(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Homework Help: Absorbed dose of nuclear radioactivity

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