Mathematical Appendix — Solving the Age Equation

Sigma logo for mathematics

Uraninite Pb isotope dating relates the measured radiogenic lead-isotope ratio in Swamp Quarry uraninite to the known radioactive decay rates of uranium-235 and uranium-238.  The purpose of this appendix is to show, step by step, how the equation is solved for t, the time since the uraninite crystallized.

The age equation is:

Pb207 Pb206 = 1 137.88 × e λ235 t − 1 e λ238 t − 1

For the Swamp Quarry uraninite, the measured radiogenic lead-isotope ratio is:

207Pb / 206Pb = 0.0526

The objective is to determine t, the time elapsed since the uraninite crystallized.

Step 1 — Determine the decay constants

The radioactive decay constant, λ, describes the probability of radioactive decay per unit time. It is related to the half-life of an isotope by:

λ = ln ( 2 ) T 1/2

For uranium-235, the half-life is approximately 703.8 million years. Using millions of years (Ma) as the time unit:

T1/2 = 703.8 Ma

For uranium-238, the half-life is approximately 4,468 million years:

T1/2 = 4,468 Ma

Substituting these half-lives into the decay-constant equation gives the decay constants for the two uranium isotopes.

λ235 = ln(2) 703.8 = 9.849 × 10−4 Ma−1

λ238 = ln(2) 4468 = 1.551 × 10−4 Ma−1

Step 2 — Substitute the numerical values into the age equation

We now substitute the measured lead-isotope ratio and the two calculated decay constants into the age equation. Because the decay constants are expressed in Ma−1, the resulting value of t will be in millions of years.

0.0526 = 1 137.88 × e ( 9.849 × 10−4 t ) −1 e ( 1.551 × 10−4 t ) −1

 

Step 3 — Why the equation must be solved numerically

At this point all quantities in the equation are known except t, the elapsed time since the uraninite crystallized. The objective is therefore to solve the equation for t.

If t occurred in only one exponential term, a natural logarithm could be used to remove the exponential and isolate t. Here, however, t occurs in two exponential terms with different decay constants: one for uranium-235 and one for uranium-238.

We can simplify and rearrange the equation algebraically, but taking the logarithm of both sides does not isolate t. There is no simple closed-form algebraic solution for t. The equation must therefore be solved numerically.

The procedure is to rearrange the equation so that all terms are on one side and define a function f(t). The required age is the value of t for which:

f ( t ) = 0

In mathematical terms, we are now looking for the root of the function. A numerical root-finding procedure tests values of t, evaluates the equation, and progressively narrows the range until the value of t that makes f(t) equal to zero is found.

The following steps carry the algebra as far as possible, define f(t) explicitly, and then determine its root numerically.

Step 4 — Carry the algebra as far as possible

Starting with the numerical age equation, we first multiply both sides by 137.88. This removes the factor 1/137.88 and leaves only the ratio of the two exponential terms.

0.0526 × 137.88 = e 9.849 × 10 −4 t − 1 e 1.551 × 10 −4 t − 1

The left side can now be evaluated directly:

0.0526 × 137.88 = 7.2525

The equation therefore becomes:

7.2525 = e 9.849 × 10 −4 t − 1 e 1.551 × 10 −4 t − 1

Next, multiply both sides by the denominator. This removes the fraction and gives:

7.2525 ( e 1.551 × 10 −4 t − 1 ) = e 9.849 × 10 −4 t − 1

Expanding the left side gives:

7.2525 e 1.551 × 10−4 t − 7.2525 = e 9.849 × 10−4 t − 1

We now move all terms to one side of the equation and define the resulting expression as f(t):

f(t) = e 9.849 × 10−4 t − 7.2525 e 1.551 × 10−4 t + 6.2525 = 0
Step 5 — Solve f(t) = 0 numerically

We now evaluate f(t) at selected values of t. If two trial values produce results with opposite signs, a root lies between them. We can then progressively narrow that interval until the value of t is determined to the desired precision.

Recall that:

f(t) = e 9.849× 10−4 t − 7.2525 e 1.551× 10−4 t +6.2525

As a first trial, evaluate the function at t = 300 Ma and t = 325 Ma. These values bracket the expected geological age while remaining far enough apart to demonstrate the numerical method clearly.

Evaluating the function at the two trial ages gives:

f(300) ≈ −0.001675

f(325) ≈ +0.002304

The function is negative at 300 Ma and positive at 325 Ma. Because f(t) is continuous over this interval, it must pass through zero somewhere between these two ages. The root is therefore bracketed:

300 < t < 325 Ma

We now test a much narrower interval around the expected root:

f(310) ≈ −0.000169

f(315) ≈ +0.000627

The sign again changes from negative to positive, so the root has now been narrowed to:

310 < t < 315 Ma

Refining the interval once more:

f(311) ≈ −0.00001184

f(312) ≈ +0.00014616

The root therefore lies between 311 and 312 Ma. Continuing the numerical refinement gives a root of approximately 311.08 Ma when the rounded values displayed in this appendix are used.

t ≈ 311.08 Ma

The calculation above uses rounded values for the isotope ratio, half-lives, decay constants, and intermediate numerical quantities. With these displayed values, the numerical solution is approximately 311.08 Ma, which is consistent with the reported age of approximately 312 Ma when expressed to the precision appropriate for this calculation.

What the calculation shows

The uranium–lead age equation cannot be solved for t by ordinary algebra because the unknown time appears in two exponential terms with different decay constants. Instead, the equation is rearranged into a function, f(t), and the value of t that makes f(t) = 0 is found numerically.

For the Swamp Quarry uraninite, this procedure gives an age of approximately 312 million years. This represents the time when uranium became incorporated into the newly formed uraninite crystal and its uranium–lead geological clock began to run.

The same procedure can be applied to another uranium-bearing mineral when the appropriate radiogenic lead-isotope ratio is known. The measured isotope ratio and applicable decay constants are inserted into the age equation, and the resulting equation is solved numerically for t.

↑ Back to Uranium Menu