Milk
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The following quantities appear in the problem:
Zeit \(t\) / Masse \(m\) / Energie \(E\) / Aktivität \(A\) / Organdosis \(H\) / Radius \(r\) / Oberfläche \(S\) / Zerfallskonstante \(\lambda\) / Energiedosis \(D\) / Strahlungswichtungsfaktor \(w_R\) /
The following formulas must be used to solve the exercise:
\(D = \dfrac{E}{m} \quad \) \(S = 4 \pi r^2 \quad \) \(A_t = A_0 \cdot \text{e}^{-\lambda t} \quad \) \(H = wD \quad \)
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Exercise:
Ase a liter of milk typically has an activity of pqBq due to the upbeta-emitter isotopeK. If a person drinks two glasses pq.l per day estimate the total effective dose in Sievert received in a year. As a crude model ase the milk stays in the stomach for hours and is then released. Ase also that very roughly % of the pq.MeV released per decay is absorbed by the body. Compare your result to the normal allowed dose of pqmSv. Make your estimate for a pqkg adult.
Solution:
Each decay deposits % of its energy pq.MeV in the body i.e. pq.MeV. Since the source's activity is pqBq its deposited energy is E_ pq.MeV pqBq pq.J pqs^- pq.J/s. In nine hours the body absorbes E_ E_ t pq.J. In a year this is E E_ pq.J. The absorbed energy dose of a pqkg person corresponding to this energy is D fracEm fracpq.Jpqkg pq.Gy. Since isotopeK is a upbeta-emitter both upbeta^+ and upbeta^- occur the q-factor for its radiation is electrons and photons. The equivalent dose hence is H qD pq.Sv. The absorbed dose in a year is pq.upmu Sv and is negligible to pqmSv the normal allowed dose.
Ase a liter of milk typically has an activity of pqBq due to the upbeta-emitter isotopeK. If a person drinks two glasses pq.l per day estimate the total effective dose in Sievert received in a year. As a crude model ase the milk stays in the stomach for hours and is then released. Ase also that very roughly % of the pq.MeV released per decay is absorbed by the body. Compare your result to the normal allowed dose of pqmSv. Make your estimate for a pqkg adult.
Solution:
Each decay deposits % of its energy pq.MeV in the body i.e. pq.MeV. Since the source's activity is pqBq its deposited energy is E_ pq.MeV pqBq pq.J pqs^- pq.J/s. In nine hours the body absorbes E_ E_ t pq.J. In a year this is E E_ pq.J. The absorbed energy dose of a pqkg person corresponding to this energy is D fracEm fracpq.Jpqkg pq.Gy. Since isotopeK is a upbeta-emitter both upbeta^+ and upbeta^- occur the q-factor for its radiation is electrons and photons. The equivalent dose hence is H qD pq.Sv. The absorbed dose in a year is pq.upmu Sv and is negligible to pqmSv the normal allowed dose.