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May 20th, 2014, 05:58 PM   #1
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logarithms

There are two primary equations used when dealing with standard radioactive decay.

1.) N = N0e^-kt

2.) K = ln2/r

N represents the final amount remaining
N0 represents the initial amount
e is the exponential constant
k is the constant exponent for that sample
t represents the amount of time that passed
r is the half-life of that sample
ln stands for natural logarithm

the half-life of radium-226 is 1600 years. Suppose you have a 22mg sample. After how long will only 18mg of the sample remain?

Last edited by skipjack; May 21st, 2014 at 01:21 PM.
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May 20th, 2014, 11:34 PM   #2
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If I understood it correctly

$\displaystyle n=n_0 \cdot e^{-kt } \\ k= \frac{\ln 2}{r}$

Therefore

$\displaystyle k= \frac{\ln 2}{1600} \\ 18=22 \cdot e^{-kt} \\ -kt =\ln {(9/11)}$

$\displaystyle t=-1600 \cdot \log_{2} (9/11)$
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May 21st, 2014, 01:25 PM   #3
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I can't be certain which is expected, but a value obtained by use of a calculator would often be acceptable for this type of question.
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May 21st, 2014, 05:52 PM   #4
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I think you are supposed to understand how the answer is arrived at rather than simply typing some stuff into a calculator. It's possible to train monkeys to push buttons, after all.
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May 22nd, 2014, 01:11 AM   #5
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$\displaystyle \log_2 (9/11) \approx -0.289475$, therefor $\displaystyle t = -1600 \cdot (-0.289475) = 463.16$
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May 22nd, 2014, 06:59 AM   #6
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Quote:
Originally Posted by ron246 View Post
oh no...i can use a calculator, i was just asking if that last line was the official answer or do i need to type that into the calcuator and press =
Do you have a calculator which calculates log base 2???
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