Question #2720a

Question

1295 views around the world

You can reuse this answer
Creative Commons License

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#