Saving a young man's life
Finally lets have fun with Physics!!!
Lets say a wonderful young man weighting 80 Kg is diagnosed with an early stage cancer, goes into the doctors office and receives 1.0 mSv of the radioactive isotope 137^Cs. Where 137^Cs radiates .66 Mev gamma ray photons. Lets find out how many gamma rays are taken in by the young man.
Note:
Radiation type RBE
X rays 1
Gamma rays 1
Beta particles 1-2
Neutrons 5-20
Alpha particles 20
Step one: Lets state what we know
The RBE for Gamma rays is = 1
Dose equivalent (Sv) = 1 mSv x .001 because its in millisievert mSv
.66 Mev = energy of each proton taken in by the man
Dose equivalent (Sv) = absorbed dose (Gy) x RBE so lets rearrange the equation to solve for absorbed dose (Gy)
Absorbed dose (Gy) = dose equivalent (Sv) / RBE
Absorbed dose (Gy) = .001 Gy = .001 J/kg
So what this means is that for a whole body exposure of a 80 Kg man the energy ejected from the radioactive isotope is .001 J/Kg x 80 Kg = .08 J
Understanding that .08 J are absorbed by the body of the man, now we need find the number of protons in the .08 J.
By solving for joules we get N = .08 J / ((6.6 x 10^5 eV/photon)(1.6 x 10^-19 J/eV))
We get that N = 7.57 x 10^11 photons
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