If 24% of a certain radioisotope decays in 6.5 years, what is the half-life of this isotope

Question

Quizplus · Accepted Answer

If 24% of the isotope decays in 6.5 years, then 76% of the isotope remains after 6.5 years. 
Using the formula for exponential decay, we can determine the half-life of the isotope:
N(t) = N0*(1/2)^(t/T)
where N(t) is the amount remaining after time t, N0 is the initial amount, and T is the half-life.
If we let t = 6.5 years and N(t) = 0.76N0 (i.e. 76% remains), then we can solve for T:
0.76N0 = N0*(1/2)^(6.5/T)
Simplifying, we get:
(1/2)^(6.5/T) = 0.76
Taking the logarithm of both sides, we get:
(6.5/T)*log(1/2) = log(0.76)
Solving for T, we get:
T = 16 years
Therefore, the half-life of the isotope is 16 years.

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