Question

Element X decays radioactivity with a half-life in 10 minutes if there are 660 g of element X how long to the nearest 10th of a minute would it take for the element to decay to 58 g

Answer 1

`\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( (1)/(2) \right)^{(t)/(h)}\qquad \begin{cases} A=\textit{current amount}\dotfill & 58\\ P=\textit{initial amount}\dotfill &660\\ t=minutes\dotfill &t\\ h=\textit{half-life}\dotfill &10 \end{cases} \\\\\\ 58 = 660\left( \cfrac{1}{2} \right)^{(t)/(10)} \implies \cfrac{58}{660}=\left( \cfrac{1}{2} \right)^{(t)/(10)}\implies \cfrac{29}{330}=\left( \cfrac{1}{2} \right)^{(1)/(10)\cdot t}`

`\log\left( \cfrac{29}{330} \right)=\log\left[ \left( \cfrac{1}{2} \right)^{(1)/(10)\cdot t} \right]\implies \log\left( \cfrac{29}{330} \right)=t\log\left[ \left( \cfrac{1}{2} \right)^{(1)/(10)} \right] \\\\\\ \cfrac{ ~~ \log\left( (29)/(330) \right) ~~ }{\log\left[ \left( (1)/(2) \right)^{(1)/(10)} \right]}=t\implies 35.1\approx t`

Answer 2

Explanation:

58 g = 660 g * ( 1/2)^n

58 / 660 = (1/2) ^n LOG both sides

-1.056 = n log 1/2

n = 3.508 half lives

3. 508 x 10 min = ~ 35.1 min