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anonymous

  • 5 years ago

What is the half life of a radioative substance that decays by 6% in 7 years?

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  1. anonymous
    • 5 years ago
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    half life is the period of time it takes to decay 50% of its original value. so what ration of 50% is 6% and apply that ratio to 7

  2. anonymous
    • 5 years ago
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    ummmm 50/6?

  3. anonymous
    • 5 years ago
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    So I make it about 58.3333

  4. anonymous
    • 5 years ago
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    yep 50/6 * 7

  5. anonymous
    • 5 years ago
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    Ohh! Thanks!

  6. anonymous
    • 5 years ago
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    hmm that is not what i got. let me try again.

  7. anonymous
    • 5 years ago
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    i got 78.4

  8. anonymous
    • 5 years ago
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    i think this problem is a little more complicated that ratios. the general formula for decay is \[A=A_0e^{rt}\] where r is the rate of decay (r will be negative) t is time, \[A_0\] is the initial amount and \[A\] is what you have after t years. we do not know r, but we know that when t = 7 you have decreased by 6%. decreasing by 6% means you have 94% =.94 of the original amount so set \[.94=e^{7r}\] and solve for r \[7r = ln(.94)\] \[r=\frac{ln.94}{7}=-.008839\] rounded now half life means you have half the original amount so set \[\frac{1}{2}=e^{-.00839t}\] or better still (to avoid rounding error) \[\frac{1}{2}=e^{\frac{ln(.94)}{7}t}\] and solve for t \[ln(\frac{1}{2})=\frac{ln(.94)t}{7}\] \[t=\frac{7ln(\frac{1}{2})}{ln(.94)}=78.4\] rounded

  9. anonymous
    • 5 years ago
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    i could be wrong of course, but i am fairly sure this is correct.

  10. anonymous
    • 5 years ago
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    ah yes what was I thinking. but isn't decay \[e^{-rt}\] and not \[e^{rt}\] because that's growth not decay so taking the log of both sides should give \[-7r=\ln(0.94)\] the rest follows?

  11. anonymous
    • 5 years ago
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    I agree with the answer given. here is a doc to clarify the above comment. sorry for the misunderstanding

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  12. anonymous
    • 5 years ago
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    you can write \[A=A_0e^{-rt}\] it is just a matter of preference. my 'r' was negative. if you write it your way r is positive. it makes no difference.

  13. anonymous
    • 5 years ago
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    there is another way to solve this problem that amounts to the same thing. since you know it decreases by 6% in 7 years you could write the formula is \[A=A_0(.94)^{\frac{t}{7}}\] then set \[\frac{1}{2}=(.94)^{\frac{t}{7}}\] and solve for t, via \[ln(\frac{1}{2})=ln((.94)^{\frac{t}{7}})=\frac{t}{7}ln(.94)\] and still get \[t=\frac{7ln(\frac{1}{2})}{ln(.94)}\]

  14. anonymous
    • 5 years ago
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    Nice :D

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