In a college laboratory in Pune, students count the activity of a short-lived source every hour. They get \(1000\), \(741\), \(549\), \(407\), \(301\) counts per minute, and so on. Plotted, the points lie on a smooth curve that bends. A curve is hard to read: its slope changes everywhere, and a ruler cannot follow it. This lesson can be read on its own, whenever a physics lesson sends you here.
Taking the logarithm of every point
The decay law is \(A = A_0e^{-\lambda t}\). Take the natural logarithm of both sides. The logarithm of a product is a sum, and the logarithm of \(e^{-\lambda t}\) is its power. So \(\ln A = \ln A_0 - \lambda t\). This has the form \(y = c + mx\) of a straight line. The variable up the page is \(\ln A\), the slope is \(-\lambda\), and the intercept is \(\ln A_0\).
A graph with a logarithm on one axis and an ordinary scale on the other is a semi-log graph. On it, equal falls of \(\ln A\) in equal times mean equal ratios of \(A\). That is exactly what the decay law says. The straight line is also a test. If the points of real data do not lie on a line, the decay is not a single exponential.
Reading the line
Choose two points far apart on the line. The slope is the change in \(\ln A\) divided by the change in \(t\). In the figure, \(\ln A\) falls from \(6.31\) at \(2\,\text{h}\) to \(5.11\) at \(6\,\text{h}\). The slope is \(-1.2/4 = -0.30\) per hour, so \(\lambda = 0.30\) per hour. The intercept at \(t = 0\) is \(6.91\), and \(e^{6.91} = 1000\) counts per minute.
For a sample, \(\ln N\) is plotted against \(t\) in hours. The points lie on a straight line from \(\ln N = 18.0\) at \(t = 0\) to \(\ln N = 16.5\) at \(t = 5\,\text{h}\). Find \(\lambda\) and \(N_0\).
The line is \(\ln N = \ln N_0 - \lambda t\). Its slope is \(-\lambda\), and its intercept on the \(\ln N\) axis is \(\ln N_0\).
Slope \(= (16.5 - 18.0)/(5 - 0) = -0.3\) per hour.
So \(\lambda = 0.3\) per hour. The intercept is \(\ln N_0 = 18.0\), so \(N_0 = e^{18} = 6.6 \times 10^7\).
\(\lambda = 0.3\) per hour and \(N_0 = 6.6 \times 10^7\) nuclei. The half-life follows as \(0.693/0.3 = 2.3\,\text{h}\).
Two cautions make the reading honest. Many laboratories use common logarithms. Since \(\log A = \ln A/2.303\), the slope of \(\log A\) against \(t\) is \(-\lambda/2.303\), so multiply it by \(2.303\). And for cooling, the law \(T - T_0 = (T_i - T_0)e^{-kt}\) decays towards the room temperature, not towards zero. So plot \(\ln(T - T_0)\), the logarithm of the excess, never \(\ln T\).
A semi-log graph of a sample's activity shows \(\ln A = 7.60\) at \(t = 0\) and \(\ln A = 5.60\) at \(t = 10\,\text{min}\), on a straight line. What is the decay constant, in per minute?
A student plots \(\log_{10}A\) against \(t\) and finds a slope of \(-0.13\) per hour. What is the decay constant, in per hour?
Chai cools in a room at \(30\,°\text{C}\). To get a straight line whose slope is \(-k\), what should be plotted against time?
Power laws on log-log axes
Kepler's third law says the period of a planet grows as the power \(3/2\) of its orbit, \(T = kr^{3/2}\). Take the common logarithm of both sides: \(\log T = \log k + 1.5\log r\). So a graph of \(\log T\) against \(\log r\) is a straight line of slope \(1.5\). Such a graph, with logarithms on both axes, is a log-log graph. Any power law \(y = kx^n\) becomes a line of slope \(n\) on it.
Two points give the power. Jupiter has \(r = 5.20\,\text{AU}\) and \(T = 11.86\) years; the Earth has \(1\) and \(1\). So \(n = (\log 11.86 - 0)/(\log 5.20 - 0) = 1.074/0.716 = 1.50\). A log-log graph finds the power of a law from data, even when the law is not known in advance.
A quantity is suspected to follow \(y = kx^n\) with an unknown \(n\). Which graph will be a straight line if the suspicion is right?
For a simple pendulum, a student plots \(\log T\) against \(\log L\). Two points on the line are \((-1.00, -0.20)\) and \((0.00, 0.30)\). What is the power \(n\) in \(T \propto L^n\)?
The students' curve has now revealed its number. Taking \(\ln A\) of each count straightened the curve, and the slope of the line gave \(\lambda = 0.30\) per hour with a ruler. The same line tested the law, since real points that bend away would have shown a second process.
A semi-log line is straight only for a single exponential with nothing added. A steady background of cosmic-ray counts makes the late points flatten. Two nuclides decaying together give a line that bends from one slope to another. And only a positive number has a logarithm: a cooling excess or a count must be above zero. Subtract the background first, then take the logarithm.
Semi-log graphs give the decay constant of a radioactive sample and test the exponential law, in the chapter Radioactivity. The logarithm of the excess temperature against time tests Newton's law of cooling and gives the cooling constant. Log-log graphs find the power of a law from data, as for Kepler's third law of the planets or the period of a pendulum. Wherever a straight line can be drawn through logarithms, a ruler replaces a guess.
- Plotting \(\ln T\) for a cooling body.Plot \(\ln(T - T_0)\). The excess over the room decays to zero; the temperature does not.
- Forgetting the factor 2.303 on a common-logarithm graph.The slope of \(\log y\) against \(t\) is \(-\lambda/2.303\). Multiply it by 2.303.
- Reading the intercept as \(y_0\) itself.The intercept is \(\ln y_0\). Take \(e\) to that power to get \(y_0\).