Absorbed dose of nuclear radioactivity


by planesinspace
Tags: absorbed, dose, nuclear, radioactivity
planesinspace
planesinspace is offline
#1
May31-08, 10:42 AM
P: 21
1. The problem statement, all variables and given/known data
One kg of human body contains about 0.2% potassium of which 0.0117% is potassium-
40 (40K). 40K is radio active and in 89% of the time the product of the decay is a gamma ray of energy
1.46 MeV. If we assume that all of these gamma rays deposit their energy
in the body calculate the following:
1. Absorbed dose in the body



2. Relevant equations
Absorbed dose is equal to the amount of energy absorbed per unit mass, and equivalent dose = RBE * absorbed dose


3. The attempt at a solution
Well, I worked out that 0.00234% of the body contains the radioactive substance.(0.2*.0117). Then I converted the 1.46MeV to Joules, giving me 2.339x10^-13 J. THen I multiplied this by 42736 (which is 100/0.00234) to see how much is acting on the whole unit mass of the body (1kg) and I got absorbed dose = 1.02x10^-6 Gy.
Is my working way off?
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dynamicsolo
dynamicsolo is offline
#2
May31-08, 11:26 AM
HW Helper
P: 1,664
There is one more step that you may have done, although I don't see it here. The absorbed dosage is measured in terms of unit mass of tissue, rather than unit mass of isotope. You have found the mass of K-40 present in 1 kg. of tissue and the amount of energy released per decay of one K-40 nucleus. You will also need to find the number of K-40 nuclei present in 1 kg. of tissue.

You would make the standard calculation of the number of moles of K-40 present and then multiply by Avogadro's number to obtain the number of nuclei present. You can then find

(number of K-40 nuclei/1 kg. of tissue) x (J/K-40 nucleus) = (J/1 kg. of tissue) ,

which can then be converted into rads or Grays.

(There is a further factor which estimates the biological effect of various forms of ionizing radiation, which you would multiply the absorbed dose by in order to find the effective dosage in rems or Sieverts. For gamma-radiation, though, this factor is just 1. See, for instance, http://www.radiation-scott.org/radsource/2-0.htm .)
planesinspace
planesinspace is offline
#3
Jun4-08, 05:28 AM
P: 21
I have forgotten how to work out the number of moles, is it right to go n=M/MM so that M=1kg and MM=0.0117?
in that case I got the number of nucleii to be 5.147e25,
and then multiplied that by my previous answer for energy, and got a wopping 5.245e19 J.

dynamicsolo
dynamicsolo is offline
#4
Jun4-08, 11:20 PM
HW Helper
P: 1,664

Absorbed dose of nuclear radioactivity


Sorry I haven't gotten back to you sooner, but I've been on the road (1600+ miles in the past 3 days!) and have had only intermittent 'Net access...

Remember that the fractions you have been given are percentages. So the 0.2% of the 1 kg. of tissue which is potassium is 0.002 kg. = 2.0 gm. Of that, only 0.0117% = 1.1710^-4 is K-40. So the mass of K-40 in 1 kg. of tissue is 2.3410^-4 gm.

The molar weight of K-40 will be 40 gm. per mole, so this mass of K-40 represents

2.3410^-4 gm / (40 gm/mole) = 5.8510^-6 mole ,

which is

(5.8510^-6 mole) (6.0210^23 nuclei/mole) = 3.5210^18 nuclei .

Of these, we are told 89% = 0.89 of them decay by gamma emission, which is 3.1310^18 nuclei.

This is the number of decays that are releasing 2.3410^-13 J each. You can now use this to find the total gamma emission energy release from the K-40 resident in the 1 kg. of tissue.
(It looks large, though nowhere near 10^19 J...)

It should also be kept in mind that all this energy is not released at once, but in an exponentially declining fashion, roughly over an interval of about 10 half-lives.
planesinspace
planesinspace is offline
#5
Jun5-08, 12:36 AM
P: 21
You are great at explaining without doing to much of it yourself, thankyou have you been very helpful!


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