# Av=15.5 MeV ac=0.72 MeV ap=34.0 MeV as=16.8 MeV asym=23.0 MeV.

## Presentation on theme: "Av=15.5 MeV ac=0.72 MeV ap=34.0 MeV as=16.8 MeV asym=23.0 MeV."— Presentation transcript:

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

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

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

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

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

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

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

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

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

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

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

so: so: and: Note: as 2→0 as seen before. As N1(t) and N2(t) may reach the state so that:

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.

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

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

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:

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

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)

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

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.

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

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.

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).

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?

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

= 3900 years

1896 1899 a, b g 1912

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.

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