Nuclear Decay kinetics : Transient and Secular Equilibrium

What can we say about the plot to the right?

IV. Parent-Daughter Relationships

Key point: The Rate-Determining Step

A. Case of Radioactive Daughter

A (Parent)

B

(Daughter)

t1/2(A)

C

(Grand-

daughter)

t1/2(B)

210Bi

210Po

5.01d -

206Pb

(stable)

138.4d

Same problem as a stepwise chemical reaction!

B. Mathematics of the Problem

1. Parent: A

Decay: t

AAAAA AeNtNN

dt

dN 0)( Nothing

new here

2. Daughter: B

Formation: dNB/dt = ANA

Decay: dNB/dt = BNB

NET: BB

t

AABBAAB NeNNN

dt

dNA

0

This is a linear first-order differential equation

3. Solution

t

B

tt

AB

AAB

BBA eNeeN

N

0

0

If the sample is pure A initially, second term vanishes since 00 BN

4. Daughter: C

CBAA NNNN 0 Conservation of atoms

5. Special Cases: Long time solutions

classification time relationship Rate-determining step

a. No Equilibrium t > tB > tA B C

b. Transient Equilibrium t > tA > tB A B

c. Secular Equilibrium tA >> t > tB A B

(c. is case of U-Th decay series)

C. No Equilibrium

Daughter is rate-determining step

t1/2 (B) > t1/2 (A)

a. Curve a: Total activity of initially PURE

sample of A

NOTE: resembles two-component independent

decay curve. NOT !!!

A(total) = A(A) + A(B)

b. Curve b: Parent activity, A(A)

c. Curve d: Daughter activity, A(B) ; note

growth curve

1. Figure 3-4 (on right)

2. Curve c: Long-term behavior: t > > t1/2 (A)

Therefore, t/t1/2(A) =

0 tAe

t

AB

AAB

BeN

N

0

Note: B < A ; ()() = +

3. Long term curve:All of A has disappeared ; this defines t1/2(B)

D. Transient Equilibrium

Parent decay is rate-determining step

t1/2 (A) > t1/2 (B)

1. Fig. 3-2 (on right)

a. Curve a: Total activity of initially pure

sample of A

A decay curve that initially increases with

time is a signature of transient or secular

equilibrium.

A(total) = A(A) + A(B)

b. Curve b: Parent Activity: A(A)

c. Curve d: Daughter Activity: A(B)

d. Initial growth of A(total) and then decay

according to the half-life of the parent.

2. Maximum Daughter Activity

a. Maximum occurs when dNB/dt = 0

Therefore, differentiate equation for NB and set this equal to zero; this

defines tmax as

NB is maximum when t = tmax

AB

ABt

)/ln(max

b. Importance

Medical isotopes – Milking a cow – how long must one wait before

extracting the daughter activity again?

c. Example: 55137

Cs 1

50137m

Ba

m30 y

2 6.137

Ba (stable)

min6.2/693.0

]min/103.5min)6.2/30ln[(

30

693.0

6.2

693.0

)](/)(ln[)/ln( 5

2/12/1max

yy

ym

BtAtt

AB

AB

tmax = 58 min

3. Long-Time Solution: Curve e

a. For initially pure A, t > > t1/2(B)

0 tBe

t

AB

AAB

AeN

N

0

AB

AAB

NN

AB

A

A

B

N

N

i.e., at long time, ratio NB/NA is CONSTANT with time SYSTEM APPEARS TO BE IN EQUILIBRIUM

4. Consequences

a. Long term decay is governed by parent

b. Activity: multiply equation in 3a. above by cB

cBNB = = AB =

B

A B

AA

cBANA

B - A

c. Half-life of B can be determined by

combining:

long term behavior – t1/2(A)

activity ratio above

E. Secular Equilibrium (Fig. 3-3)

t1/2(A) > > t > > t1/2 (B)

Special Case of Transient Equilibrium; e.g., U-Th Decay Series

Rn gas problem; natural background

from U and Th decay products; dating

1.

a. Curve a – Total activity of

initially PURE Sample

NOTE: at long time, ACTIVITY IS

CONSTANT ; signifies special

case

b. Curve b – Parent activity ; since

t/t1/2 ~ 0 , N/t = N0 = constant

c. Curve d – Daughter activity growing in

2. General Solution

a. Assume NB0 = 0 (i.e., pure A) ; t1/2(A) > > t

10

eetA

BABAB

t

B

AAB

BeN

N

1

0

Growth curve

b. Long-time solution

t >> t1/2(B) ; eBt e = 0

00

AABB

B

AAB NN

NN

c. If cA = cB AB = AA = Atotal/2

d. Example:

How many atoms of 222Rn are present in an initially pure sample

of 226Ra after 3 months? Assume 226 mg of 226Ra;

What is the activity?

t1/2(222Rn) = 3.82 d ; t1/2(

226Ra) = 1620 y

NRn =

Ra

Rn 1/2

N (Ra)Rn)

t Ra)N (Ra) =

(3.82 d)

1620 y (365 d / y)

226 10-3

g

226 g / mole

01 2

0t / (

(

NRn = 3.94 1015

atoms = 6.54 109

moles = 1.46 104

mL @ STP

dNdt

N =3.82 d

3.94 10 atoms

1440 m/ dRn

15

0693.

≅ 5.0 1011

d/min

Solution :

3. Points to keep in mind

a.

b.

c. Half-life of A

.0 constAA AA 0

AtotalB AAA

Weigh and determine from AA = cNA0

weighMeasure

counts

d. Half-life of B

at t = t1/2(B) ; e Bt= 1/2

NN

= ANA0

2 or AB AB =

A A

0

B BA0

( / ) ,1 1 2

1

2

corresponds to t1/2 (B)

Since A(total) = A AA0

B , t1/2 (B) is also time at which

A(total) = (3/2) AA0

F. Several Successive Decays

A B C D, etc.

1. Bateman Solutions

dN

dtc = BNB cNc

IV. Branching Decay

Competitive Decay Modes

for the same nucleus

EC

2NO+O2

N2O4

A. Total Probability = = 1 + 2 + 3 + …

or 1

t 1

t 1

t 1

t1/2 1/2(1) 1/2(2) 1/2(3)

... ; ti = partial half-lives

B. Partial Half-Life

Definition: The half-life a nucleus would have if the competing decay modes were switched off. (but NO switch).

C. Determination of Partial half-lives

1. Branching Ratio: BR (Assume two branches, 1 & 2)

BR =

1 1 1 2

1 2 1total total

t (total)

t /

/ ( )

2. Measurement; if c1 = c2

BR =

1 1 1

c N

c N

A

Atotal total

3. Example: 40

K ; t1/2 = 1.28 109 y

BR(EC) = (0.107) = t

tEC

1 2/ ; tEC = 1.19 1010

y

BR() = 0.893) =

t

t1 2/

; t

= 1.43 10

9 y

NOTE: t1/2 < t1/2 () < t1/2 (EC)

V. Determination of Half-lives

Measurable range: 1023

s to ~ 1030

y (60 orders of magnitude)

A. T1/2 ≳ 1 year: Specific Activity

1. A = cN A = counts/unit time 0.693

N = weight of sample Measure =

c = detection coefficient

e.g., 238 mg of 238

U has a specific activity of ~300 dps c = 1

2. Limit: Natural background radiation; when A(sample – A(bkg) 0,

large errors

B. Decay Curves: 10 y ≳ t1/2 ≳ 1 s

C. Electronic Techniques

D. Doppler Shift

E. Channeling in Crystals

F. Heisenberg Uncertainty Principle

G. Angular Distributions of Emitted Particles

H. Small Angle Correlations

t1/2

1

0.5 t1/2

t

1n A