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Suppose a given particle has a 0.01 probability of decaying in any given  sec .

Suppose a given particle has a 0.01 probability of decaying in any given  sec. Does this mean if we wait 100 sec it will definitely have decayed?. If we observe a large sample N of such particles, within 1 sec how many can we expect to have decayed?.

josephherbert
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Suppose a given particle has a 0.01 probability of decaying in any given  sec .

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  1. Suppose a given particle has a 0.01 probability of decaying in any given sec. Does this mean if we wait 100 sec it will definitely have decayed? If we observe a large sample N of such particles, within 1 sec how many can we expect to have decayed? Even a tiny speck of material can include well over trillions and trillions of atoms!

  2. # decaysN (counted by a geiger counter) the size of the sample studied t time interval of the measurement each decay represents a loss in the original number of radioactive particles fraction of particles lost Note: for 1 particle this must be interpreted as the probability of decaying. -constant This argues that: this is what themeans!

  3. If events occur randomly in time, (like the decay of a nucleus) the probability that the next event occurs during the very next second is as likely as it not occurring until 10 seconds from now. T) True.F) False.

  4. P(1)Probability of the first count occurring in in 1st second P(10)Probability of the first count occurring in in 10th second i.e., it won’t happen until the 10th second ??? P(1) = P(10) ??? = P(100) ??? = P(1000) ??? = P(10000) ???

  5. Imagine flipping a coin until you get a head. Is the probability of needing to flip just once the same as the probability of needing to flip 10 times? Probability of a head on your 1st try, P(1) = Probability of 1st head on your 2nd try, P(2) = Probability of 1st head on your 3rd try, P(3) = 1/2 1/4 1/8 Probability of 1st head on your 10th try, P(10) = (1/2)10 = 1/1024

  6. What is the total probability of ALL OCCURRENCES? P(1) + P(2) + P(3) + P(4) + P(5) + ••• =1/2+1/4 + 1/8 + 1/16 + 1/32 + ••• ?

  7. A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is enough? What is the probabilitythatone rollis enough? 1/6 What is the probability that it will take exactly 2 rolls? (probability of miss,1st try)(probability of hit)= What is the probability that exactly 3 rolls will be needed?

  8. imagine the probability of decaying within any single second is p = 0.10 the probability of surviving that same single second is p = 0.90 P(N) probability that it decays in the Nth second (but not the preceeding N-1 seconds) P(1) = 0.10 = P(2) = 0.90  0.10 = P(3) = 0.902 0.10 = P(4) = 0.903 0.10 = P(5) = 0.904 0.10 = P(6) = 0.905 0.10 = P(7) = 0.906 0.10 = P(8) = 0.907 0.10 = P(9) = 0.908 0.10 =

  9. Probability of living to time t=N sec, but decaying in the next second (1-p)Np Probability of decaying instantly (t=0) is? Probability of living forever (t ) is? 0 0

  10. We can calculated an “average” lifetime fromS (N sec)×P(N) N=1 (1 sec)×P(1)= (2 sec)×P(2)= (3 sec)×P(3)= (4 sec)×P(4)= (5 sec)×P(5)= sum=3.026431 sum=9.690773 sum=8.3043 sum=6.082530 sum=9.260956 sum=9.874209

  11. the probability of decaying within any single second p = 0.10 = 1/10 = 1/t where of course t is the average lifetime (which in this example was 10, remember?)

  12. The probability that the particle has decayed after waiting N seconds must be cumulative, i.e. Probability has decayed after2seconds:P(1)+P(2) after3seconds: P(1)+P(2)+P(3) after4seconds: P(1)+P(2)+P(3)+P(4) P(1)= P(1)+P(2)= P(1)+P(2)+P(3)= P(1)+P(2)+P(3)+P(4)=

  13. This exponential behavior can be summarized by the rules for our imagined sample of particles fraction surviving until timet= e-lt fraction decaying by timet= (1 - e-lt ) where  = 1/t (and tis the average lifetime)

  14. or probability of surviving through to time t then decaying that moment (within t and t)

  15. Number surviving Radioactive atoms logN time

  16. probability of still surviving by the time t # decaying within dt Notice: this varies with time! This is the number that survive until t but then decay within the interval t and t + dt Thus the probability of surviving until t but decaying within t and t + dt

  17. the probability of surviving until t but decaying within t and t + dt If then gives the average “lifetime” 0 - 0 = 1/

  18. What is directly measured is a “disintegration rate” or ACTIVITY but only over some interval t of observation:

  19. If t <<  then and i.e., the Activity  N and N(t) doesn’t vary noticeably Your measured count is just

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