{a-x}\end{aligned} \\ & \begin{aligned} & a=\text { Initial concentration }=200 \mathrm{mg} \\ & a-x=\text { Concentration at time } t=25 \mathrm{mg}\end{aligned}\end{aligned}$
$\begin{aligned} t & =\frac{2.303}{k} \log \frac{200}{25} \\ & =\frac{2.303}{1.203 \times 10^{-4}} \log 8 \\ & =1.914 \times 10^4 \log (2)^3 \\ & =1.914 \times 3 \times 10^4 \times 0.301 \\ & =1.7280 \times 10^4 \\ & =17280 \text { years. }\end{aligned}$
It is a first order reaction.
$\begin{aligned} & t \frac{1}{2}=\frac{0.693}{k} \\ & \begin{aligned} & k=\frac{0.693}{5760} \\ &=1.203 \times 10^{-4} \text { years }^{-1} \\ & k=\frac{2.303}{t} \log \frac{a}{a-x}\end{aligned} \\ & \begin{aligned} & a=\text { Initial concentration }=200 \mathrm{mg} \\ & a-x=\text { Concentration at time } t=25 \mathrm{mg}\end{aligned}\end{aligned}$
$\begin{aligned} t & =\frac{2.303}{k} \log \frac{200}{25} \\ & =\frac{2.303}{1.203 \times 10^{-4}} \log 8 \\ & =1.914 \times 10^4 \log (2)^3 \\ & =1.914 \times 3 \times 10^4 \times 0.301 \\ & =1.7280 \times 10^4 \\ & =17280 \text { years. }\end{aligned}$
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