218 po 214 pb 214 Bi 214 po 0.69 0.44 0.25 0.25 39, A room measures 10 m × 8 m ×

Question

218 po 214 pb 214 Bi 214 po 0.69 0.44 0.25 0.25 39, A room measures 10 m × 8 m × 3 m. It contains 80 pCi L-1 of 218Po, 60 pCi L-1 of 214Pb, and 25 pCi L-1 each of 2i4 Bi and 214Po. (a) Calculate the WL concentration in the room. (b) Calculate the total potential alpha-particle energy in the room. (c) What is the concentration of 214Po atoms in the air? (d) If secular equilibrium existed at this working-level concentration, what would be the activity concentration of 214 pb atoms What would be the exposure in WLM of an individual who occupied the room 12 hours per day, 6 days per week, for one year? (e)

Explanation / Answer

(a)

According to the rule of thumb 1WL of radon daughters is often associated with the radon concentration of 200pCi/L. Given 80pCi/L = 218Po, 60pCi/L = 214Pb , 25pCi/L = 214Bi = 214Po

therefore the WL in the room = (80/200)+(60/200)+(25/200)+(25/200) = 0.4+0.3+0.125+0.125 = 0.95 WL

The WL in the is 0.95WL

(b)

The total potential alpha-particle energy in the room = 218Po + 214Pb as other two will produce.i.e.214Bi ,214Po beta not alpha in the room

radon decays ->218Po = 5.49 MeV alpha -> 214Pb = 6.00MeV alpha ->214Bi =0.7 MeV beta -> 214Po = 3.3 MeV beta. ( Literature value of Radon decays)

Therefore The total potential alpha-particle energy in the room = 218Po + 214Pb = (5.49+6.00 ) MeV = 11.49 MeV

The total potential alpha-particle energy in the room =11.49 MeV

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