The half-life of Molybdenum-93 is 4000 years. A sample of Molybdenum-93 has a mass of 10 mg. When will the mass be reduced to 1 mg?

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###### Solution.

The mass of a radioactive element decays at a rate proportional to the mass.

If m(t) is the mass of the radioactive element at time t then

dm/dt = -λm (1)

where λ is a decay constant. The minus sign means that the amount of the radioactive material decreases over time.

The only solutions of the differential equation (1) are the exponential functions

m(t) = m(0)e^{-λt} (2)

We know that half-life is the time it takes for half a given mass of an element to decay. In order to determine the value of λ, we use the fact that

m(4000) = 5

Thus,

5 = 10e^{-λ4000}

or

1/2 = e^{-λ4000}

Now we take the natural logarithm of both sides

ln (1/2) = -λ4000

or

– ln2 = -λ4000

λ = ln2 /4000.

Therefore, we can rewrite the equation (2) as

m(t) = m(0)e^{-t*(ln2)/4000}

We want to find the value of time such that m(t) = 1, so

1 = 10e^{-t*(ln2)/4000}

or

0.1 = e^{-t*(ln2)/4000}

We solve this equation for t by taking the natural logarithm of both sides

ln0.1 = -t*(ln2)/4000

or

t = -4000*ln0.1/ln2 ≈ 13287.7 years