Units: 37 GBq = 37×10^9Bq = 1 Curie

Lead 210, 22.3 years, 207.2 g/mol… Q = 0.064 MeV (beta)

Polonium, 137.376 days, 210 g/mol… Q = 5.407 MeV (alpha)

Let’s assume the amount of lead and Po to be 0.5 micro-gram each in 1 cigarette.

SO

(0.5/207.2)x10^(-6) mol of Pb

and

(0.5/210)x10^(-6) mol of Po.

1 mol = 6.022×10^(23) “elements”

Implies 6.022×10^(17)x0.5/207.2 nuclei of of Pb = 1.45×10^15/cigarette

Implies 6.022×10^(17)x0.5/210 nuclei of Po = 1.43×10^15/cigarette

Activity:

Pb: (10^15)x(1.45×0.693)/(22.3x365x24x3600)

= 1.43×10^6 decays/second = 1.43MBq

= 0.00143/37 Curies = 38.65 microCurie

Po: (10^15)x(1.43×0.693)x(137.376x24x3600)

= 8.35×10^7 decays/second = 83.5MBq

= 0.0835/37 Curies = 2.26 milliCuries

SO the beta energy of the Pb(210) makes it

1.43(MBq)x0.064MeV = 91.52GeV

= 91.52GeV/cigarette

= 91.52×10^(9)x1.6×10^(-19)J = 14.64 nanoJ/cigarette.

Or 14.64 nanoGray/cigarette.

(Considering 1Kg of absorbing material)

= 1.76nSv/cigarette

Considering 20 cigarettes a day (a pack) and

regular smoking through a year this would be

20x365x1.76nSv=12.83 microSv/year

(Lung has a weighing factor of 0.12 and Beta emission has a weighing factor of 1)

SO Po(210) with alpha emission would be

83.5(MBq)x5.407MeV which is actually 4.51×10^5GeV

=4.51×10^(14)x1.6×10^(-19)J = 72.2microJ/cigarette.

Or 72.2 microGray/cigarette.

=8.664microSv/cigarette

SO with a pack of 20 cigarettes and 365 days of smoking one would collect a dose of

20x365x8.664microSv = 63.247 milliSv/year

(Again considering 1Kg of absorbing material, lung’s radiation damage factor and a simple alpha emission effect not considering alpha emission nuclear recoil. Also I have not multiplied the factor 20 for Alpha-particles which is done since I already calculated the energy I am not sure its necessary or not. In any case it’s dangerous enough. Po is more dangerous than Pb for lung cancer)

If the cigarette would indeed contain 0.5 microgram of Pb and Po, which is not known a priori, the values from experiment performed, 2 examples shown below shows different value, which just means the cigarettes contain correspondingly different radioactive contents, nonetheless harmful enough to kill smokers by lung cancer.

Since from the experiments of “Khater” and “Papastefanou” below, we have an annual dose of 100 microSv on an average for radio-active Pb as well as Po, it just means that we have more than 0.5 microgram of Pb but less than 0.5 microgram of Po in a single cigarette. Po is highly radio-active (Po-210 has a much smaller half-life) and emits alpha-rays, therefore much more dangerous than Pb-210.

References: (only the abstract available to me)

Ashraf E. M. Khaterl , mailto:khater_ashraf@yahoo.com

National Center for Nuclear Safety and Radiation Control, Atomic Energy Authority, P.O. Box 7551, Nasr City, Cairo 11762, Egypt

Available online 3 October 2003.

Abstract

“Due to the relatively high activity concentrations of 210Po and 210Pb that are found in tobacco and its products, cigarette smoking highly increases the internal intake of both radionuclides and their concentrations in the lung tissues. …(partially removed the abstract content)…. The results of this work indicate that the average (range) activity concentration of 210Po in cigarette tobacco was 16.6 (9.7–22.5) mBq/cigarette. The average percentages of 210Po content in fresh tobacco plus wrapping paper that were recovered by post-smoking filters, ash and smoke were 4.6, 20.7 and 74.7, respectively. Cigarette smokers, who are smoking one pack (20 cigarettes) per day, are inhaling on average 123 mBq/d of 210Po and 210Pb each. The annual effective doses were calculated on the basis of 210Po and 210Pb intake with the cigarette smoke. The mean values of the annual effective dose for smokers (one pack per day) were estimated to be 193 and 251 μSv from 210Po and 210Pb, respectively.”

C. Papastefanou

Author Affiliations

Aristotle University of Thessaloniki, Atomic and Nuclear Physics Laboratory, Thessaloniki 54124, Greece

Corresponding author: papastefanou@physics.auth.gr