Differential Equations

First Order Equations

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Radioactive Decay

Solved Problems

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 \(5\,\text{g}?\)

Solution.

Using the radioactive decay law, we can write:

\[N\left( t \right) = {N_0}{e^{ - \lambda t}},\;\;\text{where}\;\; \lambda = \frac{{\ln 2}}{T}.\]

Solve this equation for \({N_0}:\)

\[{N_0} = \frac{{N\left( t \right)}}{{{e^{ - \lambda t}}}} = N\left( t \right){e^{\lambda t}} = N\left( t \right){e^{\frac{{t\ln 2}}{T}}}.\]

Substituting the known values \(T = 2.8\) days, \(T = 14\) days, and \(N\left( {t = 14} \right) = 5\,\text{g},\) we have:

\[{N_0} = 5{e^{\frac{{14\ln 2}}{{2.8}}}} \approx 5{e^{3.47}} \approx 5 \cdot 32 = 160\,\text{g}.\]

Example 4.

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

Solution.

Activity of an isotope is measured by the number of nuclei decaying for a time unit. Suppose that \(d{N_d}\) nuclei decay for a short period of time \(dt.\) Then the isotope activity \(A\) is expressed by the formula

\[A = \frac{{d{N_d}}}{{dt}}.\]

It follows from the radioactive decay law that

\[N\left( t \right) = {N_0}{e^{ - \lambda t}},\]

where \(N\left( t \right)\) is the quantity of the remaining nuclei. Therefore,

\[{N_d}\left( t \right) = {N_0} - N\left( t \right) = {N_0} - {N_0}{e^{ - \lambda t}} = {N_0}\left( {1 - {e^{ - \lambda t}}} \right).\]

By differentiating with respect to \(t,\) we find the expression for activity:

\[A\left( t \right) = \frac{{d{N_d}}}{{dt}} = {N_0}\lambda {e^{ - \lambda t}}.\]

The initial isotope activity is equal to

\[A\left( {t = 0} \right) = {A_0} = {N_0}\lambda .\]

Hence,

\[A\left( t \right) = {A_0}{e^{ - \lambda t}}.\]

As it can be seen, the activity decreases over time by the same law as the amount of undecayed material. Substituting the expression for the half life \(T = \frac{{\ln 2}}{\lambda }\) in the last formula, we can write:

\[A\left( t \right) = {A_0}{e^{ - \frac{{t\ln 2}}{T}}}.\]

The value of \(T\) can be found from the last expression:

\[e^{ - \frac{{t\ln 2}}{T}} = \frac{A}{{{A_0}}},\;\; \Rightarrow - \frac{{t\ln 2}}{T} = \ln \frac{A}{{{A_0}}},\;\; \Rightarrow \frac{{t\ln 2}}{T} = \ln \frac{{{A_0}}}{A},\;\; \Rightarrow T = \frac{{t\ln 2}}{{\ln \frac{{{A_0}}}{A}}}.\]

In our case, the half life period of the given isotope is

\[T = \frac{{t\ln 2}}{{\ln \frac{{{A_0}}}{A}}} = \frac{{30\ln 2}}{{\ln \frac{{100}}{{90}}}} \approx \frac{{30 \cdot 0.93}}{{\ln 1.11}} \approx 197.3\,\text{days}.\]
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