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

Explore the concept of particle decay probability in radioactive samples over time intervals and learn how to calculate the average lifetime. Discover the rules and exponential behavior governing decay rates.

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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