Whole-body dose
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
Need help? Yes, please!
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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Don't forget to subscribe to our channel, like the videos and leave comments!
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
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
Meta Information
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
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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