Calculating single-grain ages:
many ways to skin a cat

Ten naturally-occuring long-lived $\alpha$-emitting radionuclides exist on Earth: $^{144}$Nd, $^{148}$Sm, $^{147}$Sm, $^{152}$Gd, $^{174}$Hf, $^{186}$Os, $^{190}$Pt, $^{232}$Th, $^{235}$U and $^{238}$U. For the purpose of helium-thermochronology, all but the heaviest three of these nuclides can often be neglected because of their low abundance and low helium-yield. For example, only one $\alpha$-particle is produced per $^{147}$Sm, whereas six to eight are formed in the Th and U decay series. Further simplification is possible because the present-day $^{238}$U/$^{235}$U-ratio is constant in the solar system (= 137.88; Steiger and Jäger, 1977). Therefore, the ingrowth of helium with time (t) can be written as a function of the elemental U, Th and He abundances or concentrations:

$$[He] = \left( 8\frac{137.88}{138.88}\left(e^{\lambda_{238}t}-1\ri... ...^{\lambda_{235}t}-1\right) \right) [U]+ 6\left(e^{\lambda_{232}t}-1\right)[Th]$$ (2)

with $\lambda_{232}$, $\lambda_{235}$ and $\lambda_{238}$ the decay constants of $^{232}$Th, $^{235}$U and $^{238}$U, respectively. Equation 2 has no analytical solution but is easy to solve iteratively. However, for young ages (t $\ll$ 1/ $\lambda_{235}$), a reasonably accurate linear approximation also exists:

$$t = \frac{[He]}{P}$$ (3)

with P the present-day helium production rate:

$$P = \left( 8\frac{137.88}{138.88}\lambda_{238} + \frac{7}{138.88}\lambda_{235} \right) [U]+ 6\lambda_{232}[Th]$$ (4)

The accuracy of this solution will be discussed in Section 4. Besides being easy to implement, the linear age equation is useful for illustrative purposes and opens up some new applications which will be discussed in Section 3.

Meesters and Dunai (2005) introduced an alternative direct solution to the (U-Th)/He age equation:

$$t = \frac{1}{\lambda_{wm}}~ ln\left(1+\frac{\lambda_{wm}}{P}[He]\right)$$ (5)

with $\lambda_{wm}$ the weighted mean decay constant:

$$\lambda_{wm} = \frac{\left( 8\frac{137.88}{138.88}\lambda_{238}^2 + \frac{7}{138.88}\lambda_{235}^2 \right) [U]+ 6\lambda_{232}^2[Th]}{P}$$ (6)

As shown by Meesters and Dunai (2005) and in Section 4, this solution is remarkably accurate for all practical applications.