Question 117972
A certain radioactive isotope has a half-life of approximately 1900 years. How many years to the nearest year would be required for a given amount of this isotope to decay to 80% of that amount.
;
The half-life equation that I am familiar with:
A = Ao(2^(-t/h))
where:
A = resulting amt
Ao = initial amt
t = time
h = half life of the substance
:
Let our Ao = 1 and A = .8
h = given as 1900 yrs
find t
:
1(2^(-t/1900)) = .8
Find the nat log of both sides
ln(2^-t/1900)) = ln(.8)
:
Log equivalent of exponents
(-t/1900)*ln(2) = ln(.8)
:
(-t/1900)*.693 = -.223
{{{-t/1900}}} = {{{-.223/.693}}}
:
{{{-t/1900}}} = -.322
:
Multiply both sides by -1900
t = -1900 * -.322
:
t = 611.8 years
:
You can check this on your calc: enter: 2^(-611.8/1900) = .79996 = .8 or 80%
:
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