Answer 1

During irradiation, decay competes with production

      R = Nλ(1 – e^–λt)

R is the rate at which radioactive nuclei are produced (GBq µA^-1). The term (1 – e^–λt) is called the “saturation factor” and accounts for decay over the period of irradiation t of the target.

Saturation occurs when t >> T½ so that e^–λt ≈ 0, the saturation factor is 1, and R = Nλ (that is, the rate of decay equals the rate of production).

T½ for ¹⁸F is 1.83 hours, so λ = 0.693/1.83 = 0.38 h^-1

Saturation factor is 1 – e^{-0.38 × 3} = 0.68

200 µA × 14 GBq µA^-1 × 0.68 = 1904 GBq

Answer 2

R is also the rate at which neutrons are being produced!

One n for each (p,n) reaction, so the production rate is 200 µA × 14 GBq µA^-1 = 2.8 × 10 ¹² n sec^-1

Assuming isotropic distribution, at 3 metres from target, fluence rate is

• Ф_n =2.8 × 10¹² n sec^-1 / 4 π (300 cm)²
• Ф_n = 2.48 × 10⁶ n sec^-1 cm²

Neutron energy is up to several MeV and the fluence to dose conversion factor varies widely with both energy and exposure orientation. Assuming the worst case, the conversion factor (various ICRP references) would be roughly 400 pSv n^-1 cm^-2.

• 2.48 × 10⁶ n sec^-1 cm² × (4 × 10^-10 Sv n^-1 cm^-2)
• 0.99 mSv s^-1 or 3.6 Sv h^-1