anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
your formula is \[y=y_0e^{kt}\]
anonymous
  • anonymous
\[y_0=30\] and you know that when t=10, y = 20
anonymous
  • anonymous
so set \[20=30e^{10k}\]

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anonymous
  • anonymous
and solve for k: \[20=30e^{10k}\] \[\frac{2}{3}=e^{10k}\] \[ln(\frac{2}{3})=10k\] \[k=\frac{ln(\frac{2}{3})}{10}\]
anonymous
  • anonymous
about -4.055
anonymous
  • anonymous
thank you thank youi thank you
anonymous
  • anonymous
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 \[\frac{1}{2}=e^{-4.055t}\]
anonymous
  • anonymous
and now solve for t: \[ln(.5)=-4.055t\] \[t=\frac{ln(.5)}{-4.055}=\]
anonymous
  • anonymous
wait wait wait
anonymous
  • anonymous
big mistake on my part
anonymous
  • anonymous
\[\frac{ln(\frac{2}{3}}{10}=-.04055\]
anonymous
  • anonymous
so wrong from there on in. it is \[\frac{ln(.5)}{-.04055}=15.32\]\]
anonymous
  • anonymous
very sorry i was off by two decimal places
anonymous
  • anonymous
thats fine :)
anonymous
  • anonymous
hope you are still here to see my correction
anonymous
  • anonymous
oh good.

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