Nuclear kinetics Transient and Secular Equilibrium...
Transcript of Nuclear kinetics Transient and Secular Equilibrium...
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)