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Project PHYSNET Physics Bldg. Michigan State University East Lansing, MI

MISN-0-311

EXPONENTIAL DECAY:

OBSERVATION, DERIVATION

ln P(t)

time

number

of

decays

t

(min)

0

-1

-2

-3 0 2 4 6

e- tl

1

EXPONENTIAL DECAY: OBSERVATION, DERIVATION

by

Peter Signell

1. Nuclear Decay: Exponential

a. The Exponential Decay Law (EDL) . . . . . . . . . . . . . . . . . . . . . .1b. Decay-Constant and Mean-Life Values . . . . . . . . . . . . . . . . . . .2c. Explanation of the EDL: The 3-Line Derivation . . . . . . . 2

2. Comparison to Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3a. Result s of One Experim ent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3b. Repetitions of the Same Experiment .....................3c. N and R Equations Demand N, R . . . . . . . . . . . . . . . . . 3

3. The Probabilistic EDLa. Probability Applied to Decay .. . . . . . . . . . . . . . . . . . . . . . . . . . .4b. T he Probabilistic EDL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4c. Constant : No Internal Clock . . . . . . . . . . . . . . . . . . . . . . . . . . . 5d. The No-Aging Assumption and QM ...................5

4. Deducing Mean Life From Data . . . . . . . . . . . . . . . . . . . . . . . . 5a. T he Semi-Log Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5b. Mean Life from Number Data . . . . . . . . . . . . . . . . . . . . . . . . . . . 6c. Mean Life from Rate Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

d. Dealing wit h Dat a Scat t er . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6

Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

A. Semi-Log Plots and Exponential Behavior . . . . . . . . . . . . 7

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MISN-0-311 7

former can usually be reduced through better and/or more costly experi-mental design, but the effect of quantum fluctuations can be reduced onlyby increasing the number of data. Methods for dealing with such data-

scatter are presented elsewhere,

8

but a simple result is that the deducedmean life value is made uncertain. Graphically speaking, the randomcharacter of the fluctuations make uncertain the location of the line onthe semi-log plot. That uncertainty is transferred to the lines slope andhence to the mean life. In many experimental situations, however, thedata are so large in number that they make and t accurate enough forpractical applications.

Acknowledgments

Preparation of this module was supported in part by the NationalScience Foundation, Division of Science Education Development and Re-search, through Grant #SED 74-20088 to Michigan State University.

A. Semi-Log Plots and Exponential Behavior

An exponential decay function produces a straight line on a semi-log

plot, and the slope of that line is the negative of the decay constant where, for example,

f(x) = f(0) exp(x) .

To prove these relationships, we take the natural logarithm of the aboveequation, getting:

n f(x) = n f(0) x . Help: [S-11]

We now define a new dependent variable,

y(x) n f(x) ,

which linearizes the relationship between the new dependent and inde-pendent variables:

y(x) = y(0) x .

Thus plotting y vs. x on a graph produces a straight line whose slope is.

8See Linear Least Squares Fits to Data (MISN-0-162, UC).

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MISN-0-311 AS-1

SPECIAL ASSISTANCE SUPPLEMENT

QUESTIONS:

1. By direct substitution of t = 0 in Eq. 1, determine and check themeaning of N(0).

2. Why does Eq. 2 have a minus sign on the right hand side?

3. Verify that Eq.3 follows from Eqs. 1 and 2.

4. Explain the correspondence between data and the exponential decaylaw, including time-bin fluctuations and the changes in their signifi-cance as the number of systems is increased.

5. Justify Eq. 4 qualitatively.

6. Show that Eq. 5 follows from the preceding equations.

7. Verify Eq.6.

8. Verify Eq.7.

9. Verify Eq.8.

10. Sketch a curve, on a linear plot, which is an exponential function,and a curve which is not. On a semi-log plot, sketch a curve whichrepresents an exponential function, and one which does not.

11. From the observations shown, determine the mean life for the speciesbeing observed. Help: [S-6]

ln N(t)

t

0

-0.10

-0.20

-0.30

1s 2s 3s

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MISN-0-311 AS-2

12. From the three rate observations reported below, show graphicallythat the data are in agreement with exponential decay. Determinethe speciess mean life. [S-8]

t R(t)0 s 2638/sec1 s 755/sec2 s 216/sec

Brief Answers:

11. 10s

12. 0.799 s

SEQUENCES:

S-1 (from TX, 1c and [S-2])

R(t) dN(t)

dt=

d

dtN(0)et = N(0) d

dtet

= N(0)()et = N(0)et = N(t)

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MISN-0-311 AS-3

S-2 (from TX,1c)

Input:

(i) Assumption: Decay rate is proportional to number remaining.Stated as an equation: R(t) = N(t). const. of propor-tionality.

(ii) Rate-number relationship: R(t) dN(t)

dt

3-Line Derivation:Combining the input, dN(t)/dt = N(t), solution is: N(t) =c exp(t) where c is the single constant of integration. Proof: by

differentiation. Help: [S-1]Substitute t = 0 and find c = N(0) hence: N(t) = N(0)exp(t)

Alternative 6-Line Derivation:dN(t)

dt= N(t)

dN

N= dt

n N(t) n N(0) = (t 0)

n

N(t)N(0)

= t

N(t)

N90)= et

N(t) = N(0)et

S-3 (from TX, 3b)

Use P(t) = N(t)N(0)

and r dPdt

in the Input and Derivation steps of

[S-2], converting all Ns and Rs to Ps and rs. Finally, use the 100%probabilty of being undecayed at time zero.

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MISN-0-311 AS-4

S-4 (from TX, 4c)

For decay, dN/dt is negative whereas the decay rate, the number of

decays per unit time, is positive.Mathematically: N(t) is the number of undecayed particles at time t.Then the number of decayed particles is: ND(t) = N(0) N(t). Therate of decay is:

R(t) =dND(t)

dt=

dN(t)

dt

S-5 (from TX, 1b)

The mean value of a speciess lifetime is produced by weighting eachtime t with the number of systems N(t) enjoying that time as part oftheir undecayed lives:

t =

0tN(t) dt

0N(t) dt

Thus t = 1/ when N(t) = N(0) exp(t). Help: [S-7]

S-6 (from AS, [Q-11])

Slope =(n N)

t=.10

1s= 0.10/sec= . t =

1

= 10s.

S-7 (from AS, [S-5])

Substitute N(t) into the integrals and integrate. Note that:

0

tet dt = d

d

0

et dt = d

d

1

= 2

You should remember this trick as well as the value of

0exp(t) dt.

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MISN-0-311 AS-5

S-8 (from AS, [Q-12])

n 2638 = 7.878

> difference = 1.251n 755 = 6.627> difference = 1.251

n 216 = 5.375

slope =(n R)

t=1.251

1 s= t = 1/ = 0.799 s.

S-9 (from TX, 4b)

slope = (n P)t

= 35min

= t = 1/ = 1.67min

S-10 (from TX, 4c)

N(t) = N(0) exp(t)R(t) = dN(t)/dt = N(0)exp(t)Then: R(0) = N(0)Hence: R(t) = R(0)exp(t)

which is indeed the form shown for N(t) in the top line above.

S-11 (from Appendix A)

ln (AB) = ln(A) + ln(B); ln [exp(C)] = C

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MISN-0-311 ME-1

MODEL EXAM

1. Which of the curves below could not correspond to y(x) being a de-caying exponential function of x? Justify your choice(s).

ln f(x)f(x)

x x

A

B

C

E

F

G

HD

2. From the three observations reported below, show graphically that thedata are in agreement with exponential decay and determine the meanlife of the species.

t N(t)

0 min. 6392 min. 2174 min. 73

3. Describe the relationship between the exponential decay law and ex-perimental data for a finite number of decaying systems.

4. Derive the exponential decay law for the number of undecayed systems,starting from the no aging (constancy of decay constant) assumption.

Brief Answers:

1. A, B, C, E, F, H

2. 1.85 min, with a small experimental fluctuation.

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