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Particle Nature of The Nucleus -All nuclei have mass that is multiple of specific number. -t.f. nucleus made of smaller particles: protons, neutrons. 1.

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Presentation on theme: "Particle Nature of The Nucleus -All nuclei have mass that is multiple of specific number. -t.f. nucleus made of smaller particles: protons, neutrons. 1."— Presentation transcript:

1 Particle Nature of The Nucleus -All nuclei have mass that is multiple of specific number. -t.f. nucleus made of smaller particles: protons, neutrons. 1

2 Nuclides – elements same p+ Isotopes – specific types of nuclides Mass numberA X atomic proton# Z Proton Z number identifies Z type of element. Ex: all H has one proton. Oxygen has 8 protons in nucleus. Proton number is equal to number of e- in neutral element. 2

3 Mass or Nucleon Number Nucleons live in nucleus -protons and neutrons. Mass number (A) is equal to the total all neutrons and protons in the nucleus. If p+ and n o considered as atomic weight 1, the mass number gives the atomic weight in amu - u. 3

4 Isotopes Elements come in different isotopes. Two isotopes of the same substance will have equal proton numbers, but different numbers of neutrons. For the same element, the Z number must always be the same but the A number may vary. FYI for the most part dif isotopes of the same element behave the same way chemically. 4

5 Nuclides Nuclide is a specifically defined isotope. Nuclides are defined by many different aspects, such as half life, mode of decay, percent abundance, and so on. More specific than isotope. Isotopes are sets of nuclides having the same number of p+, but different number of n o -. Each individual isotope is a separate nuclide. 5

6 Calculations of nucleons: For a given isotope: protons (Z) + neutrons = mass number (A) Its easy to find the number of n o in isotope. A – Z is total neutrons. Find the number of n o in 37 Cl 17 6

7 Mass Spectrometer separates isotopes by mass 7 Ionization (thermal) Accelerate ions E field Velocity detector Magnetic Deflector Detector.

8 Velocity Selector Crossed E and B fields Velocity selector F elc = F mag Eq = qvBv = E/B 8

9 Entering Magnetic Field Heavier ions are deflected less than lighter ones. F mag = F c. qvB = mv 2 /r. qB = mv/r. qBr = mv r = mv qB

10 Detectors Count How many land at each point. 10

11 Read Hamper 7.2 and the Mass Spectrometer pg 255-256 Do Hwk worksheet w reading. 11

12 Nuclear Force 12

13 The strong nuclear force holds nucleons together. Heating metals or shining EM radiation can strip e- from atom. Not so to pull apart nucleus. 13

14 Binding E (BE) = work (J or eV) needed to pull apart nucleus. When work is added to nucleus to split it up, where does the E go? If they are not in motion no KE, or PE chm or PE elc what happens to work put in? Einstein -separate nucleons have greater mass than when they are bound in nucleus. E from W to split atom was converted to mass! E = mc 2. 14

15 The higher the BE between nucleons, the more work needed to split it up, the more stable the nucleus – the less likely to decay (fall apart) 15

16 Binding E per Nucleon. Total E of Nucleus Num. Nucleons. 16

17 Higher binding E per nucleon = stable elements. 17

18 Mass Defect/Deficit The difference between the mass of the atom and the mass of the individual constituents. Since the nucleus has less mass than the sum of its parts, the difference is called mass defect and it equals the BE. Look at the table, which are the most stable? 18

19 What happens when nucleus in not stable? Strive for stability. 19

20 20 Mass Defect 20

21 Because E and mass are equivalent it can be shown that 1 u = 931.5 MeV. We can use the above relationship to calculate the BE per nucleon for each element. 21

22 Calculation of BE per nucleon 1. Calculate the BE of 54 Fe which has an average mass of 53.9396 u. Mass p+ = 1.00782 u Mass n o = 1.00866 u

23 Calculate mass defect & BE/nucleon 26 p + x 1.00782 u = 26.20332 28 n o x 1.00866 u = 28.24248 26 e- x 0.0005490.014274 Mass constituents54.4458 u mass Fe nucl53.9396 u subtract Mass defect0.5062 u energyx 931.5 MeV/u BE = 471.5 MeV ÷ 54 nucl = 8.7 MeV/nucl 23

24 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 2n o. Find the sum of the parts in u. Look up in table: e- rest mass p+ rest mass n o rest mass

25 25 2 x (0.000549u)0.001098 +2 x (1.007277u)2.014554 +2 x (1.008665u)2.017300 constituents4.032982 u nucleus4.002602 u defect0.3038 u x 931.5 MeV/nucl 28.29897 MeV ÷ 4 nucl = 7.0747 MeV/nucl 25 Mass Defect

26 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. 26

27 27 Binding Energy The amount of work needed to pull apart nucleus. The nucleus converts the missing mass to BE, it’s like a glue to hold nucleus together. When nucleus transitions to lower energy state, mass defect E is released! 27

28 BE and atomic size 3 min http://www.youtube.com/watch?v=UkLkiXi OCWUhttp://www.youtube.com/watch?v=UkLkiXi OCWU

29 29 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.

30 Do Hamper pg 154 #9. and addl prob’s Holt 30

31 Radioactive Decay Radioactivity – emission of energy. 1896 Antoine Becquerel discovered that certain U salts emitted rays that could penetrate dark paper to expose a photographic plate. Called penetrating rays radiation. Marie Curie followed up & discovered other elements with the same properties (Th). 31

32 32 Two types of Transmutation Spontaneous Unstable Nucleus Artificial/Induced Bombard Nucleus with a particle.

33 Emitted From Nucleus Alpha Beta Positron Gamma Neutrino Antineutrino 33

34 34 Spontaneous Radioactive Decay is random process The type of decay & rate of decays do not depend on pressure, temperature, chemical bonds.

35 1899 Rutherford found U emits two dif types of “radiation”  and .  rays were more penetrating. Later, Villard found another type of radiation with even more penetrating power - gamma . 35

36 Alpha Rays  rays are helium nuclei, (2p + and 2n o ), that are emitted from nucleus. They are positively charged since e- missing. When nucleus is too large, it may emit alpha particle. 36

37 37 Parent Daughter Alpha Decay of Americium-241 to Neptunium-237 He nucleus

38  particles are easily stopped by skin or thin sheet of paper. Likely to knock e- from orbits if they hit them.  loses all its KE at once when it’s stopped. Charge = +2e. Mass = 4 units. Energy is KE = ½ mv 2. 38

39 Alpha  Decay 39 Alpha  have KE ~5 MeV

40 Beta  Particles Fast moving electrons emitted from nucleus. More penetrating than alpha because they are smaller. Think of B as a n o that emits e-, becomes a p+. 40

41 Less capable of ionizing (knocking out e-) - Charge = -1 or -e. mass = e, but considered massless in mass # calcs. KE = ½ mv 2. v can be sig portion of c. Need a few mm of Al to stop them. 41

42 42

43 Positron Decay like beta decay but p+ loses  + (positive electron )becomes a neutron:

44 A highly penetrating type of nuclear radiation, similar to x-rays and light, except that it comes from within the nucleus of an atom, and, has a shorter. Gamma ray emission is a decay mode by which excited state of a nucleus de-excite to lower (more stable) state in the same nucleus. In diagrams, a gamma ray is represented by this: In equations = . Can pass thru human body, concrete, and lead. Gamma  Radiation 44

45 Gamma Radiation 45

46 Gamma rays are EM waves. They have: Lowest ionizing power. No charge. No mass. Energy described by E = hf. Travel with vel of light in vacuum. No maximum stopping range. 46

47 Neutrinos are particles that are emitted with beta and positron decay. No charge. Almost no mass. They do not interact with matter. 47

48 Which emitter would be safer to swallow? 48

49 How could we distinguish the different types of radiation? What could we observe? 49

50 50

51 51

52 52

53 7.2.4 Biological Effects of Ionizing Radiation Outline them.

54 Becquerel Rays 9 min. https://www.youtube.com/watch?v=INF9y1 54EZAhttps://www.youtube.com/watch?v=INF9y1 54EZA 54

55 55 Products of Decay When a parent nucleus decays, a daughter product, is produced, a new element could be produced. We can identify products by balancing mass and atomic numbers.

56 56 1. U – 232 decays by alpha emission. Write the nuclear equation and determine the daughter of the decay.

57 57 alpha 2324A U He + X 922Z X will have 4 less nucleons than U-232. X will have 2 less protons than U-232.

58 58 Daughter will be:228 X 90 58 Look up on periodic table element with 90 p+ (Thorium). Answer is: 228 Th 90

59 59 Positron/Beta Decay During beta decay a beta,  - (e-) or a positron  + (+e) is emitted from the nucleus. When an e- (  -) emitted a n o changes to a p +, the atomic number increases by 1. When an e+ (  +) emitted a p + changes to a n o, the atomic number decreases by 1.

60 60 A neutrino v or antineutrino v is also emitted in positron or beta decay.  + decay emits neutrino v.  decay emits an antineutrino v.

61 61 Positron Decay like beta decay but: Protonn o +  + + 90 Ru  + + 90 Tc +. 44+1 43 A positron particle  + is the antimatter e-. neutrino 90 Ru  + + ___ +. 44+1

62 62 2. C-14 undergoes beta decay. What will be the daughter. Remember a beta particle is an e- ejected from the nucleus so that a neutron becomes a proton.

63 63 14 0A C e + X + v. 6-1Z A beta particle/e- has atomic number –1 and is considered massless.

64 64 Balance – make right side add up to left side. 14 0A C e + X + v. 6-1Z A will be 14. Z will be ??

65 65 Z will be 7 since 7 and -1 = 6. So: 14 014 C e + X + v 6 -17 Z=7 is nitrogen. So daughter is : 14 N 7

66 66 3. Radium-226 decays by alpha emission. What is the resulting daughter element?

67 67 226 Ra X + 4 He 882 Mass X = 226 – 4 = 222. Atomic number X 88 – 2 = 86. 222 Rn 86

68 68 4. Sulfur 35 emits  - particles when it decays. Write the equation. What will be the daughter product?

69 69 35 S 0 e+ A X + v. 16-1Z Atomic number X is 17 since neutron went to proton, so daughter is 17 Cl – 35.

70 70 5. Complete the equation: 23 Mg  + + __ __ + . 12+1 __ 23 Mg  + + 23 Na+. 12+1 11

71 Nuclear Stability The more stable the nucleus, the less likely that it will decay.

72 Electrostatic repulsion in the positively charged nucleus makes it want to decay. The strong nuclear force holds the nucleus together. 72

73 The extra neutrons increase the strong force & help shield against electric repulsion. 73

74 As more p + added to nucleus more n o needed for stability (for nucleus not to decay). There seems to be a ratio between p + and n o for each element to maintain enough strong force to keep nucleus from flying apart. In general heavier elements require more n o. p+– n o ratio determines stability. 74

75 Nuclides above the band are too large - decay by . To the left  decay occurs. Nuclides below the band have too few n o, positron decay occurs. A p+ becomes a n o. 75

76 Summery: Natural Decay occurs spontaneously.  decay reduces the mass # by 4 and the atomic # by 2.   decay does not affect the mass #, but increases daughter's the atomic # by 1.   decay does not affect the mass #, but decreases the daughter's atomic # by 1.  decay affects neither the atomic nor the mass # but returns excited nucleus to ground state. 76

77 77 http://www.youtube.com/watch?v=INF9y154EZ Ahttp://www.youtube.com/watch?v=I7WTQD2x YtQ Properties of Becquerel Rays 9 min

78 Hwk Read Hamper 7.3 do #11-13 But Write out the balanced nuclear reaction for each. 78

79 Nuclear Reactions: Natural & Artificial Transmutation A process changing the nucleus =nuclear reaction. Typical reactions are nuclear fission & fusion. Reactions can occur spontaneously or be artificially induced. Transmutation – can be artificially induced nuclear reaction. 79

80 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. 80

81 1919 Rutherford bombarded gas with  particles. New particles were produced in the gas which were not  particles. From the deflection he concluded they were p+. He later discovered that the nuclei of the gas absorbed the  particle & emitted a p+. The  fused to form a larger nucleus. Fusion occurs naturally in stars. 81 Nuclear Fusion

82 Equation for Fusion Rx. Mass # of parent increases 4 He + 14 N 17 O + 1 H 278 1 Proton  82

83 Other types of particles can be used to bombard the nucleus. Neutrons, protons, and H-2 are common. Which rx is this? 83

84 16 O + 1 n A X + 2 H 80z1 Identify X if a 2 H is emitted in the reaction. Balance mass numbers & proton numbers on the left must and the right first. 84

85 16 O + 1 n A X + 2 H 80z1 16 O + 1 n 15 X + 2 H 8071 Element must have 7 p+. It will be 15 N. 7 Solve for the mass & proton number for X by balancing. 85

86 Energy Considerations in Fission & Fusion 86

87 Nuclear decay seeks to stabilize nucleus. When nucleus goes from lower to higher binding energy ratio/nucleon, energy is released in process. 87

88 Release of E occurs when element goes to more stable state. 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. 88

89 235 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! 89

90 BE Released in Rx can be converted to KE and/or heat. 90

91 3: Lithium can be bombarded with n o to induce the following rx: Mass Li-66.015126 Mass n o =1.008665 Mass H-33.016030 Mass He-44.002604 6 Li + 1 n 3 H + 4 He 3012 How much E is released?

92  m = (7.023791 – 7.018634) u  m = 0.005175 u x 931.5 MeV/u = 4.804 MeV Available as KE some may be released as heat. 92 Calculate mass for each side. (7.023791u) (7.018634) 6 Li + 1 n 3 H + 4 He 3012

93 9. Energy From Reactions 10 min http://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL0191606751B22A12& feature=results_mainhttp://www.youtube.com/watch?v=- YMgacsJyD0&playnext=1&list=PL0191606751B22A12& feature=results_main 93 8.1 Natural transmutations and half lives 10 min http://www.youtube.com/watch?v=I7WTQD2xYtQ

94 Read Hamper 7.4 and 7.5 ForHwk. Hamper pg 162 15-17, pg 163 #18 – 19, 20 and pg 166 #21,22, pg

95 Decay Series Heavier elements – even a decay often does not decrease the mass of heavy elements enough for stability. These elements go through a series of decays before reaching a stable state. 95

96 Radioactive Decay Series All elements above Z=83 are radioactive. Most of these decay through a series of transitions. The daughters are radioactive and decay a number of times before they become a stable element. Many elements decay through many  and  steps.

97 U-238 decay series 97

98 Spontaneous Random Decay & Half life Radioactive decay is random & not affected by outside factors. There is no way of knowing whether a particular nucleus will decay.

99 Changing temperature or pressure does not change chances of a decay, a chemical rx does not alter its activity. 99

100 We can only know the chances of a decay happening. 100

101 Rate of Decay is proportional to number (N) of atoms in the sample. If large number of atoms, the number decaying would be larger. Radioactive decay is an exponential process. The number, N, of parent atoms decreases exponentially over time so do the number of decays/ unit time. 101

102 As number of parent atoms decreases, the rate in units per second decreases exponentially. The daughter product grows as a mirror image. 102

103 Half Life t 1/2 defined as: time it takes for half of original parent in a sample to decay, or time for original mass to be halved, or time for activity to be halved. 103

104 Activity in counts/sec (Bq) easier to measure. It’s the decay rate of the sample. 104

105 What is the half life of this sample? 105

106 Ex 1: The half life of Quinnium is 4 days. If we start with 160 g of pure Quinnium, how much will be left after 12 days?? 106

107 160 g 80 g 40 g 20 g 4 days 8 days 12 days 3 half lives have passed 107 Exact multiple of half lives:

108 Rd Hamper 7.4 Do 15-17 pg 162 and finish #2 pg 167 108

109 Radioactive Decay Equations The activity defined av # of disint/ sec for a particular sample unit = becquerel (Bq). Activity of sample: A = activity (Bq) counts or decays/s  N =# disintegrations  T = time (s) Negative sign b/c disintegrations are decreasing as N decreases.

110 Ex 1: If a sample starts with 1000 parent atoms and decreases to 100 parent atoms in 3600 seconds, what is the activity of the sample? 900 atoms/ 3600 s = 0.25 Bq. 110

111 Activity is proportional to N (parent atoms) at any time so: is radioactive decay constant s -1. We can’t predict which nucleus will decay but we can predict the number of nuclei decaying in a specific period of time. 111

112 Ex 2: A sample has an activity of 2 x 10 3 Bq and a of 5 x 10 -5 s -1. What is the number of remaining parent atoms at this time? A/ = N 2 x 10 3 Bq/ 5 x 10 -5 s -1 = 4 x 10 7 parent atoms remain. 112

113 Derive the half life equation. 113

114 The graph below shows the fraction of original atoms remaining with the passage of ½ lives. The number of parent atoms decays exponentially over time. The Y axis is the activity of the sample. N

115 The equation of the graph is: N = N o e - t. N o original # atoms (parent) = decay constant. N # (parent)atoms remaininge is natural log Radioactive Decay Law 115

116 Relating ½ life to decay constant N = N o e - t. When t is t 1/2, then N = N o. 2 Rearrange: 2 = N o. N 116

117 But when t = t 1/2 2 = N o. N Take ln both sides: ln 2 = T 1/2. T 1/2 = ln2 so flip

118 N = N o e - t. can be written in terms of original sample activity rather than number of atoms. 118 A =  o e - t

119 119 Since: N = N o e - t. Multiply both sides by. N = N o e - t. & A = N o e - t A =  o e - t

120 120 Ex 3: A sample of radioactive isotope originally contains 1.0 x 10 24 atoms and has a half-life of 6.0 hours. Find the: a. decay constant b. initial activity c. number of atoms remaining after 12 h. d. number of atoms remaining after 30 min.

121 How to determine the half life. 1. Measure the activity using ionizing properties (Geiger Counter) 121

122 Short Lived Nuclei – read it off 122

123 Longer Lived Isotopes (long half life) Plot N vs t on semi log graph Find gradient 123

124 Detection Devices Electroscope - charged electroscope can be discharged by oppositely charged ions. Rate of discharge related to greater radioactivity. 124

125 Other Detection Methods Ionization Chamber Geiger-Muller Detector Gas filled tubes. Use the fact that radiation can ionize other atoms thereby giving them a charge. Current is generated and detected. 125


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