Transcript Radioactive Dating Techniques

M ( A, Z )  Zm p  ( A  Z )mn  B( A, Z )c 2
B( A, Z )  av A  as A
2/3
1 / 3
 ac Z ( Z  1) A
 asym ( A / 2  Z ) / A   ( A, Z )
2
av=15.5 MeV
ac=0.72 MeV
ap=34.0 MeV
as=16.8 MeV
asym=23.0 MeV
For a fixed value of A :
1 / 3
B( A, Z )  aCv 1A  as A
C2  ac Z ( Z  1) AC3
2/3
2
 asym
(
A
/
2

Z
)
/ A   ( A, Z )
C4 D
B(A,Z) is clearly a quadratic function of Z
42Mo
A = 104 isobars
43Tc
103.912
48Cd
Mass, u
103.910
47Ag
103.908
45Rh
103.906
44Ru
46Pd
103.904
42
44
46
Atomic Number, Z
48
For a fixed value of A :
1 / 3
B( A, Z )  aCv 1A  as A
C2  ac Z ( Z  1) AC3
2/3
2
 asym
(
A
/
2

Z
)
/ A   ( A, Z )
C4 D
B(A,Z) is clearly a quadratic function of Z

ap
A3 / 4
 ( A)  0

ap
A3 / 4
even Z, N
Z=N=0
odd Z, N
42Mo
A = 104 isobars

43Tc
103.912
?A

XN  Z+1
Y
+

?
? N-1
A
A
e-capture: ZX N + e  Z-1YN+1
A
-decay: Z
Mass, u
103.910
Odd Z
47Ag
103.908
45Rh
Even Z
103.906
48Cd
44Ru

46Pd
103.904
42
44
46
Atomic Number, Z
48
Peaks at ~8.795 MeV near A=60
Binding energy per nuclear
particle (nucleon) in MeV
for A>50
~constant
8-9 MeV
Mass Number, A
Is 236
94 Pu unstable to -decay?
Pu 
U+ 
236
94
232
92
236
94
M(
M(
Pu)c2
4
2
232
92
4
2
U)c + M( )c2 +
2
Q
Q = (MPu – MU  M)c2
= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u
= 5.87 MeV > 0
URANIUM DECAY SERIES

238 
92U

234 
Th
90

234 
Pa
91

234
U


230 
Th
90

226 
Ra
88
92
234
92U
222 
 84Po218
86Rn

214 
Pb
82
214
83Bi



214 
Po
84
210
82Pb

210 
Pb
82
210
83Bi



210 
Po
84
206
82Pb
“Uranium I”
“Uranium II”
“Radium B”
“Radium G”
4.5109 years
2.5105 years
U238
U234
radioactive Pb214
stable
Pb206


214
82Pb
Radioactive parent isotopes & their stable daughter products
Radioactive Parent
Potassium 40
Stable Daughter
Argon 40
Rubidium 87
Thorium 232
Uranium 235
Uranium 238
Strontium 87
Lead 208
Lead 207
Lead 206
Carbon 14
Nitrogen 14
Half Lives for Radioactive Elements
Radioactive Parent
Stable Daughter
Half life
1.25 billion yrs
Potassium 40
Argon 40
Rubidium 87
Thorium 232
Uranium 235
Strontium 87
Lead 208
Lead 207
48.8 billion yrs
14 billion yrs
704 million yrs
Uranium 238
Carbon 14
Lead 206
Nitrogen 14
4.47 billion yrs
5730 years
Growth of Radioactive Daughter Products
dN 2   1 N1dt  2 N 2dt
where decay of parent
(original) nucleus:
 1t
N1 (t )  N1,0e
if it’s radioactive
itself!
try:
N 2 (t )  Ae
 1t
 Be
note:
and if:
N 2, 0  0
N 2 (t )  A(e
dN 2
dt

 A  2e
 2t
 2t
e
 1e
 2t
N 2, 0  A  B
 1t
 1t
as we’ve seen before
)
  N  N
1
1
2
2
 2 A(e
 2t
e
 1t
)
Growth of Radioactive Daughter Products
dN 2   1 N1dt  2 N 2dt
so:
A1e
Ae
and:
 1t
 1t
 1N1  2 Ae
      N e
1
N 2 (t ) 
2
1
1
1  2
 1t
 2t
 1e
2→0
A
so:
0
N 0 (e
Note: as
 1t
 1t
1
1  2
N0
)
N 2 (t )  N 0 (1  e
 1t
)
as seen before.
As N1(t)
and N2(t)
may reach the state
so that:
dN 2  0
1 N1  2 N 2
Note:
some elements have both radioactive
and non-radioactive isotopes.
Examples: carbon, potassium.
Just saw: 3 isotopes of uranium.
238U the most abundant (99.2739%)
Radioactive elements tend to become concentrated
in the residual melt that forms during the crystallization
of igneous rocks.
More common in
SIALIC rocks
(granite, granite pegmatite)
and continental crust.
Radioactive isotopes don't tell much about the age of
sedimentary rocks (or fossils).
radioactive minerals in sedimentary rocks
derived from the weathering of igneous rocks
Thus: dating sedimentary rock gives
the time of cooling of the magma
that formed the original igneous rock.
tells us nothing about when the sedimentary rock formed.
To date a sedimentary rock, it is necessary to isolate
a few unusual minerals (if present) which formed on
the seafloor as the rock was cemented.
Glauconite is a good example. It crystallizes under reducing
conditions that cause precipitation of minerals into sediments
Glauconite contains potassium,
so it can be dated using the
potassium-argon technique.
Minerals you can date
Most minerals containing radioactive isotopes are in igneous rocks.
The dates they give indicate the time the magma cooled.
•Potassium 40 is found in:
•potassium feldspar (orthoclase)
•muscovite
•amphibole
•glauconite (greensand; found in some
sedimentary rocks; rare)
•Uranium may be found in:
•zircon
•urananite
•monazite
•apatite
•sphene
2 different rock samples have ratios of
238U to 206Pb atoms of 1.2 and 1.8.
Compute the age of each sample.
238U
atoms remaining today:
N 0e t
The number 238U atoms decayed  number of 206Pb atoms today:
N0 (1  et )
So:
R=
238U
number
atoms
number of 206Pb atoms
1
1
t  ln(  1)
 R
 t
1
N 0e
= N (1  et )  t
e 1
0
4.468 109
1
t
ln(  1)
ln 2
R
R=1.2: 3.907109 years
R=1.8: 2.848109 years
What if there were initially some daughter products
already there when the rock was formed?
N daughter(t )  N parent(t )  N 0,daughter  N 0, parent
unknown!
Remember: elements come in many isotopes
(some even tag a specific decay series!)
If other (stable) isotopes of the daughter are also present
N d  (t )  N 0,d 
Then look at the ratio:
N daughter(t )  N parent(t )
N d  (t )

N 0,daughter  N 0, parent
N 0,d 
N daughter(t )  N parent(t )

N d  (t )
N 0,daughter  N 0, parent
N 0,d 
Which we can rewrite as:
N daughter (t )

N d  (t )
y
N d  (t )

N parent (t )
N(t)=N0et
N daughter(t )
N d  (t )
N 0, parent

=
N0=N(t)et
N parent(t )
N d  (t )
x
N d  (t )

e

 ( t t 0 )
m

1 
+
N 0,daughter
N 0 ,d 
N 0,daughter
N 0,d 
b
Rb-Sr dating method
Allows for
the presence
of initial 87Sr
G.W. Wetherill, Ann. Rev.
Nucl. Sci. 25, 283 (1975)
Age = 4.53  109 y
2 = 0.04  109 y
All samples lie along the
same line, so formed from
the same batch of magma
Some concentration ratios measured for Eagle Peak Pluton
of the Sierra Nevada Batholith
0.7248-0.7076 = 0.001213  e  ( t t )  1
14
ln 1.001213  0.001212265   (t  t0 )
t  t0  0.001212265/ 

0

t  t0  0.001212265/ 
Rubidium half-life=48.8 By = 48,800,000,000 years
t 1  ln2
2
  0.693147/ 48800000000 years
t  t0  8,530,000 years
The Sierra Nevada pluton was formed by subduction
(one tectonic plate driven beneath another) remelting
continental crust and forming volcanic rock called
basalt 8,530,000 years ago. Sure enough west of
this fault (and deeper below the basalts) are 4.55 By
samples of continental crust.
How does Carbon-14 dating work?
Cosmic rays strike Nitrogen 14 atoms in the atmosphere
and form (radioactive) Carbon 14 which
combines with oxygen to form radioactive carbon dioxide
14
14

6 C8  7 N 7  
with a half-life of 5730 years
The steady barrage (for at least 10s of thousands of years) of
cosmic cays gives the atmosphere equilibrium concentrations
12C
98.89%
13C
1.11%
14C 1 atom for every 1012 atoms of 12C
How does Carbon-14 dating work?
radioactive carbon dioxide is absorbed and used by plants.
enters the food chain & the carbon cycle.
Living things are in equilibrium with the atmosphere.
All living things contain a constant ratio of 14C to 12C
(1 in a trillion).
At death, 14C exchange ceases and any 14C in the
tissues of the organism begins to decay to Nitrogen 14,
and is not replenished by new 14C.
The change in the 14C to 12C ratio is the basis for dating.
The half-life is so short (5730 years) that this method
can only be used on materials less than 70,000 years
old. Archaeological dating uses this method.) Also useful
for dating the Pleistocene Epoch (Ice Ages).
Assumes that the rate of Carbon 14 production
(and hence the amount of cosmic rays striking the Earth)
has been constant (through the past 70,000 years).
a)An old wood fragment is burned to release CO2
which is collected in a 200.0 cc vessel to a
pressure of 2.00104 Pa (N/m2) at 295 K.
In one week, 1420  decays are counted.
b) An atmospheric sample of carbon dioxide is
placed into the same size vessel under the
same P and T for comparison purposes.
What is the age of this fragment of wood?
What number of counts does sample (b) give us?
PV (2.00 104 N / m 2 )( 200.0cm3 )
n

 1.63 103 mole
RT
(8.314 J / mole  K )( 295K )
N  (1.63 103 mole)(6.02 1023 molecules / mole)
 9.82 1020 molecules
Only 1 in a trillion are radioactive:
NC14  9.82 10 molecules 10
20
A = N
12
 9.82 10 molecules
8
0.693147
1 yr
= 5730 yr  31,557,600 sec/yr  9.82  108
A = 3.76  104/sec
In 1 week expect
(3.76 10 4 / sec)(3.895111011 sec)  2277
But the CO2 collected from the wood
fragment is only 1420/2277 as active
1420
 t
e
2277
1
1420
t  ln
 2277
5730
1420

ln
0.693147 2277
= 3900 years
1896
1899
1912
, 
g
Fission Track Dating
Charged particles from radioactive decay
(spontaneous fission of uranium)
pass through mineral's crystal lattice
leave trails of damage called FISSION TRACKS.
Procedures to study: Enlarge tracks by etching in acid
(so visible with light under microscope)
More readily seen with electron microscope
Count the etched tracks (or note track density in an area)
Useful in dating:
Micas (up to 50,000 tracks per cm2)
Tektites
Natural and synthetic (manmade) glass
Reheating "anneals" or heals the tracks.
The number of tracks per unit area is a function of age and uranium concentration.