2. av=15.5 MeV
ac=0.72 MeV
ap=34.0 MeV
as=16.8 MeV
asym=23.0 MeV

3. B(A,Z) is clearly a quadratic function of Z
For a fixed value of A : C1 C2 C3 C4 D
B(A,Z) is clearly a quadratic function of Z

4. A = 104 isobars 42Mo 103.912 43Tc 48Cd 103.910 47Ag Mass, u 103.908
45Rh
44Ru
46Pd 42 44 46 48
Atomic Number, Z

5. B(A,Z) is clearly a quadratic function of Z
For a fixed value of A : C1 C2 C3 C4 D
B(A,Z) is clearly a quadratic function of Z
even Z, N
Z = N = 0
odd Z, N

6. A = 104 isobars e-capture: X + e  Y 42Mo  103.912 43Tc
-decay: X  Y +  - A Z A
Z+1 ? N
N-1 ?
e-capture: X + e  Y A Z A
Z-1
48Cd N
N+1
Odd Z
47Ag
Mass, u
45Rh
Even Z
44Ru 
46Pd 42 44 46 48
Atomic Number, Z

7. for A>50 ~constant 8-9 MeV Peaks at ~8.795 MeV near A=60
Binding energy per nuclear
particle (nucleon) in MeV
Mass Number, A

8. Is Pu unstable to -decay?
236 94
236 94
Pu  U + 
232 92 4 2
M( )c M( )c2 M( )c2 + Q
236 94
Pu  U 
232 92 4 2
Q = (MPu – MU - M)c2
= ( u – u – u)931.5MeV/u
= 5.87 MeV > 0

9. URANIUM DECAY SERIES 92U238  90Th234  91Pa234  92U234  
92U238  90Th234
 91Pa234
 92U234   
92U234  90Th230
 88Ra226
 86Rn222  
 84Po218
 82Pb214   
82Pb214  83Bi214
 84Po214
 82Pb210   
82Pb210  83Bi210
 84Po210
 82Pb206
“Uranium I” 109 years U238
“Uranium II” 105 years U234
“Radium B” radioactive Pb214
“Radium G” stable Pb206

10. Radioactive parent isotopes & their stable daughter products
Potassium 40
Argon 40
Rubidium 87
Strontium 87
Thorium 232
Lead 208
Uranium 235
Lead 207
Uranium 238
Lead 206
Carbon 14
Nitrogen 14

12. Half Lives for Radioactive Elements
Radioactive Parent
Stable Daughter
Half life
Potassium 40
Argon 40
1.25 billion yrs
Rubidium 87
Strontium 87
48.8 billion yrs
Thorium 232
Lead 208
14 billion yrs
Uranium 235
Lead 207
704 million yrs
Uranium 238
Lead 206
4.47 billion yrs
Carbon 14
Nitrogen 14
5730 years

13. Growth of Radioactive Daughter Products
where decay of parent
(original) nucleus:
if it’s radioactive
itself!
as we’ve seen before
try:
note:
and if:

14. Growth of Radioactive Daughter Products
so:
so:
and:
Note: as 2→0
as seen before.
As N1(t)
and N2(t)
may reach the state
so that:

15. Note: some elements have both radioactive
and non-radioactive isotopes.
Examples: carbon, potassium.
Just saw: 3 isotopes of uranium.
238U the most abundant ( %)
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.

16. 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.

17. 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.

18. 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

19. R = = 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:
The number 238U atoms decayed  number of 206Pb atoms today:
number 238U atoms
number of 206Pb atoms
So:
R = =
R=1.2: 109 years
R=1.8: 109 years

20. If other (stable) isotopes of the daughter are also present
What if there were initially some daughter products
already there when the rock was formed?
unknown!
Remember: elements come in many isotopes
(some even tag a specific decay series!)
If other (stable) isotopes of the daughter are also present
Then look at the ratio:

21. Which we can rewrite as:
N(t)=N0e-t
N0=N(t)e+t
y = x  m b

22. Rb-Sr dating method Allows for the presence of initial 87Sr
Age = 4.53  109 y
2 = 0.04  109 y
G.W. Wetherill, Ann. Rev.
Nucl. Sci. 25, 283 (1975)

23. Some concentration ratios measured for Eagle Peak Pluton
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 = 14

24. Rubidium half-life=48.8 By = 48,800,000,000 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.

25. 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
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 %
13C %
14C 1 atom for every 1012 atoms of 12C

26. 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.

27. Assumes that the rate of Carbon 14 production
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).

29. What is the age of this fragment of wood?
An old wood fragment is burned to release CO2
which is collected in a cc vessel to a
pressure of 2.00104 Pa (N/m2) at 295 K.
In one week, 1420 b- 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?

30. A = lN = A = 3.76  10-4/sec Only 1 in a trillion are radioactive:
5730 yr
1 yr
31,557,600 sec/yr
A = lN =  
9.82  108
A = 3.76  10-4/sec
In 1 week expect
But the CO2 collected from the wood
fragment is only 1420/2277 as active

31. = 3900 years

32. 1896
1899
a, b g
1912

33. 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.