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Sagot :
Answer:
332.19 years
Explanation:
The weight, W of the substance after n years is given by:
[tex]W=W_o\mleft(\frac{1}{2}\mright)^{\frac{n}{100}}[/tex]Let the initial weight = 100%
If the substance loses to 10% of its initial weight, then:
• Wo = 100%
,• W= 10%
Substitute these into the formula:
[tex]\begin{gathered} \frac{10}{100}=\frac{100}{100}\mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \\ \implies0.1=\mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \end{gathered}[/tex]We then solve the equation for the value of n.
Take the logarithm of both sides.
[tex]\begin{gathered} \log (0.1)=\log \mleft(\frac{1}{2}\mright)^{\frac{n}{100}} \\ \implies\log (0.1)=\frac{n}{100}\log (\frac{1}{2})^{} \end{gathered}[/tex]Then divide both sides by log(1/2):
[tex]\begin{gathered} \frac{\log (0.1)}{\log (\frac{1}{2})}=\frac{\frac{n}{100}\log(\frac{1}{2})^{}}{\log(\frac{1}{2})} \\ \implies\frac{n}{100}=\frac{\log (0.1)}{\log (\frac{1}{2})} \end{gathered}[/tex]Finally, multiply both sides by 100:
[tex]\begin{gathered} 100\times\frac{n}{100}=100\times\frac{\log (0.1)}{\log (\frac{1}{2})} \\ n=332.19\text{ years} \end{gathered}[/tex]It will take at least 332.19 years for the radioactive substance to lose to 10% of its initial weight.
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