Question #2720a

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Answer 1 · 2017-03-01T23:57:14

Decay constant: $lambda approx 2.3 times 10^(-4)$

If the half-life is 3000 years, we can say that the amount of activity follows this rule:

$A(t) = A_o (1/2)^(t/3000)$

So for $t = 3000$ we have $A(3000) = A_o (1/2)^(1) = A_o/2$

This is the most literal formulation of the idea of a half-life .

The decay constant is slightly more abstract, it requires us to move to using natural logs:

$A(t) = A_o (1/2)^(t/3000) = A_o (2)^(color(red)(-) t/3000) = color(blue)(A_o e^(- lambda t))$ , where $lambda$ is the decay constant.

So we have now calculus-friendly exponential decay.

We can then say that:

$2^(-t/3000) =e^(- lambda t)$

Or:

$ln 2^(-t/3000) = ln e^(- lambda t)$

$implies -t/3000 ln (2 ) = - lambda t$

$implies lambda = (ln (2 ))/(3000 ) approx 2.3 times 10^(-4)$

So we test this again using our new definition:

$A(3000) = A_o e^(-2.3 times 10^(-4) * 3000) = 0.501 approx 1/2$