If a human was exposed to 20 nm Ti02 particles at the ambien

If a human was exposed to 20 nm Ti02 particles at the ambient concentration of 1.0/mu g/hr daily, calculate how many years will be required for his/her epithelial cells in the lungs to be damaged by the particles. You are given that alveolar area is 75 m^2 and epithelial cells account for 3% of that. Particles caused oxidative DNA damage to lungs in an in vitro experiment in the lab at a concentration of 25 mu g/9.6 cm^2. How many particles per hour need to be inhaled to cause that damage in that duration? Density of Ti02 is 4.23 g/cm^3.

Solution

A1)

Total area of epithelial cells occupied in alveolar= 75*3/100

                                                                                                = 2.25m² = 225cm²

But the deposition takes place on all the area of alveolar including epithelial cells. So we cannot use only the area of the epithelial cells. We have to consider total surface area of the alveolar.

Time required to deposit lethal amount of chemical on 1cm² = 25µg/ (9.6cm² *1.67 µghr^-1) = 1.55 hr cm^-1

Time required depositing lethal amount of                                 = 25µg/ (9.6cm² *1.67 µghr^-1) * 7500cm²

chemical on alveolar evenly including epithelial cells                = 1169535.93 hr

No of years                                                                                                  = 1169535.93 hr/(24*365 hr yr^-1) = 133 yrs.

A2)

Amount of TiO₂ in inhaling air                                                             = 25 µg

Volume of TiO₂ in inhaling air                                                               = 25*10^-6 g/ 4.23g cm^-3

(Since there is no mention about the particle volume, let’s assume the one particle is 20 nm³)

No of particles inhale by a person in an hour                                = (25*10^-6 g/ 4.23g cm^-3) / (20*10^-21 nm³)

                                                                                                                         = 2.95* 10¹⁴