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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
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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
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Tags
inpp, particle-physics, poisson
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Difficulty
(2, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Decoration