Decay Chain
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
\(\LaTeX\)
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Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
A radioactive isotope with a decay constant caO decays o a second radioative isotope with a decay constant cbO. In the ning there are mO of each nuclide. abcliste abc Calculate the remaining masses after tO. abc Graph the masses as a function of time. abcliste
Solution:
abcliste abc The system is defined by the matrix bf A leftmatrix-c_ & c_ & -c_matrixright with eigenvalues lambda_c_ and lambda_c_ and eigenvectors bf v_ leftmatrix fracc_c_-c_matrixright bf v_ leftmatrix matrixright where c_ and c_ are the decay constants of the two isotopes see notes. For the general solution we have to solve the leftmatrixm_ m_matrixright a_ bf v_+a_ bf v_ leftmatrixa_ fraca_ c_c_-c_+a_matrixright The obvious solution is a_ m_ a_ m_left-fracc_c_-c_right so the general solution for the decay can be written as m_t maF m_t mbF The numerical results are m_ mtimes e^-catimes t resultmaS m_ mtimes fraccacb-ca e^-ca times t+mtimesleft-fraccacb-caright e^-cbtimes t resultmbP abc The graph confirms the results in a. center includegraphicswidthtextwidth#image_path:two-stagdecay-chain# center abcliste
A radioactive isotope with a decay constant caO decays o a second radioative isotope with a decay constant cbO. In the ning there are mO of each nuclide. abcliste abc Calculate the remaining masses after tO. abc Graph the masses as a function of time. abcliste
Solution:
abcliste abc The system is defined by the matrix bf A leftmatrix-c_ & c_ & -c_matrixright with eigenvalues lambda_c_ and lambda_c_ and eigenvectors bf v_ leftmatrix fracc_c_-c_matrixright bf v_ leftmatrix matrixright where c_ and c_ are the decay constants of the two isotopes see notes. For the general solution we have to solve the leftmatrixm_ m_matrixright a_ bf v_+a_ bf v_ leftmatrixa_ fraca_ c_c_-c_+a_matrixright The obvious solution is a_ m_ a_ m_left-fracc_c_-c_right so the general solution for the decay can be written as m_t maF m_t mbF The numerical results are m_ mtimes e^-catimes t resultmaS m_ mtimes fraccacb-ca e^-ca times t+mtimesleft-fraccacb-caright e^-cbtimes t resultmbP abc The graph confirms the results in a. center includegraphicswidthtextwidth#image_path:two-stagdecay-chain# center abcliste
Meta Information
Exercise:
A radioactive isotope with a decay constant caO decays o a second radioative isotope with a decay constant cbO. In the ning there are mO of each nuclide. abcliste abc Calculate the remaining masses after tO. abc Graph the masses as a function of time. abcliste
Solution:
abcliste abc The system is defined by the matrix bf A leftmatrix-c_ & c_ & -c_matrixright with eigenvalues lambda_c_ and lambda_c_ and eigenvectors bf v_ leftmatrix fracc_c_-c_matrixright bf v_ leftmatrix matrixright where c_ and c_ are the decay constants of the two isotopes see notes. For the general solution we have to solve the leftmatrixm_ m_matrixright a_ bf v_+a_ bf v_ leftmatrixa_ fraca_ c_c_-c_+a_matrixright The obvious solution is a_ m_ a_ m_left-fracc_c_-c_right so the general solution for the decay can be written as m_t maF m_t mbF The numerical results are m_ mtimes e^-catimes t resultmaS m_ mtimes fraccacb-ca e^-ca times t+mtimesleft-fraccacb-caright e^-cbtimes t resultmbP abc The graph confirms the results in a. center includegraphicswidthtextwidth#image_path:two-stagdecay-chain# center abcliste
A radioactive isotope with a decay constant caO decays o a second radioative isotope with a decay constant cbO. In the ning there are mO of each nuclide. abcliste abc Calculate the remaining masses after tO. abc Graph the masses as a function of time. abcliste
Solution:
abcliste abc The system is defined by the matrix bf A leftmatrix-c_ & c_ & -c_matrixright with eigenvalues lambda_c_ and lambda_c_ and eigenvectors bf v_ leftmatrix fracc_c_-c_matrixright bf v_ leftmatrix matrixright where c_ and c_ are the decay constants of the two isotopes see notes. For the general solution we have to solve the leftmatrixm_ m_matrixright a_ bf v_+a_ bf v_ leftmatrixa_ fraca_ c_c_-c_+a_matrixright The obvious solution is a_ m_ a_ m_left-fracc_c_-c_right so the general solution for the decay can be written as m_t maF m_t mbF The numerical results are m_ mtimes e^-catimes t resultmaS m_ mtimes fraccacb-ca e^-ca times t+mtimesleft-fraccacb-caright e^-cbtimes t resultmbP abc The graph confirms the results in a. center includegraphicswidthtextwidth#image_path:two-stagdecay-chain# center abcliste
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