How do I solve this half-life/Exponential decay problem
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How do I solve this half-life/Exponential decay problem

[From: ] [author: ] [Date: 11-05-04] [Hit: ]
Call the initial amount (M(0)) 1, then when t = 693 years, the remaining amount (M(693)) = 0.0.5 = 1.ln(0.......
An isotope has a half-life of 693 years. Determine the annual decay rate, r. There is more to the problem, but I just need to know how to find r.

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The proportion of the active constituent diminishes to ½ of its initial amount after one half-life period.

The formula is of the form

M(t) = M(0)e^(-kt)

Call the initial amount (M(0)) 1, then when t = 693 years, the remaining amount (M(693)) = 0.5

Therefore

0.5 = 1.e^(-693k)

ln(0.5) = - 693k

k = ln(0.5) / -693

. .= -0.69315 / -693 = 0.00100

which is the annual decay rate.
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