Radioactive Decay

Home → Radioactive Decay

In nature, there are a large number of atomic nuclei that can spontaneously emit elementary particles or nuclear fragments. Such a phenomenon is called radioactive decay. This effect was studied at the turn of $19 - 20$ centuries by Antoine Becquerel, Marie and Pierre Curie, Frederick Soddy, Ernest Rutherford, and other scientists. As a result of the experiments, F.Soddy and E.Rutherford derived the radioactive decay law, which is given by the differential equation:

\frac{d N}{d t} = - λ N,

where $N$ is the amount of a radioactive material, $λ$ is a positive constant depending on the radioactive substance. The minus sign in the right side means that the amount of the radioactive material $N (t)$ decreases over time (Figure $1$ ).

The given equation is easy to solve, and the solution has the form:

N (t) = C e^{- λ t} .

To determine the constant $C,$ it is necessary to indicate an initial value. If the amount of the material at the moment $t = 0$ was $N_{0},$ then the radioactive decay law is written as

N (t) = N_{0} e^{- λ t} .

Further, we introduce two useful parameters that follow from the given law.

The half life or half life period $T$ of a radioactive material is the time reguired to decay to one-half of the initial value of the material. Hence, at the moment $T :$

N (T) = \frac{N_{0}}{2} = N_{0} e^{- λ T} .

The formula for the half life follows from here:

e^{- λ T} = \frac{1}{2}, \Rightarrow - λ T = \ln \frac{1}{2} = - \ln 2, \Rightarrow T = \frac{1}{λ} \ln 2.

The average lifetime $τ$ of a radioactive atom is given by

τ = \frac{1}{λ} .

As it can be seen, the half life $T$ and the average lifetime $τ$ are related to each other by the formula:

T = τ \ln 2 \approx 0.693 τ

These $2$ parameters vary widely for different substances. For example, the half life of Polonium- $212$ is less than $1$ microseconds, but the half life of Thorium- $232$ is more than 1 billion years. A wide range of isotopes with different half lives was thrown from the atomic reactors and cooling pools in Chernobyl and Fukushima disasters (Figure $2$ ).

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the mass of a radioactive isotope if $3$ half lives occurred. The initial mass of the material was $80 g .$

Example 2

The initial mass of an Iodine isotope was $200 g .$ Determine the Iodine mass after $30$ days if the half life of the isotope is $8$ days.

Example 1.

Find the mass of a radioactive isotope if $3$ half lives occurred. The initial mass of the material was $80 g .$

Solution.

The mass of a radioactive material decreases as a result of decay twice after each half life. So, after $3$ half lives the quantity of the material will be ${(\frac{1}{2})}^{3} = \frac{1}{8}$ of the initial amount. Hence, the mass after decay is $80 g \cdot \frac{1}{8} = 10 g .$

Example 2.

The initial mass of an Iodine isotope was $200 g .$ Determine the Iodine mass after $30$ days if the half life of the isotope is $8$ days.

Solution.

According to the radioactive decay law the mass of an isotope depends on time as follows:

N (t) = N_{0} e^{- λ t} .

Here the decay constant $λ$ is equal to

λ = \frac{\ln 2}{T}, where T = 8 days .

Calculate the mass of the Iodine isotope in $30$ days:

N (t = 30) = 200 e^{- \frac{30 \ln 2}{8}} = 200 e^{- \frac{30 \cdot 0.693}{8}} \approx 200 e^{- 2.6} \approx 200 \times 0.074 = 14.9 g .