# Differential Equations

## First Order Equations 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:

$\frac{{dN}}{{dt}} = - \lambda N,$

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

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

$N\left( t \right) = C{e^{ - \lambda t}}.$

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

$N\left( t \right) = {N_0}{e^{ - \lambda t}}.$

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 $$T:$$

$N\left( T \right) = \frac{{{N_0}}}{2} = {N_0}{e^{ - \lambda T}}.$

The formula for the half life follows from here:

${e^{ - \lambda T}} = \frac{1}{2},\;\; \Rightarrow - \lambda T = \ln \frac{1}{2} = - \ln 2,\;\; \Rightarrow T = \frac{1}{\lambda }\ln 2.$

The average lifetime $$\tau$$ of a radioactive atom is given by

$\tau = \frac{1}{\lambda }.$

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

$T = \tau \ln 2 \approx 0.693\,\tau$

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

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### Example 1

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

### Example 2

The initial mass of an Iodine isotope was $$200\,\text{g}.$$ 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 $$80\,\text{g}.$$

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 $${\left( {\frac{1}{2}} \right)^3} = {\frac{1}{8}}$$ of the initial amount. Hence, the mass after decay is $$80\,\text{g}\cdot {\frac{1}{8}} = 10\,\text{g}.$$

### Example 2.

The initial mass of an Iodine isotope was $$200\,\text{g}.$$ 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:

$N\left( t \right) = {N_0}{e^{ - \lambda t}}.$

Here the decay constant $$\lambda$$ is equal to

$\lambda = \frac{{\ln 2}}{T},\;\;\text{where}\;\; T = 8\;\text{days}.$

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

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

See more problems on Page 2.