My purpose in this letter is to note that the link between Rutherford's observations and the Avogadro constant can be expressed succinctly and can be clarified through differential expressions not presented in Leenson's article. The lengthy quotation of Rutherford's work, Part 1 of the article, describes determination of his defined quantity x = the initial rate of He gas production from a-decay by ²²⁶Ra. The overall treatment recognizes that He gas is also produced by three decay products of ²²⁶Ra. His result is x = 0.107 mm³ day^-1 g^-1 where the unit "mm³" is the volume of He gas at 0 °C and 1 atm. Suggested here is that x be translated into rate terms with n_He and n_Ra denoting the macroscopic amounts (e.g., moles) of the species indicated for a given mass m of ²²⁶Ra:

x = (1/m)(dn_He/dt)V _He °

where V_He ° is the molar volume of He gas at 0 °C and 1 atm, or

(dn_He/dt) = mx/V_He ° = -(dn_Ra/dt) = ln_Ra = lm/M_Ra

Here, M_Ra is the molar mass of ²²⁶Ra. The macroscopic information from the quoted results from Rutherford's laboratory relates to the decay constant l but not to the Avogadro constant! Solving, one finds

l = x M_Ra/V_He ° =
(0.107 mm³ day^-1 g^-1)(226 g mol^-1)/(22.4 x 10⁶ mm³ mol^-1) =
(1.079 x 10^-6 day^-1)(365.25 day year^-1) = 3.94 x 10^-4 year^-1

the corresponding half-life t_1/2 being 1759 years, a value consistent with the presently accepted 1622 years in view of the precision of mass measurement involved in determination of x. Therefore, macroscopic information on He formation upon decay of Ra does not allow estimation of the Avogadro constant, N_A.

To estimate N_A requires some further observations, which must include the particle nature of a-decay. Only later, under Question 2 of "Part 2. Questions for Students", does Leenson's article offer as "additional information" results of observations reported in 1908 by Rutherford and Geiger. The information provided expresses quantitatively the rate of a-particle emission per unit mass of Ra. Over an interval of 10.0 minutes an average of 49.0 a-decays were detected by scintillations on a fluorescent screen of d = 1.25 mm located at l = 1.50 m from a Ra sample of mass = 0.055 mg. The total rate of Ra decay for this mass can be written in terms of this observation and the geometry of the experiment:

-(dN_Ra/dt) = [4pl ²/p(d/2)²](49.0/10.0 min) = 1.13 x 10⁸ min^-1

By first-order decay, one can express this same rate in terms of the decay constant l and N_Ra, the number of Ra atoms in the given mass:

-(dN_Ra/dt) = lN_Ra = ln_Ra N_A = a(m/M_Ra )N_A

This last expression does include N_A. All the other quantities are known from Leenson's Part 1 and the "additional information" given under his Question 2. Solving for N_A one has

N_A = (M_Ra/lm) [-(dN_Ra/dt)] =
(226 g mol^-1)(24 h day^-1)(60 min h^-1)(1.13 x 10⁸ min^-1)/
(1.079 x 10^-6 day^-1)(0.055 x 10^-3g) =
6.2 x 10²³ mol^-1

This is a reasonable value for N_A in view of the implied precision of Rutherford's experiments--especially indirect determination of the very small masses of the samples studied.