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Modern Physics By Neil Bronks Atoms C 12 6 Mass Number Mass Number - Number of protons + Neutrons. Atomic Number Atomic Number - Number of protons In.

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Presentation on theme: "Modern Physics By Neil Bronks Atoms C 12 6 Mass Number Mass Number - Number of protons + Neutrons. Atomic Number Atomic Number - Number of protons In."— Presentation transcript:

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2 Modern Physics By Neil Bronks

3 Atoms C 12 6 Mass Number Mass Number - Number of protons + Neutrons. Atomic Number Atomic Number - Number of protons In a neutral atom the number of electrons and protons are the same. In Carbon it is……… 6

4 Hydrogen Electron Proton The simplest atom has one negative electron orbiting one positive proton. The electron is very light compared to the proton.

5 Helium Neutron In this atom we see two neutrons and two protons forming the nucleus. The Neutron has no charge but is the same mass as the proton. Electron Proton

6 4 Forces of Nature (Order of strength) Gravitational - Only Positive - Very long range Weak Nuclear- Associated with beta decay Strong Nuclear – Holds nucleus together - Very Short Range Electromagnetic – Positive and negative

7 Radiation Decay of nucleus by the emission of a particle or a ray. Discovered by Henri Becquerel Units 1 Bq is one decay per second Natural happens without outside bombardment Artificial happens due to bombardment J’ai fais ça ! Dodgy Beard

8 Safety Wear Gloves or Apron of lead Don’t point at anyone Don’t eat!!!

9 Nuclear Equations Top and bottom must add up Top is mass number Bottom is atomic number Proton H Neutron electron And Alpha

10 Alpha Particles  Helium Nuclei Positive Charge Heavy so not very penetrating Very Ionizing Very Ionizing 14 7 N He  17 8 O H

11 Beta Particle ß Fast electron from the nucleus Negative charge Moderately Penetrating Moderately ionizing 14 7 N  14 8 O ß

12 Gamma Ray  High energy e-m wave (A Photon) No charge - not deflected by field Very penetrating – Need lead to stop most of them Not very ionizing Release energy after reaction

13 Penetrating Power  Paper Al foil Concrete

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16 H/W LC Ord 2007 Q11

17 Particles in Fields    Charged particles move in a circular path as the force is always at right angles to the direction of motion- Fleming's Left Hand Rule Radioactive Source Cloud Chamber

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19 Click here for internet demo

20 Ionisation We can prove that radiation creates ions as we bring a source close to a charge electroscope 

21 Ionising Power Alpha is heaviest and so does most damage – poison with Polonium Beta is only moderately ionising Gamma is only slightly ionising but difficult to stop

22 Solid State Detector PN This a P-N junction in reverse bias. This creates a huge depletion layer. + - A piece of radiation passes through the depletion layer and creates enough carriers to carry one pulse of current.

23 Geiger Muller Tube

24 H/W LC Ord 2004 Q10

25 Experiments All experiments the same stick a DETECTOR in front of a source and count the decays. Move it away for distance and plot Time for half life and plot Put things in front for penetration

26 Penetration A Gieger Muller Tube and Counter. Plot the activity against the thickness or the type of barrier

27 Distance A Gieger Muller Tube and Counter. Plot the activity against the distance r. r

28 Half Life A Gieger Muller Tube and Counter. Plot the activity against the time Time it takes for half the atoms to decay

29 Half-Life – time it takes for half the radioactive particles to decay Atoms Not Decayed Time 1234

30 Half life demo from internet click here

31 Half Life Calculations 1000 particlestime=6s 2 half-life 500 particles time=9s 3 half-life 250 particlestime=12s 4 half-life 125 particlestime=15s 5 half-life 4000 particlestime= particlestime=3s 1 half-life

32 Calculations – we use the decay constant λ in our calculations. =0.693/T ½ =0.693/3s =0.231s -1

33 Activity Calculations Rate of Decay = x number you started with dN/dt = - x N Start with 4000 particles and =0.231 Activity = 4000 x 0.231=924 Bq

34 Calculations 1) You start with 100 grams of sulfur-35, which has a half life of days. How much time will it take until only 12.5 grams remain? How many half lives? 100>50>25>12.5 so 3 half lives Time = 3 x = days

35 Calculations 2) You measure the radioactivity of a substance, then when measuring it 120 days later, you find that it only has 25 % of the radioactivity it had when you first measured it. What is the half life of that substance? How many half lives 100%>50%>25% 2 half lives =120 days 1 half life = 60 days

36 Calculations 3) Your professor gives you 64g of phosphorus- 32 (half life = days). (a) What is its decay constant ? (b) What is its activity (Rate of Decay)? (a) Using the formula =0.693/T ½ =0.693/(14.263x24x60x60) = 6.62 x s -1

37 Calculations 3) Your professor gives you 64g of phosphorus- 32 (half life = days). (a) What is its decay constant ? (b) What is its activity (Rate of Decay)? = 6.62 x s -1 (b) Using Activity =dN/dt = - N N= Moles x 6x10 23 = 2 x 6x10 23 Activity = 6.62 x x 12x10 23 = = 7.3 x10 17 Bq

38 Isotopes Same atomic number different mass number

39 Isotope pp

40 Uses of Radioactive Isotopes Medicine – treatment and imaging Smoke detectors Food Irradiation Carbon-14 Dating

41 Isotopes Same Atomic number different Mass number

42 Carbon-14 Dating At death all animals contain the same ratio of C-14 to C-12 The rate of decay of C-14 is fixed The C-14 left tells us how long ago it died %C-14 time

43 H/W LC Ord 2005 Q12(d) LC Higher 2003 Q11 LC H (d)

44 Rutherford Scattering Do I look like Freddie?

45 Rutherford on internet

46 Rutherford Scattering – alpha particles fired at gold foil. very small Most pass unaffected - So the nucleus is very small

47 Rutherford Scattering – alpha particles fired at gold foil. Nucleus Deflected A small number of high energy alphas are Deflected totally positive. Some reflected completely back - Nucleus totally positive.

48 Rutherford Scattering – alpha particles fired at gold foil. Nucleus Deflected A small number of high energy alphas are Deflected very small More pass unaffected - So the nucleus is very small totally positive. Some reflected completely back - Nucleus totally positive.

49 Cockcroft and Walton Alpha pha  Lithium Target Alpha  Hydrogen discharge tube Accelerated by An huge electric Field (700000v) Proton Alpha strikes the screen Producing a flash that Is seen with the microscope

50 Internet explanation

51 Nobel Prize for Physics Proton + Lithium  2xAlpha + Energy Proves Einstein ’ s Law E=mc 2 First Transmutation by artificial Bombardment of an element Ernest Walton

52 Binding Energy The total nucleus weighs less than all its parts Mass Defect Difference is Mass Defect Converted to energy to hold the nucleus together Converted to energy to hold the nucleus together E=mc 2 E=mc 2

53 As Iron is the most stable as you go towards it you release energy So Carbon-12 is lighter than 12 protons The difference is the binding energy

54 Binding Energy of a Deuteron A deuteron is the nucleus of a deuterium atom, and consists of one proton and one neutron. The masses of the constituents are:deuteriumprotonneutron m proton = u (u is Atomic mass unit ) Atomic mass unit m neutron = u m proton + m neutron = = u The mass of the deuteron is: Atomic mass 2 H = u The mass difference = = u

55 Convert to Kg Multiply by conversion factor 1u = 1.66x Kg Mass = m = ( ) x 1.66x Mass = m = 3.96x Kg

56 Use Famous Formula E=mc 2 E=mc 2 E= x 2 E= 3.96x Kg x (3x10 8 m/s) 2 E = E = 3.56x Joules

57 Fusion – The sun and the stars Fusion is the joining together of 2 light nuclei to make one nucleus with release of energy. Fusion is the joining together of 2 light nuclei to make one nucleus with release of energy. Caused by a super fast collision at high temperature in a magnetic bottle. Caused by a super fast collision at high temperature in a magnetic bottle. 21H21H 21H21H

58 Fission The breaking apart of a heavy nucleus to form smaller nucleus with release of energy. The breaking apart of a heavy nucleus to form smaller nucleus with release of energy. Caused artificially by the bombardment of the right speed of neutron. MASS DEFECT In both fusion and fission the products are lighter than the reactants and the MASS DEFECT is turned into Energy E=mc 2

59 1.Subtract mass in a.m.u. 2.Convert to kg 3.Use E=mc 2 Also produced 3 fast neutrons that can cause another fission and so a chain reaction Uranium-235

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62 Nuclear Equation In the isotope U-238 the neutrons must be slowed down by a moderator - Graphite

63 Fuel rods contain the Uranium-235 (Enriched to ensure chain reaction) Moderators (Graphite) slow down the neutrons to the right speed Control Rods (Boron Steel) absorb neutrons to stop the reaction and prevent meltdown Heat to heat exchanger prevents Radiation escaping steam to turbine

64 H/W LC Ord 2006 Q 9

65 Leptons Fundamental particles 1/1846 of an a.m.u. Does not feel the strong nuclear force Matter – Electron, Muon, Tau, …… Anti-matter – Positron, Anti-Tau Anti-matter first suggested by Paul Dirac

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67 Annihilation e+e+ e-e- Matter combining with anti-matter to form energy in the form of e-m radiation 2 photons conserve momentum ? e + + e - 2hf (  )

68 Annihilation e+e+ e-e- An electron and a positron collide to make energy. All the mass of the electrons gets turned into gamma waves So Energy E=mc 2 To find frequency of wave E = 2h.f Matter turns Into energy Matter combining with anti-matter to form energy in the form of e-m radiation 2 photons conserve momentum

69 Wave made by Anihilation A proton and a Anti proton. The masses of the constituents are:protonAnti m proton = u (u is Atomic mass unit)Atomic mass unit m proton + m anti = = u The mass difference = u To use this in a calculation we covert to kg

70 Convert to Kg Multiply by convertion factor 1u = 1.66x Mass = m = ( ) x 1.66x Mass = m = 3.34x Kg

71 Use Famous Formula E=mc 2 E=mc 2 E= x 2 E= 3.34x Kg x (3x10 8 m/s) 2 E = E = 3.01x Joules

72 Use Planks Equation E=hf h= planks constant 3.01x Joules = (6.6x js)x(f) f= 3.01x Joules / 6.6x js = 4.56x10 23 Hz In practice this is low as KE from particles increases this.

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75 Pair Production e+e+ e-e- An electron and a positron are created from a gamma ray. (We can also get a proton and an anti-proton) We do the calculation in reverse To find energy of wave E = h.f As we get 2 electrons E = 2mc 2 A matter and anti-matter pair being created by energy from an e-m wave

76 Annihilation and Production p+p+ p-p- New particles are produced from the KE of the colliding protons They must conserve charge If we carry in 4Gev (1.6x x10 9 = 6x J) As Energy to make 3 Pions is E=mc 2 =(3x x xcxc) =6.7x J Subtracting we find the KE after collision. p+p+ p-p- +0-+0-

77 H/W LC Higher (a)

78 Quarks - Inside the Hadrons 6 Quarks 6 Anti-Quarks – Opposite Signs UP +2/3 STRANGE -1/3 TOP +2/3 DOWN -1/3 CHARMED +2/3 BOTTOM -1/3

79 Hadrons BaryonsMesons 3 quarksQuark+anti-quark ProtonPion uudud Feels strong nuclear force

80 Baryon Meson

81 Hadrons on internet

82 Subject to all forces Particle Zoo Hadrons Leptons Fundamental particles Do not feel Strong Nuclear Force Baryons 3 Quarks Proton uud Mesons Quark + Anti-quark Pion ud

83 Ghost Particle Mystery By 1930 most of the particle physics world was understood However the decay of the neutron to a proton producing a beta particle did not obey Einstein's Law n 0 → p + + e - Pauli said there must be a new particle called a neutrino

84 Beta decay In β− decay, the weak force converts a neutron into a proton while emitting an electron and an antineutrino n 0 → p + + e - + ν e This explains loss in energy and momentum. Pauli proposed it’s existence in 1930 but was not discovered until 1956 as it is so weakly interacting with other particles.

85 Nuclear Formula

86 Particle Accelerators-Linear Very high Voltage electric fields Electro-magnetic attraction pulls particles down.

87 Circular Accelerators Particles spiral in fields (Flemings Left hand rule) Cyclotron- We put the field at right angles e more power with oscillating field

88 CERN Particles can travel in opposite directions and double the collision energy

89 Magnets force particles in circular path so stay in tube Circular more compact High velocity needed to overcome repulsion Vacuum to avoid collisions increase mean free path More velocity more KE so more new particles made

90 Detectors

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92 H/W LC Ord 2002 Q11 LC Higher (a)

93 H/W Roundup (Yeh har) 2007 q q q q12(d) 2006 q (a)


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