Question

as we saw in Section 3.4, a radioactive substance decays exponentially: The mass at time t is \(m(t) = m(0){e^{kt}},\) where m(0) is the initial mass and k is a negative constant. The mean life Mof an atom in the substance is

\(M = - k\int_0^\infty t {e^{kt}}dt\)

For the radioactive carbon isotope,\(^{14}C\), used in radio-carbon dating the value of k is -0.000121. Find the mean life of a \(^{14}C\)atom.

Short answer

The mean life of the carbon isotope \(^{14}C\) atom \(8264.5\)years

Step-by-step solution

Given Data

The equation of mean life\(M\)of an atom is\(M = - k\int_0^\infty t {e^{kt}}dt\).

The value of \(k\) is -\(0.000121\)

Find the Mean life of the carbon isotope \(^{14}C\) atom

The given integral equation is\(M = - k\int_0^\infty t {e^{kt}}dt\).

Substitute\(u = t,dv = {e^{kt}}dt,v = \frac{1}{k}{e^{kt}}\), and\(du = dt\).

Using the above definition, rewrite the given integral as shown below:

\(\begin{aligned}{c}M &= - k\int_0^\infty t {e^{kt}}dt\\ &= - k\mathop {\lim }\limits_{x \to {0^ + }} \int_0^x t {e^{kt}}dt\end{aligned}\)

Using integration by parts, obtain the value of the above integral as shown below:

\(\begin{aligned}{c}M &= - k\mathop {\lim }\limits_{x \to {0^ + }} \int_0^x t {e^{kt}}dt\\ &= - k\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {\left( {t\left( {\frac{1}{k}} \right){e^{kt}}} \right)_0^x - \int_0^\infty {\frac{1}{k}} {e^{kt}}dt} \right\}\\ &= - k\mathop {\lim }\limits_{x \to {0^ + }} \left( {t\left( {\frac{1}{k}} \right){e^{kt}} - \frac{1}{{{k^2}}}{e^{kt}}} \right)_0^x\\ &= - k\mathop {\lim }\limits_{x \to {0^ + }} \left( {\left( {t - \frac{1}{k}} \right)\frac{{{e^{kt}}}}{k}} \right)_0^x\\ &= - k\mathop {\lim }\limits_{x \to {0^ + }} \left( {\left( {t - \frac{1}{k}} \right)\frac{{{e^{ - (0.000121\;t)}}}}{k}} \right)_0^x\end{aligned}\)

The above equation can be further simplified as shown below:

Therefore, the mean life of the carbon isotope\(^{14}C\)atom is 8264.5 years.