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11. More Applications

Dated: 10-11-2024

Radioactive Decay

Many radioactive materials disintegrate at a rate proportional to the amount present.
if \(A(t)\) is the amount of radioactive substance present at time \(t\) then

\[\frac{dA}{dt}=kA\]

Where \(k\) is the constant of proportionality.
If \(A_0\) is the initial amount then using the population growth rate model,1 we have

\[A(t) = A_0e^{kt}\]

The half life of radio active substance is the time it takes for half of the atoms present to decay.
if \(T\) is the half life then

\[A(T) = \frac {A_0} 2\]
\[\frac{A_{0}}{2}=A_{0}e^{kt}\]
\[kT=-\ln 2\]

Example

A radio active isotope has half life of \(16\) days.
We have \(30g\) of it remaining at the end of \(30\) days.
How much of it was initially present?

Solution

Let the amount present at time \(t\) be \(A(t)\) and initial amount be \(A_0\).

\[\frac{dA}{dt}=kA, \quad A(30)=30\]
\[A(t)=A_{0}e^{kt}\]
\[kT = - \ln 2\]
\[k = - \frac {\ln 2} T\]
\[k = - \frac {\ln 2}{16}\]
\[30=A_{0}e^{30k}\]
\[A_{0}=30e^{-30k}=30e^{\frac{30\ln 2}{16}}=110.04 g\]

Newton's Law of Cooling

Temperature \(T(t)\) of a body changes at a rate proportional to the difference between the temperature in the body and the temperature \(T_m\) of its surroundings.
This is called Newton's law of cooling.

\[\frac{dT}{dt}=k(T-T_{m}), \quad T(0)=T_{0}\]
\[\int\frac{dT}{T-T_{m}}=\int k~dt\]
\[\ln|T-T_{m}|=kt+C\]
\[T-T_{m}=e^{kt+C}\]
\[T(t)=T_{m}+C_{1}e^{kt}\]
\[\because T(0) = T_0\]
\[\therefore C_1 = T_0 - T_m\]
\[T(t)=T_{m}+(T_{0}-T_{m})e^{kt}\]

If we have temperature known at \(t_1\) and \(t_2\) then

\[\because T(t)=T_{m}+(T_{0}-T_{m})e^{kt}\]
\[\implies T(t_{1})-T_{m}=(T_{0}-T_{m})e^{kt_{1}} \text{ and } T(t_{2})-T_{m}=(T_{0}-T_{m})e^{kt_{2}}\]
\[\frac{T(t_{1})-T_{m}}{T(t_{2})-T_{m}}=e^{k(t_{1}-t_{2})}\]

Example

Suppose that a dead body was discovered at midnight when its temperature was \(80^\circ F\).
The room temperature is kept constant at \(60^\circ F\).
Two hours later, the body temperature drops to \(75^\circ F\).
Find the time of death.

Solution

Assuming that the person was not sick. Therefore,

\[T(0) = 98.6^\circ F = T_0\]
\[T_m = 60^\circ F\]
\[\frac{dT}{dt}=k(T-60), \quad T(0)=98.6\]
\[T(t)=T_{m}+(T_{0}-T_{m})e^{kt}\]
\[\frac{T(t_{1})-T_{m}}{T(t_{2})-T_{m}}=e^{k(t_{1}-t_{2})}\]
\[T(t_{1})=80^{\circ}F \text{ and } T(t_{2})=75^{\circ}F\]
\[\because t_1 - t_2 = 2 \text{ hours}\]
\[\frac{80-60}{75-60}=e^{2k}\]
\[k=\frac{1}{2}\ln\frac{4}{3}=0.1438\]

For the time of death, we need to find the interval2 \(t_1 - t_2 = t_d\).

\[\frac{T(t_{1})-T_{m}}{T(t_{2})-T_{m}}=e^{k(t_{1}-t_{2})} \implies \frac{98.6-60}{80-60}=e^{kt_{d}}\]
\[t_{d}=\frac{1}{k}\ln\frac{38.6}{20} \approx 4.573\]

Hence the time of death is \(7:42 PM\).

Carbon Dating

Carbon-14 is produced in the atmosphere through cosmic radiation acting on nitrogen, maintaining a constant ratio with ordinary carbon.
Living organisms contain the same proportion of C-14 as the atmosphere, but once they die, they stop absorbing it.
By comparing C-14 levels in a fossil to atmospheric levels, scientists can estimate its age.
This method, which relies on C-14's half-life of about 5600 years, has been used to date artifacts, such as ancient Egyptian furniture.

Example

A fossilized bone is found to contain \(\frac 1 {1000}\) of the original amount of C-14.
Determine the age of the fissile.

Solution

\[\frac{dA}{dt}=kA, \quad A(0)=A_{0}\]
\[A(t)=A_{0}e^{kt}\]

The half life of carbon isotope is \(5600\) years.

\[A(5600)=\frac{A_{0}}{2}\]
\[\frac{A_{0}}{2}=A_{0}e^{5600k}\]
\[k = −0.00012378\]
\[A(t)=A_{0}e^{-(0.00012378)t}\]
\[\because A(t) = \frac{A_0}{1000}\]
\[\frac{A_{0}}{1000}=A_{0}e^{-(0.00012378)t} \implies -0.00012378t=-\ln 1000\]
\[t=\frac{\ln (1000)}{0.00012378}=55,800 \text{years}\]

References

Read more about notations and symbols.

  1. Read more about population growth rate model. ↩

  2. Read more about intervals. ↩