2),()(),( cZABmZAZmZAM np

),(/)2/(

)1(),(2

3/13/2

ZAAZAa

AZZaAaAaZAB

sym

csv

av=15.5 MeV

as=16.8 MeV

ac=0.72 MeV

asym=23.0 MeVap=34.0 MeV

),(/)2/(

)1(),(2

3/13/2

ZAAZAa

AZZaAaAaZAB

sym

csv

C1 C2 C3

C4

For a fixed value of A :

B(A,Z) is clearly a quadratic function of Z

D

103.912

103.910

103.908

103.906

103.904

Mas

s, u

42 44 46 48

Atomic Number, Z

A = 104 isobars42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

),(/)2/(

)1(),(2

3/13/2

ZAAZAa

AZZaAaAaZAB

sym

csv

C1 C2 C3

C4

For a fixed value of A :

B(A,Z) is clearly a quadratic function of Z

4/3

4/3

0)(

Aa

Aa

p

p

A

even Z, N

odd Z, N

Z = N = 0

D

103.912

103.910

103.908

103.906

103.904

Mas

s, u

42 44 46 48

Atomic Number, Z

A = 104 isobars42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

Odd Z

Even Z

-decay: X Y + AZ N ?

?? N-1A

Z+1

e-capture: X + e Y N N+1A

Z-1 AZ

Peaks at ~8.795 MeV near A=60

for A>50~constant8-9 MeV

Mass Number, A

Bin

din

g e

nerg

y p

er

nu

clear

part

icle

(n

ucl

eon

) in

MeV

Is Pu unstable to -decay?23694

Pu U + 23694

23292

42

Q = (MPu – MU M)c2

= (236.046071u – 232.037168u – 4.002603u)931.5MeV/u

= 5.87 MeV > 0

Pu U + 23694

23292

42M( )c2 M( )c2 M( )c2 + Q

92U238 90Th234 91Pa234 92U234

92U234 90Th230 88Ra226 86Rn222 84Po218 82Pb214

82Pb214 83Bi214 84Po214 82Pb210

82Pb210 83Bi210 84Po210 82Pb206

“Uranium I” 4.5109 years U238

“Uranium II” 2.5105 years U234

“Radium B” radioactive Pb214

“Radium G” stable Pb206

URANIUM DECAY SERIES

Radioactive parent isotopes & their stable daughter products

Radioactive Parent Stable Daughter

Potassium 40 Argon 40

Rubidium 87 Strontium 87

Thorium 232 Lead 208

Uranium 235 Lead 207

Uranium 238 Lead 206

Carbon 14 Nitrogen 14

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

Growth of Radioactive Daughter Products

dtNdtNdN22112

if it’s radioactive

itself!

where decay of parent(original) nucleus:

teNtN 1

0,11)(

as we’ve seen beforetry:

ttBeAetN 21)(

2

note: BAN

0,2and if: 0

0,2N

)()( 12

2

tteeAtN

tteeA

dt

dN12

122

2211NN

)( 12

2

tteeA

Growth of Radioactive Daughter Products

dtNdtNdN22112

so:tt

AeNeA 11

2111

so:0

21

1 NA

)()( 12

1021

12

tteeNtN

)1()( 1

02

teNtN

tteNAe 11

0121

and:

Note: as 2→0as seen before.

As N1(t) and N2(t) may reach the state2211

NN so that: 0

2dN

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: teN 0

The number 238U atoms decayed number of 206Pb atoms today:

)1(0teN

So: R = number 238U atoms number of 206Pb atoms )1(0

teN

teN 0= 1

1

te

)11

ln(1

R

t )1

1ln(

2ln

10468.4 9

R

t

R=1.2: 3.907109 yearsR=1.8: 2.848109 years

What if there were initially some daughter products already there when the rock was formed?

parentdaughterparentdaughter NNtNtN ,0,0)()(

unknown!

Remember: elements come in many isotopes (some even tag a specific decay series!)

If other (stable) isotopes of the daughter are also present

dd NtN ,0)(Then look at the ratio:

d

parentdaughter

d

parentdaughter

N

NN

tN

tNtN

,0

,0,0

)(

)()(

d

parentdaughter

d

parentdaughter

N

NN

tN

tNtN

,0

,0,0

)(

)()(

d

daughtertt

d

parent

d

daughter

N

Ne

tN

tN

tN

tN

,0

,0)( 1)(

)(

)(

)(0

Which we can rewrite as:

y = x m + b

d

daughter

d

parent

d

parent

d

daughter

N

N

tN

tN

tN

N

tN

tN

,0

,0,0

)(

)(

)()(

)(

N(t)=N0et N0=N(t)et

Rb-Sr dating method

Allows forthe presence of initial 87Sr

G.W. Wetherill, Ann. Rev. Nucl. Sci. 25, 283 (1975)

Age = 4.53 109 y 2 = 0.04 109 y

Some concentration ratios measured for Eagle Peak Plutonof the Sierra Nevada Batholith

0.7248-0.707614

= 0.001213 1)(

0 tte

)(001212265.0001213.1ln0

tt /001212265.00 tt

All samples lie along thesame line, so formed fromthe same batch of magma

/001212265.00 tt

Rubidium half-life=48.8 By = 48,800,000,000 years

2ln

21

t

years04880000000/693147.0

yearstt 000,530,80

The Sierra Nevada pluton was formed by subduction(one tectonic plate driven beneath another) remeltingcontinental crust and forming volcanic rock called basalt 8,530,000 years ago. Sure enough west ofthis fault (and deeper below the basalts) are 4.55 Bysamples 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 7

1478

146 NC

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 2.00104 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?

moleKKmoleJ

cmmN

RT

PVn 3

324

1063.1)295)(/314.8(

)0.200)(/1000.2(

molecules

molemoleculesmoleN20

233

1082.9

)/1002.6)(1063.1(

moleculesmoleculesN 8122014 1082.9101082.9 C

Only 1 in a trillion are radioactive:

A =N =0.6931475730 yr

1 yr31,557,600 sec/yr

9.82 108

A = 3.76 104/sec

In 1 week expect

2277sec)1089511.3sec)(/1076.3( 114

But the CO2 collected from the wood fragment is only 1420/2277 as active

te 2277

1420

2277

1420ln

1

t

2277

1420ln

693147.0

5730

= 3900 years

1896

1899

1912

Fission Track DatingCharged 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.