A radioactive substance used in nuclear weapons decays at the rate of 2.8% per year. What would be the half-life of the radioactive substance.

Answer 1

Hi Daniel.

Most radioactive decay problems take two "passes". The first pass is to find "k" and the second pass is to actually answer the question.

The formula for radioactive decay is:

y = Ae^kt where y is the amount you have now, A is the amount you started with, k is a constant, and t is time.

Our first pass we know in 1 year (t), we started with A, and after a year we have 2.8% less or A(1-.028)

If that is confusing, just assume you start with 100, and find 2.8% of 100, and subtract it from 100 to find how much you have now. In the formula, I am assuming we start with 100 units.

Putting these into the formula:

97.2 = 100 e^k(1)

0.972 = e^k

ln(0.972) = k = -0.028399

Now that we know k, we start over for the second pass, but now we have a value of k to put into our formula. Looking for a half-life... If we start with 100, at the end of the half life we will have 50

50 = 100 e^{(-0.028399)(t) <---- this "t" is the time it takes for half of it to decay = HALF LIFE}

0.5 = e^(-0.028399)t

ln(0.5) = (-0.028399)t

-0.69314718 = -0.028399 t

Divide both sides by -0.028399 and find that t = 24.41 years