the rate at which a radioactive isotope disntegrates is porportional to the amount present. if 30g sample will contain only 20g after 10 minutes, what is the half life of the isotope, correct to the nearest 10th. y=yoe^kt. :)

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the rate at which a radioactive isotope disntegrates is porportional to the amount present. if 30g sample will contain only 20g after 10 minutes, what is the half life of the isotope, correct to the nearest 10th. y=yoe^kt. :)

Mathematics
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your formula is \[y=y_0e^{kt}\]
\[y_0=30\] and you know that when t=10, y = 20
so set \[20=30e^{10k}\]

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and solve for k: \[20=30e^{10k}\] \[\frac{2}{3}=e^{10k}\] \[ln(\frac{2}{3})=10k\] \[k=\frac{ln(\frac{2}{3})}{10}\]
about -4.055
thank you thank youi thank you
so now you know your formula is \[y=30e^{-4.055t}\]and you want to know when you have half of what you started with. you started with 30 and half of that is 15 so if you want you can write \[15=30e^{-4.055t}\] or just cut to the chase and write \[\frac{1}{2}=e^{-4.055t}\]
and now solve for t: \[ln(.5)=-4.055t\] \[t=\frac{ln(.5)}{-4.055}=\]
wait wait wait
big mistake on my part
\[\frac{ln(\frac{2}{3}}{10}=-.04055\]
so wrong from there on in. it is \[\frac{ln(.5)}{-.04055}=15.32\]\]
very sorry i was off by two decimal places
thats fine :)
hope you are still here to see my correction
oh good.

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