n  p + e   e    e   e +  Ne *  Ne +  N  C + e   e Pu  U +  Fundamental particle decays Nuclear decays Some observed decays The transition rate, W (the “Golden Rule”) of initial  final is also invoked to understand a  b+c (+  ) decays How do you calculate an “overlap” between ???

It almost seems a self-evident statement: Any decay that’s possible will happen! What makes it possible? What sort of conditions must be satisfied? Total charge q conserved. J conserved.

Nuclear potential Coulomb potential finite (but small) probability of being found outside the nucleus at any time Tunneling always some probability of a piece of the nucleus escaping the nuclear potential with a STATIC POTENTIAL this probability is CONSTANT!

pdg.lbl.gov/pdgmail

probability of decaying (at any time - now or later) = constant ???? What’s this mean equally likely at any instant ???? must be expressed as a probability per unit time If we observe one, isolated nucleus it is equally likely it decays this moment  t as any other moment  t (even years from now) It either decays or it doesn’t.

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!

# 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. This argues that:  constant this is what the means!

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.

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) ???

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) = Probability of 1st head on your 10th try, P(10) = 1/2 1/4 1/8 (1/2) 10 = 1/1024

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 +

A six-sided die is rolled repeatedly until it gives a 6. What is the probability that one roll is 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?

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

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

We can calculated an “average” lifetime from  (N sec)×P(N) (1 sec)×P(1)= (2 sec)×P(2)= (3 sec)×P(3)= (4 sec)×P(4)= (5 sec)×P(5)= N=1  sum= sum= sum= sum= sum= sum=

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

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

This exponential behavior can be summarized by the rules for our imagined sample of particles fraction surviving until time t = e  t fraction decaying by time t = (1  e  t ) where = 1/  (and   is the average lifetime)

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

Number surviving Radioactive atoms time  log N

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

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

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

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