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: ADecay: t

AAAAA AeNtNN

dtdN 0)( Nothing

new here

2. Daughter: B

Formation: dNB/dt = ANA

Decay: dNB/dt = BNB

NET: BBt

AABBAAB NeNNN

dtdN

A 0

This is a linear first-order differential equation

3. Solution

tB

tt

AB

AAB

BBA eNeeNN

00

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 stepa. No Equilibrium t > tB > tA B Cb. Transient Equilibrium t > tA > tB A Bc. Secular Equilibrium tA >> t > tB A B

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

C. No Equilibrium

Daughter is rate-determining stept1/2 (B) > t1/2 (A)

a. Curve a: Total activity of initially PUREsample of ANOTE: 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

BeNN

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 stept1/2 (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 Activitya. 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. ImportanceMedical isotopes – Milking a cow – how long must one wait before

extracting the daughter activity again?

c. Example: 55137Cs 1 50

137mBa m30 y

2 6.137Ba (stable)

min6.2/693.0]min/103.5min)6.2/30ln[(

30693.0

6.2693.0

)](/)(ln[)/ln( 52/12/1

maxyy

ym

BtAttAB

AB

tmax = 58 min

3. Long‐Time Solution: Curve e

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

0 tBe

t

AB

AAB

AeNN

0

AB

AAB

NN

AB

A

A

B

NN

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

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 SampleNOTE: 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 ee tA

BABAB

t

B

AAB

BeNN

1

0

Growth curve

b. Long-time solution

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

00

AABBB

AAB NNNN

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 10-3g226 g / mole

0

1 20t / (

(

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 0AtotalB AAA

Weigh and determine from AA = cNA0

weighMeasure counts

d. Half-life of B at t = t1/2(B) ; e Bt = 1/2

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

dNdt

c = BNB cNc

N2O4

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

or 1

t 1t 1

t 1t1/2 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 NA

Atotal total

3. Example: 40K ; t1/2 = 1.28 109 y

BR(EC) = (0.107) = ttEC

1 2/ ; tEC = 1.19 1010 y

BR() = 0.893) = tt1 2/

; t = 1.43 109 y

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