First publicized/Available Online (Article-1, Article-2): Mar 28, 2011, April 1, 2011

Citation: “Radioactivity of cigarettes and banana”, Manmohan Dash, Communicating Science, April 11, 2011, publisher: Invariance Publishing House, MDash Foundation

1 Activity of Cigarettes

Units: 37 GBq = 37×109 Bq = 1 Curie, Radioactivity of Pb (210) and Po (210): Lead 210, 22.3 years, 207.2 g ⁄ mol Q = 0.064 MeV (β), Polonium 210, 137.376 days, 210 g ⁄ mol Q = 5.407 MeV (α). Let’s assume the amount of Lead and Polonium to be 0.5 micro-gram each in 1cigarette. So

mol of  Pband

mol of Po. 1 mol = 6.022 × 1023“elements”. This implies

nuclei of Pb is 1.45 × 1015 ⁄ cigarette. This in turn implies 6.022 × 1017 × (0.5)/(210) nuclei of Po is 1.43 × 1015 ⁄ cigarette. Activity: Pb: (1015 × 1.45 × 0.693)/( 22.3 × 365 × 24 × 3600) = 1.43 × 106 decays ⁄ second = 1.43 MBq = (0.00143)/(37). Units: Curies = 38.65 microCurie, Po: 1015 × 1.43 × 0.693 × 137.376 × 24 × 3600 = 8.35 × 107 decays ⁄ second = 83.5 MBq = (0.0835)/(37) Curies = 2.26 milliCuries. So the β energy of the Pb (210) makes it 1.43 MBq × 0.064 MeV = 91.52 GeV = 91.52 GeV ⁄ cigarette = 91.52 × 109 × 1.6×10 − 19J = 14.64 nanoJ ⁄ cigarette.

Or 14.64 nanoGray ⁄ cigarette, considering 1 kg of absorbing material = 1.76 nSv ⁄ cigarette. Considering 20 cigarettes a day, that is, a pack of cigarettes, and regular smoking through a year this would be 20 × 365 × 1.76 nSv = 12.83 microSv ⁄ year. Lung has a weighing factor of 0.12 and Beta (β) emission has a weighing factor of 1. So Po(210) with α emission would be 83.5 MBq × 5.407 MeV which is actually 4.51 × 105 GeV = 4.51 × 1014 × 1.6 × 10 − 19 J = 72.2 microJ ⁄ cigarette. Or 72.2 microGray ⁄ cigarette = 8.664 microSv ⁄ cigarette. So with a pack of 20 cigarettes and 365 days of smoking one would collect a dose of 20 × 365 × 8.664 microSv = 63.247 milliSv ⁄ year.

This is considering 1 kg of absorbing material, lung’s radiation damage factor and a simple α emission effect not considering α emission nuclear recoil. Also I have not multiplied the factor 20 for α − particles which is prescribed, since I already calculated the energy I am not sure its necessary or not. In any case its dangerous enough. Po is more dangerous than Pb for lung cancer. If the cigarette would indeed contain 0.5 micro − gram 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, which is nonetheless harmful enough to kill smokers by lung cancer.

Since from the experiments of “Khaterl”(REF-I↓) and “Papastefanou”(REF-II↓) 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 micro − gram of Pb but less than 0.5 micro − gram of Po in a single cigarette. Po is highly radio-active (Po − 210 has a much smaller half-life) and emits α − rays, therefore much more dangerous than Pb − 210.

2 Activity of Bananas

Excerpt/Abstract: from above article:

“Bananas are radioactive for a similar reason. The fruit contains high levels of potassium. Radioactive K-40 has an isotopic abundance of 0.01% and a half-life of 1.25 billion years. The average banana contains around 450 mg of potassium and will experience about 14 decays each second. It’s no big deal. You already have potassium in your body, 0.01% as K-40. You are fine. Your body can handle low levels of radioactivity.”
1. JOE: so the half life is 1.25 billion years and decays 14 times a second? are you sure you know what your talking about?
2. AMASA @joe: There are many, many, many, many, many molecules of potassium in 450 mg of potassium.
3. CHRIS @joe, the math looks a bit off, but not in the way you’re thinking, so let me break it down. half life of K − 40 = 1.248 × 109 years years, natural abundance of K − 40 = 0.012, molar mass of K − 40 = 39.96399848 g ⁄ mol, 450 mg × 0.012, 5.4 mg = 0.0054 g, (0.0054 g)/(39.96399848 g ⁄ mol) = 1.3512 × 10 − 4 mol, 1.3512 × 10 − 4mol × 6.022 × 1023particles ⁄ mol = 8.137 × 1019atoms, K − 40,  4.0685 × 1019 decay in the first 1.248 × 109 years, average of 3.26 × 1010 per year, average of 1033 ⁄ second
4. MARK ANDERSON @Chris Close, but its .012 which is really 0.00012 which works out to about 0.054 mg.
5. ELIOT Good work Chris and Mark. However your estimate for number of decays per second will be off because you didn’t figure them using logs. Radioactivity is an exponential decay function. So the number of decays per time unit starts higher and then slowly gets lower over time. So one second now will have almost twice as many decays as one second 1.248 × 109 years from now. Of course we are interested in the rate of decays for seconds now not 1 billion years in the future. This is what pushed the number to 14 per second.
6. MANMOHAN DASH  [independent work of Manmohan Dash] Hi Chris, Mark: you laid the ground work. Thanks. I will complete the calculation and show you this is exactly rounded to 14 decays per second. If you change your fraction of radioactive K − 40 to 0.01 as in the article rather than 0.012 etc I think it will give the exact answer. Here is the answer: I take the calculation you two have done correctly, i.e. the 450 milligram banana contains 8.137 × 1017 nuclei of K − 40 (not 1019 as corrected by Mark) Now the decay constant or transition probability as one may call it which gives the probability at any time of a single nucleus to decay at any time is ω = (0.693)/(Thalf − life). This is obtained by using Heisenberg Uncertainty principle for the nuclear decay. Here decay width which is given as a width of energy and mean life time which is given as a width or uncertainty in time, must multiply to h-cross (time – energy uncertainty). Now Mean life is related to half-life by the factor 0.693 (69.3% of nucleus would decay in the mean life where as 50% would decay in a half-life) So the decay probability = ω gives the probability or likelihood that any single nucleus will decay or not. So we know this probability because we know the half life to be 1.25 billion years that is 1.25 × 365 × 24 × 3600 × 109seconds = 1.25 × 3.15 × 1016seconds. Since the banana has450 milli − grams of Potassium it has as many radioactive K − 40 as we just counted, 8.137 × 1017 nuclei, at the time of eating. So the probability of decay is ω and you know how many nuclei you got. Multiply them to get the exact number of radioactivity of your banana. Needless to say this product is 14.3.(8.137 × 0.693)/(1.25 × 3.15 × 1017 − 16) = 1.43 × 10 = 14.3.

References:

I haven’t read these papers, only the abstract available to me as these are paid for publications