[tex]\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &\stackrel{grams}{0.67}\\ t=\textit{elapsed time}\dotfill &\stackrel{years}{460}\\ h=\textit{half-life}\dotfill &\stackrel{years}{92} \end{cases} \\\\\\ A=0.67\left( \frac{1}{2} \right)^{\frac{460}{92}}\implies A=0.67\left( \frac{1}{2} \right)^5\implies A\approx 0.021~grams[/tex]
