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anonymous

  • 5 years ago

a certain radioactive substance decays from 35,490 gm to 650 gm in 5 days. What is its half life? .693 days .866 days 1.600 days .690 days

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  1. anonymous
    • 5 years ago
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    equation will be \[Q=35490e^{rt}\] where t is time in days. you know that when \[t=5\] \[Q=650\] so solve for r via \[650=35490e^{5r}\] \[\frac{650}{35490}=e^{5r}\] \[ln(\frac{650}{35490}=5t\] \[t=\frac{ln(\frac{650}{35490})}{5}\]

  2. anonymous
    • 5 years ago
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    or roughtly -.8. so you want to know when \[e^{-.8t}=\frac{1}{2}\] and now solve for t: \[-.8t=ln(\frac{1}{2})\] \[t=\frac{ln(.5)}{-.8}\].

  3. anonymous
    • 5 years ago
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    about .866 days from the calculator.

  4. anonymous
    • 5 years ago
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    Wow thank you. The first equation was a little easier to understand.

  5. anonymous
    • 5 years ago
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    first one?

  6. anonymous
    • 5 years ago
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    perhaps the "half life" one was confusing. formula is always \[Q(t)=Q_0e^{rt}\] and "half life" means the time it takes to get half the original amount. so if you start with \[Q_0\] then half of it is \[\frac{1}{2}Q_0\] and you would solve \[\frac{1}{2}Q_0=Q_0e^{rt}\] for t step number one is do divide both sides by \[Q_0\] so you may as well start with \[\frac{1}{2}=e^{rt}\]

  7. anonymous
    • 5 years ago
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    or you can simply remember if the rate of decay is r, then half life is \[\frac{ln(.5)}{-r}\] here i am assuming r is written as a decimal and also as positive, so say, for example, your substance decays at a rate of 3% per hour then r =.03

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