Asked by Daisy
A radioactive isotope sits unused in the lab for 10 years at which it decays to 80% of original amount. Find the half life and the addition years it takes to reduce to 15%of original amount
Answers
Answered by
oobleck
you want a function like
y = a(1/2)^(t/k)
where a is the original amount and the half-life is k years. You have
a(1/2)^(10/k) = 0.8a
(1/2)^(10/k) = 0.8
10/k ln0.5 = ln0.8
k = 10ln0.5/ln0.8 = 31.06
so now you want to solve
(1/2)^(t/31.06) = 0.15
and just finish it off (expect t to be about 2.5 half-lives, since 1/8 < 0.15 < 1/4)
y = a(1/2)^(t/k)
where a is the original amount and the half-life is k years. You have
a(1/2)^(10/k) = 0.8a
(1/2)^(10/k) = 0.8
10/k ln0.5 = ln0.8
k = 10ln0.5/ln0.8 = 31.06
so now you want to solve
(1/2)^(t/31.06) = 0.15
and just finish it off (expect t to be about 2.5 half-lives, since 1/8 < 0.15 < 1/4)
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