Poisson statistics and radioactive decay
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
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
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:
Consider a source containing e radioactive nulei having a long lifetime. The probability that one nucleus decays in one second is . abcliste abc Calculate the average number of decays per second from this source and its standard deviation. abc Calculate the probability that in one second there are no decays. abc Calculate the probability that in one second there are multiple nuclei decaying. abc Calculate the probability that in twenty seconds there are no decays. abcliste
Solution:
abcliste abc The average number of decays per second is simply given by mu pN e . The standard deviation of this distribution is thus sigma sqrtmu sqrt. . Note that because of the small statistics the ration fracsigmamuapprox . is large. abc The probability that in one second there are no decays at all can be calculated using the Poisson distribution Pn fracmu^ne^-mun! for mu. and n P frac.^e^-.! . abc The probability that in one second there are more than one decays can be calculated knowing that the total probability to have either or or ... decays is unity. Hence Pn -P-P -frac.^e^-.!-frac.^e^-.! -.-. . abc The distribution remains Poissonian but the expected number of decays increases ba a factor of to mu. The ratio fracsigmamuapprox . is smaller compared to previous situation. Therefore the probability to count zero events is P frac^e^-! . abcliste
Consider a source containing e radioactive nulei having a long lifetime. The probability that one nucleus decays in one second is . abcliste abc Calculate the average number of decays per second from this source and its standard deviation. abc Calculate the probability that in one second there are no decays. abc Calculate the probability that in one second there are multiple nuclei decaying. abc Calculate the probability that in twenty seconds there are no decays. abcliste
Solution:
abcliste abc The average number of decays per second is simply given by mu pN e . The standard deviation of this distribution is thus sigma sqrtmu sqrt. . Note that because of the small statistics the ration fracsigmamuapprox . is large. abc The probability that in one second there are no decays at all can be calculated using the Poisson distribution Pn fracmu^ne^-mun! for mu. and n P frac.^e^-.! . abc The probability that in one second there are more than one decays can be calculated knowing that the total probability to have either or or ... decays is unity. Hence Pn -P-P -frac.^e^-.!-frac.^e^-.! -.-. . abc The distribution remains Poissonian but the expected number of decays increases ba a factor of to mu. The ratio fracsigmamuapprox . is smaller compared to previous situation. Therefore the probability to count zero events is P frac^e^-! . abcliste
Meta Information
Exercise:
Consider a source containing e radioactive nulei having a long lifetime. The probability that one nucleus decays in one second is . abcliste abc Calculate the average number of decays per second from this source and its standard deviation. abc Calculate the probability that in one second there are no decays. abc Calculate the probability that in one second there are multiple nuclei decaying. abc Calculate the probability that in twenty seconds there are no decays. abcliste
Solution:
abcliste abc The average number of decays per second is simply given by mu pN e . The standard deviation of this distribution is thus sigma sqrtmu sqrt. . Note that because of the small statistics the ration fracsigmamuapprox . is large. abc The probability that in one second there are no decays at all can be calculated using the Poisson distribution Pn fracmu^ne^-mun! for mu. and n P frac.^e^-.! . abc The probability that in one second there are more than one decays can be calculated knowing that the total probability to have either or or ... decays is unity. Hence Pn -P-P -frac.^e^-.!-frac.^e^-.! -.-. . abc The distribution remains Poissonian but the expected number of decays increases ba a factor of to mu. The ratio fracsigmamuapprox . is smaller compared to previous situation. Therefore the probability to count zero events is P frac^e^-! . abcliste
Consider a source containing e radioactive nulei having a long lifetime. The probability that one nucleus decays in one second is . abcliste abc Calculate the average number of decays per second from this source and its standard deviation. abc Calculate the probability that in one second there are no decays. abc Calculate the probability that in one second there are multiple nuclei decaying. abc Calculate the probability that in twenty seconds there are no decays. abcliste
Solution:
abcliste abc The average number of decays per second is simply given by mu pN e . The standard deviation of this distribution is thus sigma sqrtmu sqrt. . Note that because of the small statistics the ration fracsigmamuapprox . is large. abc The probability that in one second there are no decays at all can be calculated using the Poisson distribution Pn fracmu^ne^-mun! for mu. and n P frac.^e^-.! . abc The probability that in one second there are more than one decays can be calculated knowing that the total probability to have either or or ... decays is unity. Hence Pn -P-P -frac.^e^-.!-frac.^e^-.! -.-. . abc The distribution remains Poissonian but the expected number of decays increases ba a factor of to mu. The ratio fracsigmamuapprox . is smaller compared to previous situation. Therefore the probability to count zero events is P frac^e^-! . abcliste
Contained in these collections:

