Problema Solution
a certain radioactive isotope is a by-product go some nuclear reactors. due to an explosion, a nuclear reactor experiences a massive leak of this radioactive isotope. fortunately, the isotope has a very short half-life of 10 days. estimate the percentage of the original amount of the isotope released by the explosion that remains 9 days after the explosion.
Answer provided by our tutors
Let A(0) denote the amount present at t = 0. Let t2 = 10 days denote the half-life.
Then we have: A(t2) = (1/2)A(0)
Using the formula A(t) = A(0) e^(kt) we have:
A(0) e^(kt2) = (1/2)A(0)
e^(10k) = 1/2
k = ln((1/2)^(1/10))
For t = 9 days we have:
A(9) = A(0) e^(9ln((1/2)^(1/10)))
A(9)/A(0) = e^(9ln((1/2)^(1/10)))
A(9)/A(0) = 1/(2^(9/10))
A(9)/A(0) = 0.5359 or 53.39%
There is 53.39% radioactive isotope remains 9 days after the explosion.