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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    your formula is \[y=y_0e^{kt}\]

  2. anonymous
    • 5 years ago
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    \[y_0=30\] and you know that when t=10, y = 20

  3. anonymous
    • 5 years ago
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    so set \[20=30e^{10k}\]

  4. anonymous
    • 5 years ago
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    and solve for k: \[20=30e^{10k}\] \[\frac{2}{3}=e^{10k}\] \[ln(\frac{2}{3})=10k\] \[k=\frac{ln(\frac{2}{3})}{10}\]

  5. anonymous
    • 5 years ago
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    about -4.055

  6. anonymous
    • 5 years ago
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    thank you thank youi thank you

  7. anonymous
    • 5 years ago
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    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 \[\frac{1}{2}=e^{-4.055t}\]

  8. anonymous
    • 5 years ago
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    and now solve for t: \[ln(.5)=-4.055t\] \[t=\frac{ln(.5)}{-4.055}=\]

  9. anonymous
    • 5 years ago
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    wait wait wait

  10. anonymous
    • 5 years ago
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    big mistake on my part

  11. anonymous
    • 5 years ago
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    \[\frac{ln(\frac{2}{3}}{10}=-.04055\]

  12. anonymous
    • 5 years ago
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    so wrong from there on in. it is \[\frac{ln(.5)}{-.04055}=15.32\]\]

  13. anonymous
    • 5 years ago
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    very sorry i was off by two decimal places

  14. anonymous
    • 5 years ago
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    thats fine :)

  15. anonymous
    • 5 years ago
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    hope you are still here to see my correction

  16. anonymous
    • 5 years ago
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    oh good.

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