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a v =15.5 MeV a s =16.8 MeV a c =0.72 MeV a sym =23.0 MeV a p =34.0 MeV.

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Presentation on theme: "a v =15.5 MeV a s =16.8 MeV a c =0.72 MeV a sym =23.0 MeV a p =34.0 MeV."— Presentation transcript:

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2 a v =15.5 MeV a s =16.8 MeV a c =0.72 MeV a sym =23.0 MeV a p =34.0 MeV

3 C1 C1 C 2 C3C3 C 4 For a fixed value of A : B(A,Z) is clearly a quadratic function of Z D

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

5 C1 C1 C 2 C3C3 C 4 For a fixed value of A : B(A,Z) is clearly a quadratic function of Z even Z, N odd Z, N Z = N = 0 D

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

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

8 Is Pu unstable to -decay? Pu U Q = (M Pu – M U M )c 2 = ( u – u – u) 931.5MeV/u = 5.87 MeV > 0 Pu U M( )c 2 M( )c 2 M( )c 2 + Q

9 92 U Th Pa U U Th Ra Rn Po Pb Pb Bi Po Pb Pb Bi Po Pb 206 Uranium I yearsU 238 Uranium II yearsU 234 Radium B radioactivePb 214 Radium G stablePb 206 URANIUM DECAY SERIES

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

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12 Half Lives for Radioactive Elements Radioactive ParentStable DaughterHalf life Potassium 40Argon billion yrs Rubidium 87Strontium billion yrs Thorium 232Lead billion yrs Uranium 235Lead million yrs Uranium 238Lead billion yrs Carbon 14Nitrogen years

13 Growth of Radioactive Daughter Products if its radioactive itself! where decay of parent (original) nucleus: as weve seen before try: note: and if:

14 Growth of Radioactive Daughter Products so: and: Note: as 2 0 as seen before. As N 1 (t) and N 2 (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. 238 U 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 2 different rock samples have ratios of 238 U to 206 Pb atoms of 1.2 and 1.8. Compute the age of each sample. 238 U atoms remaining today: The number 238 U atoms decayed number of 206 Pb atoms today: So: R = number 238 U atoms number of 206 Pb atoms = R=1.2: years R=1.8: years

20 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: y = x m + b N(t)=N 0 e t N 0 =N(t)e t

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

23 Some concentration ratios measured for Eagle Peak Pluton of the Sierra Nevada Batholith = All samples lie along the same line, so formed from the same batch of magma

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 12 C 98.89% 13 C 1.11% 14 C 1 atom for every atoms of 12 C

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 14 C to 12 C (1 in a trillion). At death, 14 C exchange ceases and any 14 C in the tissues of the organism begins to decay to Nitrogen 14, and is not replenished by new 14 C. The change in the 14 C to 12 C ratio is the basis for dating.

27 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).

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29 a)An old wood fragment is burned to release CO 2 which is collected in a cc vessel to a pressure of Pa (N/m 2 ) 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?

30 Only 1 in a trillion are radioactive: A = N = yr 1 yr 31,557,600 sec/yr A = /sec In 1 week expect But the CO 2 collected from the wood fragment is only 1420/2277 as active

31 = 3900 years

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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 cm 2 ) 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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