Exercise
https://texercises.com/exercise/whole-body-dose-2/
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The following quantities appear in the problem: Zeit \(t\) / Masse \(m\) / Energie \(E\) / Aktivität \(A\) / Äquivalentdosis \(H\) / Radius \(r\) / Oberfläche \(S\) / Zerfallskonstante \(\lambda\) / Energiedosis \(D\) /
The following formulas must be used to solve the exercise: \(D = \dfrac{E}{m} \quad \) \(S = 4 \pi r^2 \quad \) \(A_t = A_0 \cdot \text{e}^{-\lambda t} \quad \) \(H = qD \quad \)
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Exercise:
A kg laboratory worker is exposed to a isotopeCo source with .eBq activity. During .h of working he receives a wholbody dose of .mSv. Ase the worker's body has a cross-sectional area of .metersquared. isotopeCo emits upgamma rays of energy .MeV and .MeV in quick succession approximately % of which eract with the body and deposit all their energy. Calculate the average distance of the person to the source!

Solution:
If the equivalent dose is provoked by upgamma-radiation q then the absorbed dose is: D fracHq frac.Sv .Gy Since the energy is absorbed uniformely over the whole body of the worker the absorbed energy must be: E D m fracHq m .Gy kg .J One upgamma-decay leads to two rays in quick succession one of these doublrays carrying an energy of: E_ E' + E'' .MeV + .MeV .MeV .J The worker's body therefore did absorb N_w' fracEE_ fracH mq E' + E'' frac.J.J numpr.e doublrays during the hours work which is only % of the ones hitting in his direction i.e.: N_w fracN_w'eta fracH meta q E' + E'' numpr.e In the same time the source did emit N A t .eBq .es numpr.e upgamma-rays. Since N_a corresponds to the body's cross-sectional area and N corresponds to the sphere the source radaites uniformely o we can calculate the sphere's area on whose surface the worker is: A fracNN_w A_w fracN eta q E' + E''H m A_w fracnumpr.enumpr.e .metersquared .emetersquared The workers distance to the source is hence: r sqrtfracApi sqrtfraceta q N E' + E''pi m H A_w .em boxbox r sqrtfraceta q N E' + E''pi m H A_w .em
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Exercise:
A kg laboratory worker is exposed to a isotopeCo source with .eBq activity. During .h of working he receives a wholbody dose of .mSv. Ase the worker's body has a cross-sectional area of .metersquared. isotopeCo emits upgamma rays of energy .MeV and .MeV in quick succession approximately % of which eract with the body and deposit all their energy. Calculate the average distance of the person to the source!

Solution:
If the equivalent dose is provoked by upgamma-radiation q then the absorbed dose is: D fracHq frac.Sv .Gy Since the energy is absorbed uniformely over the whole body of the worker the absorbed energy must be: E D m fracHq m .Gy kg .J One upgamma-decay leads to two rays in quick succession one of these doublrays carrying an energy of: E_ E' + E'' .MeV + .MeV .MeV .J The worker's body therefore did absorb N_w' fracEE_ fracH mq E' + E'' frac.J.J numpr.e doublrays during the hours work which is only % of the ones hitting in his direction i.e.: N_w fracN_w'eta fracH meta q E' + E'' numpr.e In the same time the source did emit N A t .eBq .es numpr.e upgamma-rays. Since N_a corresponds to the body's cross-sectional area and N corresponds to the sphere the source radaites uniformely o we can calculate the sphere's area on whose surface the worker is: A fracNN_w A_w fracN eta q E' + E''H m A_w fracnumpr.enumpr.e .metersquared .emetersquared The workers distance to the source is hence: r sqrtfracApi sqrtfraceta q N E' + E''pi m H A_w .em boxbox r sqrtfraceta q N E' + E''pi m H A_w .em
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biology, cells, physics, radioactivity, radiometry
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(3, default)
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3 (default)
Language
ENG (English)
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Calculative / Quantity
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