Half-life
Half-life is a special case of exponential decay where the rate of decay is \(50\%\).
For example, if you start with \(100\) units of a material and it decays to half of its original amount in \(10\) days, you could determine how much is remaining after \(50\) days.
In \(10\) days you would have \(50\) units.
In \(20\) days you would have \(25\) units.
In \(30\) days you would have \(12.5\) units.
In \(40\) days you would have \(6.25\) units.
In \(50\) days you would have \(3.125\) units.
\(50\) days represents \(5\) full decay periods, so you could use an exponential decay formula, \(y=100\left(1-.50\right)^5\) to solve the same problem.
Another approach to solving a half-life problem is to use the formula for half-life.
Half-life is a special case of exponential decay where the rate of decay is \(50\%\).
For example, if you start with \(100\) units of a material and it decays to half of its original amount in \(10\) days, you could determine how much is remaining after \(50\) days.
In \(10\) days you would have \(50\) units.
In \(20\) days you would have \(25\) units.
In \(30\) days you would have \(12.5\) units.
In \(40\) days you would have \(6.25\) units.
In \(50\) days you would have \(3.125\) units.
\(50\) days represents \(5\) full decay periods, so you could use an exponential decay formula, \(y=100\left(1-.50\right)^5\) to solve the same problem.
Another approach to solving a half-life problem is to use the formula for half-life.
Half-Life Formula: \(y=P\left(\frac{1}{2}\right)^x\)
In the formula, \(P\) represents the initial quantity, \(x\) represents the number of decay periods given by the formula \(x=\frac{\text{total time}}{\text{half-life time}}\), and \(y\) represents the ending quantity after \(x\) decay periods.
In the formula, \(P\) represents the initial quantity, \(x\) represents the number of decay periods given by the formula \(x=\frac{\text{total time}}{\text{half-life time}}\), and \(y\) represents the ending quantity after \(x\) decay periods.
Example 1: It is believed that a Russian agent, Alexander Litvinenko, was poisoned on November 1, 2016 with Polonium-210, a highly radioactive substance. The half-life of Polonium-210 is \(138\) days and the median lethal dose by ingestion is \(50\) nanograms. Assuming Mr. Litvinenko ingested \(50\) ng of Polonium-210 on November 1st, how much of the substance would remain in his body \(23\) days later?
\(\begin{align}&y=P\left(\frac{1}{2}\right)^x\ & \ \ \ & \text{1) Write the general formula.}\\
&P=50\ \ \ x=\frac{23}{138}\ & \ \ \ & \text{2) Identify the values of the variables. Total time is 23 days and half-life time is 138 days.}\\
&y=50\left(\frac{1}{2}\right)^{\frac{23}{138}}\ & \ \ \ & \text{3) Substitute the values into the formula.}\\
&y=44.54493591\ & \ \ \ & \text{4) Enter the values into the calculator.}\end{align}\)
Approximately \(44.545\) ng of Polonium-210 would remain after \(23\) days.
Example 2: The half-life of Carbon-14 is \(5730\) years. If you start with a sample weighing \(20\) grams, in how many years will the sample weigh at most \(3\) grams?