## 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

*Link to p*ermaweblink in citation or copy/archive material as it is

**with Creative Commons**

**and other copyrights in this web-site in mind**: permaweblink

# Radioactivity of cigarettes and banana

## Manmohan Dash

# 1 Activity of Cigarettes

**Units:**37

*GBq*= 37×10

^{9 }

*Bq*= 1

*Curie*,

**Radioactivity of**Lead 210, 22.3

*Pb*(210) and*Po*(210):*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 *Pb*and

mol of *Po*. 1 *mol* = 6.022 × 10^{23}“elements”. This implies

nuclei of *Pb* is **1.45 × 10 ^{15} ⁄ cigarette**. This in turn implies

**6.022 × 10**nuclei of

^{17}× (0.5)/(210)*Po*is 1.43 × 10

^{15}⁄

*cigarette*.

**Activity:**

*Pb*: (10

^{15}× 1.45 × 0.693)/( 22.3 × 365 × 24 × 3600) = 1.43 × 10

^{6 }

*decays*⁄

*second*= 1.43

*MBq*= (0.00143)/(37).

**Units:**

*Curies*= 38.65

*microCurie*,

*Po*: 10

^{15}× 1.43 × 0.693 × 137.376 × 24 × 3600 = 8.35 × 10

^{7 }

*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 × 10

^{9}× 1.6×10

^{ − 19}

*J*= 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 × 10^{5 }*GeV* = 4.51 × 10^{14} × 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.

*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

# Bananas Are Radioactive

## Anne Marie Helmenstine [Ph.D., About.com Guide]

Source: Chemistry, About.Com

Excerpt/Abstract: from above article:

**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?**AMASA**@joe: There are many, many, many, many, many molecules of potassium in 450*mg*of potassium.**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 × 10^{9}*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^{ − 4}*mol*× 6.022 × 10^{23}*particles*⁄*mol*= 8.137 × 10^{19}*atoms*,*K*− 40, 4.0685 × 10^{19}decay in the first 1.248 × 10^{9}years, average of 3.26 × 10^{10}per year, average of 1033 ⁄*second***MARK ANDERSON**@Chris Close, but its .012 which is really 0.00012 which works out to about 0.054*mg*.**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 × 10^{9}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.**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**independent work of Manmohan Dash]***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 × 10^{17}nuclei of*K*− 40 (not 10^{19}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)/(*T*_{half − 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 × 10^{9}*seconds*= 1.25 × 3.15 × 10^{16}*seconds*. Since the banana has450*milli*−*grams*of Potassium it has as many radioactive*K*− 40 as we just counted, 8.137 × 10^{17}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 × 10^{17 − 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*

*Ref-I [License info]*

**Polonium-210 budget in cigarettes**

*Ref-II [Open Access]*

*Int. J. Environ. Res. Public Health*

**2009**,

*6*, 558-567.