Question

The half-life of cesium-137 is 30 years. Suppose we have a 100 mg sample.

(a) Find the mass that remains after t years.

(b) How much of the sample remains after 100 years?

(c) After how long will only 1 mg remain?

Short answer

(a) The mass that remains after t hours is \(y\left( t \right) = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\).

(b) \(9.92\;{\rm{mg}}\) of the sample remains after 100 years.

(c) The sample will be remaining \(199.3\;{\rm{years}}\).

Step-by-step solution

Find an answer for part (a)

Let \(y\left( t \right)\) is the mass remaining after t years, then using theorem 2:

\(\begin{aligned}y\left( t \right) & = y\left( 0 \right){e^{kt}}\\ & = 100{e^{kt}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {y\left( 0 \right) = 100} \right)\end{aligned}\)

As 30 years is the half-life of the cesium, therefore;

\(\begin{aligned}\frac{1}{2}\left( {100} \right) & = 100{e^{30k}}\\30k & = \ln \frac{1}{2}\\k & = - \frac{{\ln 2}}{{30}}\end{aligned}\)

By the equation \(y\left( t \right) = 100{e^{kt}}\).

\(y\left( t \right) = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\)

So, the mass that remains after t hours is \(y\left( t \right) = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\).

Find an answer for part (b)

Substitute 100 for t in the equation \(y\left( t \right) = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\).

\(\begin{aligned}y\left( {100} \right) & = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)\left( {100} \right)}}\\ & = 100 \cdot {2^{\left( { - \frac{{100}}{{30}}} \right)}}\\ & \approx 9.92\;{\rm{mg}}\end{aligned}\)

Thus, \(9.92\;{\rm{mg}}\) of the sample remains after 100 years.

Find an answer for part (c)

Substitute 1 for \(y\left( t \right)\) in the equation \(y\left( t \right) = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\).

\(\begin{aligned}1 & = 100{e^{\left( { - \frac{{\ln 2}}{{30}}} \right)t}}\\\ln \frac{1}{{100}} & = \left( { - \frac{{\ln 2}}{{30}}} \right)t\\t & = - 30\frac{{\ln 0.01}}{{\ln 2}}\\ & \approx 199.3\;{\rm{years}}\end{aligned}\)

So, the sample will be remaining \(199.3\;{\rm{years}}\).