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Nuclear Reaction: involves a change in the atom’s nucleus.
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Image of Becquerel's photographic plate which was exposed to radiation from a uranium salt. The shadow of a metal Maltese Cross placed between the plate and the uranium salt is clearly visible.
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Radioactive Decay: an unstable nucleus spontaneously breaks apart and emits radiation.
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Types of Radiation: alpha radiation (α) beta decay (β) gamma rays (γ)
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Alpha Radiation, α : 4 He 2+ 2 low penetrating ability. Thickness of a sheet of paper can stop it.
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Beta Decay, β : 0 e − −1 turns a neutron into a proton and emits an electron. shielded by metal foil since it’s lighter than α.
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Gamma Rays, γ: no mass and no charge dangerous radiation: emitted when nucleus undergoes alpha decay. stopped by heavy lead or concrete shielding.
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2+
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+
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Ra → α + 226 88 4242 222 86 Rn
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U → α + 238 92 4242 234 90 Th
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C → β + 14 6 0 −1 14 7 N
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U+ n→ Ba + Kr + ? 235 92 1010 1010 92 36 141 56 3 n
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Problem: Write the equation for I undergoing beta decay. 131 53
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Problems: α + N → ?+ H Rb → ? + ? 4242 14 7 87 37 87 38 1
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Problems: H + ? → He C → ? + β 2121 4242 14 6 0 -1
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Write a balanced nuclear equation for the alpha decay of the following radioisotope U 234 92
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U → Th + α 234 92 230 90 4 2
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Write the symbols for the following: neutronalpha protonbeta (electron)gamma positron
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neutron proton electron positron alpha beta gamma n p e β α β γ1 0 0 -1 0101 1010 4242 0
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The difference between the mass of an atom’s nucleus and the sum of the masses of its protons and neutrons. Mass Defect Cu-63: 29 protons(1.00728 amu/p + ) + 34 neutrons(1.00867 amu/n) = 63.50590 amu Mass defect = 63.50590 amu - 62.91367 amu = 0.59223 amu
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The energy released when a nucleus is formed from nucleons (protons & neutrons) Einstein’s equation E = mc 2 shows mass can be converted to energy, and energy to mass. Nuclear Binding Energy
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Example: The atomic mass of S-32 is 31.972070 amu. 1)What is the mass defect? 2)What is the binding energy? 3)What is the binding energy per nucleon? 1.Determine total mass of proton, neutrons & electrons 16 p x 1.007276 amu = 16.116416 amu 16 n x 1.008665 amu = 16.138640 amu 16 e ‒ x 0.0005486 amu = 0.0087776 amu Total = 32.263834 amu 2.Find the mass defect: 32.263834 amu ‒ 31.972070 amu = 0.291764 amu 1.6605 x 10 ‒ 27 kg/amu = 4.84473 x 10 ‒ 28 kg
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Example: The atomic mass of S-32 is 31.972070 amu. 2) What is the binding energy? 3) What is the binding energy per nucleon? E=mc 2 E = (4.84473 x 10 ‒ 28 kg)(3.00 × 10 8 m/s) 2 = 4.36 × 10 −11 kgm 2 /s 2 = 4.36 × 10 −11 J Per nucleon: 4.36 × 10 −11 J / 32 nucleons = 1.36 × 10 −12 J/nucleon
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Half-Life (t 1/2 ): time required for half of the atoms in a sample to decay.
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Radioactive Decay Rates
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Determining Decay: fraction of a radioisotope remaining after n half- lives is (½) n m final = m initial (½) n
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Half-Life calculations Variables: Equations: m final m initial n t 1/2 t total t total = n t ½ m final = m initial (½) n Final mass initial mass # of half lives Time length of a half life Total length of time
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Problem: N-13 has a half-life of 10 minutes. How much of 2.00 g. N-13 will exist after 3 half-lives? How long is 3 half- lives?
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Nuclear fission
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Nuclear fusion
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