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The half-life for decay of ${ }^{14} \mathrm{~C, by~} \beta$ -emission is $5730 \mathrm{yr}$. The fraction of ${ }^{14} \mathrm{C}$ decays, in a sample that is 22920 yr old, would be
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The correct answer is:
$\frac{15}{16}$
$n=\frac{\text { total time }(t)}{\text { half -lite }(t / 2)}=\frac{22920}{5730}=4$
$$
\begin{aligned}
\text { Left amount, } N=& N_{0}\left(\frac{1}{2}\right)^{n} \\
&=N_{0}\left(\frac{1}{2}\right)^{4}=\frac{N_{0}}{16} \\
\therefore \text { Decayed fraction } &=N_{0}-\frac{N_{0}}{16} \\
&=\frac{16 N_{0}-N_{0}}{16}=\frac{15 N_{0}}{16}
\end{aligned}
$$
$$
\begin{aligned}
\text { Left amount, } N=& N_{0}\left(\frac{1}{2}\right)^{n} \\
&=N_{0}\left(\frac{1}{2}\right)^{4}=\frac{N_{0}}{16} \\
\therefore \text { Decayed fraction } &=N_{0}-\frac{N_{0}}{16} \\
&=\frac{16 N_{0}-N_{0}}{16}=\frac{15 N_{0}}{16}
\end{aligned}
$$
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