7.2 Nuclear Stability and Nuclear Reactions

Nuclides above the band are too large -decay by a Nuclides above the band are too large -decay by a. To the left b- decay occurs. Nuclides below the band have too few no, positron decay occurs. A p+ becomes a no.

Nucleon Mass , u H-1 1.007825 H-2 1.00705 H-3 1.00535 He-4 1.000605 Fe-58 0.99885

Protons and Neutrons weigh less when bound in the nucleus. p+ and no fall toward each other, emit a ~1 MeV photon. Photon has mass E = mc2. E lost from nucleus.

More nucleons in nucleus emit higher E photons p+ and no fall toward each other, emit a 7 MeV photon. Higher nucleon numbers, higher PE lost on impact! 6

Binding E (BE) The amount of E emitted per nucleon = amount of E needed to put in to tear nucleus apart. BE the amount of work (eV, J) to tear apart nucleus.

Nuclear Energy Well It’s as if nucleus is a well. BE must be added to remove p+ or no from well. PE is lost as photons/mass when nucleons fall in. More nucleons = deeper well. More nucleons in nucleus have more strong F, more BE/nucl.

Binding E per Nucleon. Total E of Nucleus Num. Nucleons. Higher binding E per nucleon = more stable elements.

Can look at E like this too!

Which is the most stable? The higher the BE between nucleons, the more E/work needed to split it up, the more stable the nucleus – the less likely to decay (fall apart) Which is the most stable?

What happens beyond Fe-56 BE/nucleon decreases. Nuclei get large Felc takes over. PE released in nucl creation decreased.

Film Clip 2:45 min BE https://www.youtube.com/watch?v=UkLkiXiOCWU

1. If U-235 decays to Pb-208. a. How much BE/nucl is released? b. How much E is released for a whole U-235 nucleus? c. How many kg of mass is represented? ~ 0.2 MeV 235 x 0.2 = 47 MeV 47 x 106 eV x 1.6 x 10-19 J/eV 7.52 x 10-12 J 7.52 x 10-12 J = m (3x108 m/s)2 8 .4 x 10-29 kg.

Nuclear Physics 9 9 minutes. https://www.youtube.com/watch?v=-YMgacsJyD0&playnext=1&list=PL0191606751B22A12&feature=results_main

Do Now: A. What l of photon will be emitted for a photon released when a single 235U nucleon when the nucleus decays to form 208Pb assuming all the E is released as a EM radiation?      B . How many AMUs is 25.32 g of anything? 2.6 x 10-14 m. 1.524 x 1025 u.

Energy from the Nucleus

Mass Defect (Dm) Sum individual nucleons weigh more than the nucleus. Mass/ E difference = mass defect/deficit. The mass defect equals the BE. Find dif btw atomic mass and mass of constituents.

E and mass are equivalent. E = mc2 (joules) 1 u = 931. 5 MeV E and mass are equivalent . E = mc2 (joules) 1 u = 931.5 MeV. (remember 1 u = 1/12 mass C-12) We can calculate the BE per nucleon for each element. Find the difference in mass between the element and the sum of the individual component nucleons.

Calculation of BE per nucleon 1. Calculate the BE per nucleon of 54Fe which has an average mass of 53.9396 u. Mass p+ = 1.00782 u Mass no = 1.00866 u Mass e- = 0.000549

Calculate mass defect & BE/nucleon 26 p+ x 1.00782 u = 26.20332 28 no x 1.00866 u = 28.24248 26 e- x 0.000549 0.014274 Mass constituents 54.4458 u mass Fe nucl 53.9396 u subtract Mass defect 0.5062 u energy x 931.5 MeV/u BE = 471.5 MeV ÷ 54 nucl = 8.7 MeV/nucl

2. Find the binding energy of He-4 which has a known mass of 4 2. Find the binding energy of He-4 which has a known mass of 4.002602 u. (I’m including the e- this time). Neutral He-4 consists of 2e-, 2p+, and 2no. Find the sum of the parts in u. Look up in table: e- rest mass p+ rest mass no rest mass

Mass Defect 2 x (0.000549u) 0.001098 +2 x (1.007277u) 2.014554 +2 x (1.008665u) 2.017300 constituents 4.032982 u nucleus 4.002602 u defect 0.03038 u x 931.5 MeV/nucl 28.29897 MeV ÷ 4 nucl = 7.0747 MeV/nucl 23

3. The nucleus of a deuterium atom consists of a proton and a neutron 3. The nucleus of a deuterium atom consists of a proton and a neutron. If the mass of deuterium is 2.014102 u, calculate the BE in MeV. .Dm = 0.002378 ~ 2.215107 MeV / 2 = ~ 1.10755 MeV /nuc.

To calculate the BE/nucl: 1 To calculate the BE/nucl: 1. Find mass defect - the difference between the mass of the separate nucleons (unbound) and the mass of the bound nucleus. 2.Calculate the mass in atomic mass units. 3. Each unit has an energy equal to 931.5 MeV. 4. Multiply the mass defect by 931.5 MeV to convert the mass to energy. 5. Divide by # nucleons.

What happens when nucleus in not stable? Nuclear Reactions and Energy What happens when nucleus in not stable? Strive for stability.

Transmutation Nuclear transmutation, the conversion of one chemical element or isotope into another through nuclear reaction. Artificial/Induced Spontaneous Bombard Nucleus with a particle.

Two types of Transmutation Fission – heavy elements split to smaller fragments. Fusion – light elements fuse to heavier.

Fission We looked at fission by spontaneous radioactive decay. Fusion nucleus splits to 2 or more parts. Parent mass # decreases. material is added and “fused” to the nucleus.

4. Which type of reaction is this? Fission – parent has more mass than daughter.

5. Which type of reaction is this? Fusion - Mass of parent increases 4 He + 14 N 17 O + 1H 2 7 8 1 Proton a Fusion - Mass of parent increases

Other types of particles can be used to bombard the nucleus Other types of particles can be used to bombard the nucleus. Neutrons, protons, and H-2 are common. Which rx is this?

16O + 1n AX + 2H 8 0 z 1 Identify X if a 2H is emitted in the reaction. What type of reaction is it?

16O + 1n AX + 2H 8 0 z 1 Solve for the mass & proton number for X by balancing. 16O + 1n 15X + 2H 8 0 7 1 Element must have 7 p+. It will be 15N. 7 Fission.

Energy Considerations in Fission & Fusion

Nuclear decay seeks to stabilize nucleus Nuclear decay seeks to stabilize nucleus. When nucleus goes from lower to higher binding energy ratio/nucleon, energy is released in process.

Release of E occurs when element goes to more stable state. 7. Use BE table to predict the total energy release. 235 U 2 117 Pd fragments 92 46 Find the energy released. Use the BE/nucleon table.

235 U ~ 7.6 MeV per nucleon 117 Pd ~ 8.4 MeV per nucleon. Take difference. This E is released per nucleon. 8.4 – 7.6 = 0.8 MeV/nucleon. But 235 U has 235 nucleons so: 235 x 0.8 MeV = 188 MeV released! Compare this to ionization energies.

BE Released in Rx can be converted to KE and/or heat.

8: Lithium can be bombarded with no to induce the following rx: 6 Li + 1n 3 H + 4He 3 0 1 2 Mass Li-6 6.015126 Mass no = 1.008665 Mass H-3 3.016030 Mass He-4 4.002604 How much E is released?

6 Li + 1n 3 H + 4He 3 0 1 2 Calculate mass for each side. Dm = (7.023791 – 7.018634) u Dm = 0.005175 u x 931.5 MeV/u = 4.804 MeV Available as KE to fragments. some may be released as heat and light.

Deuterium and Tritium undergo fusion according the follow rx: 2 H + 3H 4 He + 1n + E 1 1 2 0 How many J of energy are liberated in each reaction?

Solution Mass of reactants = 2.014102 + 3.016010 = 5.030152 u. Mass Products = 4.002603 + 1.00665 = 5.011268 u. . Dm = 0.018884 u = 17.59 MeV 2.81 x 10 -12 J.

Hwk. Hamper pg 166 top #21, 22. Write the nuclear equation and then solve each.

Go to 7.2 b. Nuclear Power

9. Energy From Reactions 10 min 8.1 Natural transmutations and half lives 10 min http://www.youtube.com/watch?v=I7WTQD2xYtQ 9. Energy From Reactions 10 min http://www.youtube.com/watch?v=-YMgacsJyD0&playnext=1&list=PL0191606751B22A12&feature=results_main