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Nuclear Reaction: involves a change in the atom’s nucleus.

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Presentation on theme: "Nuclear Reaction: involves a change in the atom’s nucleus."— Presentation transcript:

1 Nuclear Reaction: involves a change in the atom’s nucleus.

2 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.

3 Radioactive Decay: an unstable nucleus spontaneously breaks apart and emits radiation.

4 Types of Radiation:  alpha radiation (α)  beta decay (β)  gamma rays (γ)

5 Alpha Radiation, α : 4 He 2+ 2  low penetrating ability. Thickness of a sheet of paper can stop it.

6 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.

7 Gamma Rays, γ:  no mass and no charge  dangerous radiation: emitted when nucleus undergoes alpha decay.  stopped by heavy lead or concrete shielding.

8 2+

9 +

10  Ra → α + 226 88 4242 222 86 Rn

11  U → α + 238 92 4242 234 90 Th

12  C → β + 14 6 0 −1 14 7 N

13 U+ n→ Ba + Kr + ? 235 92 1010 1010 92 36 141 56 3 n

14 Problem:  Write the equation for I undergoing beta decay. 131 53

15 Problems:  α + N → ?+ H  Rb → ? + ? 4242 14 7 87 37 87 38 1

16 Problems:  H + ? → He  C → ? + β 2121 4242 14 6 0 -1

17 Write a balanced nuclear equation for the alpha decay of the following radioisotope U 234 92

18 U → Th + α 234 92 230 90 4 2

19

20 Write the symbols for the following: neutronalpha protonbeta (electron)gamma positron

21 neutron proton electron positron alpha beta gamma n p e β α β γ1 0 0 -1 0101 1010 4242 0

22 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

23 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

24 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

25 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

26 Half-Life (t 1/2 ):  time required for half of the atoms in a sample to decay.

27 Radioactive Decay Rates

28 Determining Decay:  fraction of a radioisotope remaining after n half- lives is (½) n  m final = m initial (½) n

29 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

30 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?

31 Nuclear fission

32 Nuclear fusion


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