# 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

Click 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 $$80\,\text{g}.$$

### 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 3

The radioactive isotope Indium-$$111$$ is often used for diagnosis and imaging in nuclear medicine. Its half life is $$2.8$$ days. What was the initial mass of the isotope before decay, if the mass in $$2$$ weeks was $$5\,\text{g}?$$

### Example 4

Find the half life of a radioactive element, if its activity decreases for $$1$$ month by $$10\%.$$

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

### 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

${\lambda = \frac{{\ln 2}}{T},\;\;\text{where}\;\;}\kern-0.3pt{T = 8\;\text{days}.}$

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

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

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Problems 1-2
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Problems 3-4