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 sechow 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 ofthe measurement

NdN each decay represents a loss in theoriginal number of radioactive particles

NdN / fraction of particles lost

Note: for 1 particle this must be interpreted as the probability of decaying.

This argues that: dt

NdN /constant

this is what the means!

If events occur randomly in time,(like the decay of a nucleus)

the probability that the next eventoccurs during the very next second

is as likely as it not occurringuntil 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 secondi.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 oncethe 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?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)=

36

5

6

1

6

5

What is the probability that exactly 3 rolls will be needed?

0.10.09

0.0810.0729

0.065610.059049

0.05314410.04782969

0.043046721

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) = 0.902 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 =

P(N)probability

that it decaysin the Nth

second(but not thepreceeding

N-1seconds)

p = 0.90

Probability of Decaying in the Nth Second

0

0.02

0.04

0.06

0.08

0.1

0.12

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Number of Seconds

Pro

bab

ilit

y

Series1

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

0.10.18

0.2430.2916

0.328050.3542940.3720090.3826380.387420.38742

0.3835460.3765730.3671580.3558610.3431520.3294260.3150130.3001890.285180.27017

0.2553110.2407220.2264970.2127110.1994160.1866530.1744490.1628190.1517710.1413040.1314130.1220870.1133120.1050710.0973450.0901140.0833550.0770470.0711670.065693

0.0606020.0558720.0514820.0474110.04364

0.0401490.0369190.0339340.0311770.0286320.0262840.02412

0.0221250.0202880.0185980.0170420.0156120.0142970.0130890.01198

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=3.026431

sum=6.082530

sum=8.3043 sum=9.690773

sum=9.260956 sum=9.874209

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?)

0.10.19

0.2710.3439

0.409510.468559

0.52170310.56953279

0.612579511

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 secondsmust 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)

Probability of Having Decayed by the Nth Second

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Number of Seconds

Pro

bab

ilit

y

Series1

This exponential behavior can be summarized by the rules for our imagined sample of particles

fraction surviving until time t = et

fraction decaying by time t = (1 et )

where = 1/ (and is the average lifetime)

NdtdN

teNtN )0()(

tet )(Pprobability of surviving

through to time t then decaying that moment

(within t and t)

dt

NdN / or

)(

)0( 0

tN

N

tdt

N

dN

t

N

tN )0(

)(log

teNtN 0)(N

um

ber

su

rviv

ing

Rad

ioac

tive

ato

ms

time

tNN 0logloglogN

teNtN 0)( teN

tN 0

)( probability of still surviving by the time t

teNdN 0 # 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

teN

dN 0

Thus

the probability of surviving until t but decaying within t and t + dt

teN

dN 0

Ifthe probability of surviving until t

but decaying within t and t + dt

dtet t

0then gives the average “lifetime”

ttf )( tet'g )(tetg )(1)( t'f

00| dtete tt

0 0 0|

1 te

11

0 = 1/

teNtN )0()(

What is directly measured is a “disintegration rate” or ACTIVITY

NtNeNdt

dN t )(0A

but only over some interval t of observation:

ttt A

00

00dteNNdtdt t

)1(0

000

0

t

tt

eN

NeNeNt

)1(00

teNdt t A

If t << then 1

tt and te t 1

)]1[1(00

tNdt

t A

tN 0

i.e., the Activity Nand N(t) doesn’t vary noticeably

Your measured count is just tNt 0A