Lecture 14 Nuclear Decay kinetics : Transient and Secular Equilibrium

IV. Parent-Daughter Relationships ⇒ (The Rate-Determining Step) ⇐ A. Case of Radioactive Daughter A e.g. Same problem as a stepwise chemical reaction B. Mathematics of the Problem 1. Parent: A Decay: −dN/dt = λANA ; NA = NA

0 e−λAt (nothing new)

2. Daughter: B Formation: dNB/dt = λANA Decay: dNB/dt = − λBNB

NET: dN

dt ANA BNB = ANA0 e- At - BNB

B = −λ λ λ λ λ

Linear first-order differential equation 3. Solution

N = N

e- At e- t

+ N e-

BA A

0

B AB0Bλ

λ λλ λ λ

−−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

Bt

If pure A initially, N B

0 = 0 and second term vanishes 4. Daughter: C NA

0 = NA + NB + NC

B

C

210Bi210Po

206PbStable

5.01 d β−

138.4 d α

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) 1. Fig. 3-4 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)

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

Operative Word

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

∴ t/t1/2(A) = ∞ and e−λAt

⇒ 0

OR NB = λ

λ λ

λAN

A0

B - ABt

e( )−−

Note: λB < λA ; ∴ (−)(−) = +

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

a. Curve a: Total activity of initially pure sample of A A(total) = A(A) + A(B) A decay curve that initially increases with time is a signature of transient or secular equilibrium. b. Curve b: Parent Activity: A(A) c. Curve d: Daughter Activity: A(B) d. Initial growth of A(A) and then decay according to the half-life of the parent.

2. Maximum Daughter Activity a. Maximum occurs when dNB/dt = 0 ∴ differentiate equation for NB and set this equal to zero; this defines tmax as

tmax = ln( / )λ λλ λ

B A

B A− ; i.e., NB is maximum when t = tmax

b. Importance Medical isotopes – Milking a cow – how long must one wait before extracting the daughter activity again?

c. Example: 55137Cs 1 50

137mBa m30 y

β γ⎯ →⎯ ⎯ →⎯⎯

2 6.137Ba (stable)

tmax = ln(λB/λA) = = ln [ (30y/2.6min) × 5.3 ×105 min/y] 0.693 0.693 0.693 y (0.693 y/2.6m) 2.6 m 30 y 2.6 m tmax = 58 min 3. Long-Time Solution: Curve e a. For initially pure A, t > > t1/2(B)

∴ e Bt−λ = e−∞⇒ 0

or NN

BA A

B A=

−

( )λλ λ

0

e At−λ ≅

λλ λ

λλ λ

A A

B A

B

A

A

B A

N OR N

N− −=,

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

ln[t1/2(A)/t1/2(B)]

4. Consequences a. Long term decay is governed by parent

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

cλBNB = = AB = λ

λ λB

A B−

⎛

⎝⎜

⎞

⎠⎟ AA

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 , ΔN/Δt = λN0 = constant

cλBλANA

λB - λA

c. Curve d – Daughter activity growing in 2. General Solution a. Assume NB

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

∴ NN

e- tB ~ A

0A

B

Bλ

λλ( )1−

Growth curve b. Long-time solution t >> t1/2(B) ; e−λBt ⇒ e−∞ = 0

NN

NB = A A0

BB A A

0Nλλ λ λ⇒ =B

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 = λλ

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 × 10−9 moles = 1.46 × 10−4 mL @ STP

− = ×⎛

⎝⎜⎜

⎞

⎠⎟⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

dNdt N =

3.82 d 3.94 10 atoms

1440 m/ dRn

15

λ 0693. ≅ 5.0 × 1011 d/min

3. Points to keep in mind

e At−λ≈ e0 = 1

λB > > λA ; λB − λA≅ λB

a. AA = AA

0 = constant

b. AB = Atotal − AA0

c. Half-life of A Weigh and determine from AA = cλNA

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

∴ NN

= ANA0

2 or AB AB = A A

0

B BA0λ

λλ

λ( / ) ,1 1 2

12

− = ; this time

Since A(total) = A AA

0B+ , 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 = λBNB − λcNc

2. For Secular Equilibrium -- ONLY CASE WE WILL USE THIS LONG-TIME SOLUTION t1/2 (A) > > t t1/2 (A) > t1/2 (B) > t1/2 c etc. a. λ λ λA A

0B B C CN = NN = = …

& b. −dN/dt(total = nλ A A

0N , where n is number of decays in chain

counts ⇑ wt

corresponds to t1/2 (B)

c. IF cA = cB = cC etc. AA = AB = AC … d. Marie Curie: N(238U) = [t1/2(U)/t1/2(Ra)] • NRa

IV. Branching Decay Competitive Decay Modes for the same nucleus A. Total Probability = λ λ = λ1 + λ2 + λ3 + …

or 1t 1

t 1t 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 = λ

λλ

λ1 1 1 2

1 2 1total total

t (total)t= = /

/ ( )

2. Measurement; if c1 = c2

BR = λλ

1 1 1× ×× × =c N

c NA

Atotal total

A 1 2 e.g. B C

1940K

1840 Ar 20

40Ca

β EC

N2O4

2NO2 2NO+O2

3. Example: 40K ; t1/2 = 1.28 × 109 y

∴ BR(EC) = (0.107) = ttEC

1 2/ ; tEC = 1.19 × 1010 y

BR(β−) = 0.893) = tt1 2/

β −; tβ− = 1.43 × 109 y

NOTE: t1/2 < t1/2 (β−) < t1/2 (EC) V. Determination of Half-lives Measurable range: 10−23 s to ~ 1030 y (60 orders of magnitude) A. T1/2 ≳ 1 year: Specific Activity 1. A = cλN A = counts/unit time 0.693 N = weight of sample Measure ⇒ λ =

c = detection coefficient e.g., 238 mg of 238U 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