Saving a young lady's life
Lets cure another patien!
A wonderful young lady weighing 65 Kg goes in for radiation therapy and is injected with a 30 (u)Ci beta emitter. The isotope’s half-life is 10 days. The Beta particles release .35MeV, and RBE is 1.5. eighty percent of the Beta particles are absorbed and twenty percent escape. Lets find the equivalent dose the young lady receives.
Step one; just as in the previous life saving example with the young man we know that Dose equivalent (Sv) = absorbed dose (Gy) x RBE.
But now we also need to consider the half life of the isotope and the time from start to final decay.
Ro= initial radioactivity
R= nuclear decay rate
No= number of radioactive atoms
Thus Ro=R x No
And the over all relationship between the radioactive decay and the number of Beta particles emitted is given by;
No = Ro/R = T x Ro = ((t (half life)) / ln2) x Ro
Step two; Lets find initial radioactive material, half life, and total number of Beta decay over time.
Ro = (30 x 10^-6 Ci) x ((3.7 x 10^10 Bq) / ( 1 Ci))
Ro = 1.1 x 10^6 Bq = 1.1 x 10^6 decays/second
Half life:
(t half life) = 10 days x (86,400 seconds / 1 day)
(t half life) = 8.64 x 10^5 seconds
Decay over time:
No = ((t half life) / (ln2)) x Ro = (8.64 x 10^5 seconds)(1.1 x 10^6 decays/second) / (ln2)=
No = 1.37 x 10^12
Step three: lets find the eighty percent absorbed by the body relevant to the .35 MeV in terms of energy
E = (.80)(1.37 x 10^12) x ((3.5 x 10^5 eV) x (1.6 x 10^-19 J)/(1 eV)) =
E = .061 J
Note that .061 J is not a lot of energy but since it is radiation we must find if it is an acceptable absorbed dose.
Step four: lets find the Absorbed dose
Absorbed dose = .061 J/ 65 Kg = 9.3 x 10^-4 Gy
So the Equivalent dose = 1.5 x ( 9.3 x 10^-4 Gy) = 1.4 mSv
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