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From Ideas to Implementation

From Ideas to Implementation. Core Module 9.4. Introduction. A new dawn in physics had begun at the start of the twentieth century… …it appeared that the stage had been reached when all the disparate elements of scientific knowledge seemed to be coming together…

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From Ideas to Implementation

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  1. From Ideas to Implementation Core Module 9.4

  2. Introduction A new dawn in physics had begun at the start of the twentieth century… …it appeared that the stage had been reached when all the disparate elements of scientific knowledge seemed to be coming together… …many scientists felt that the total knowledge about the universe lay just beyond their horizon… …future generations of scientists, it was believed, would have nothing to discover… …this idea was about to be challenged.

  3. Until now, physicists believed that all phenomena could be broken down into fundamental interactions that were capable of being described by the laws of physics. However, early in the twentieth century, physics, or classical physics as it is now known was to be shaken to the core by two remarkable new insights about the universe. The first of these insights was was made by Albert Einstein when he proposed his now famous Theory of Special Relativity in which he showed that space and time were interrelated.

  4. The second insight came when Max Planck realised that in spite of his best efforts to make the laws of classical physics fit natural phenomena on the atomic scale, these laws were simply incapable of resulting in an explanation. These insights lead to the development of quantum physics, which was primarily developed to explain atomic phenomena and eventually proved to be more fundamental than classical physics. By extending the laws of quantum physics to large masses, distances and times, it was shown that it was possible to derive the laws of classical physics.

  5. In this unit, we will investigate some of the fundamental ideas of quantum physics. We start with the discovery and identification of cathode rays and how this ultimately led to modern television. We then look at how black body radiation and the photoelectric effect led to a new view of the nature of electromagnetic radiation with the quantum theory of Max Planck. We conclude by looking at how the study of sub-atomic particles led to the invention of the transistor and to the discovery of superconductivity.

  6. Students learn to: Students: • explain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves • explain that cathode ray tubes allowed the manipulation of a stream of charged particles • identify that moving charged particles in a magnetic field experience a force • identify that charged plates produce an electric field • describe quantitatively the force acting on a charge moving through a magnetic field • discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates • describe quantitatively the electric field due to oppositely charged parallel plates • outline Thomson’s experiment to measure the charge/mass ratio of an electron • outline the role of: • – electrodes in the electron gun • – the deflection plates or coils • – the fluorescent screenin the cathode ray tube of conventional TV displays and oscilloscopes • perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes • perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes: • – containing a maltese cross • – containing electric plates • – with a fluorescent display screen • – containing a glass wheel • – analysethe information gathered to determine the sign of the charge on cathode rays • solve problem and analyse information using: • and 1. Increased understandings of cathode rays led to the development of television

  7. In 1855, Heinrich Geissler (a glass blower and maker of scientific equipment) had refined a vacuum pump that could evacuate a glass tube to within 0.1% of normal air pressure. Discharge Tubes By the 1850’s, much was known about which solids and liquids were electric conductors or insulators, and it was thought that all gases were electric insulators.

  8. Julius Plucker (a colleague of Geissler’s) sealed a wire in each end of a glass tube and connected it to a high voltage power supply. He then steadily removed the air from inside using Geissler’s pump. To their surprise, they found that at reduced pressures, air (and all other gases) conduct electricity! Plucker showed that as the pressure is reduced in the discharge tube (as it is now known), a series of changes progressively take place. We will examine these in the following first hand investigation:

  9. ACTIVITIES • Experiment 20A: Discharge tubes

  10. Discharge Tube Sample observations: - + Cathode Anode Negative glow Positive striations Crooke’s dark space Faraday’s dark space As the pressure of the gas inside the tube was reduced, both the anode and cathode were surrounded by a luminous glow. As the pressure was reduced further, the positive glow extended about halfway along the tube (between the positive glow and the negative glow is an area called the Faraday dark space). At even lower pressure, the positive glow broke up into columns called striations. Lower still, an area called the Crooke’s dark space filled the entire tube and a green glow appeared in the glass at the end of the discharge tube opposite the cathode.

  11. By 1875 Sir William Crookes had made further discoveries. He built and experimented with a variety of different tubes (which he called Crookes tubes - variations of the tubes used by Geissler and Plucker). Crookes Tubes The discharge coming from the cathode had by now been named cathode rays. The Crookes tubes had structures built into or around them to allow the cathode rays to be manipulated. For example, electrodes were built into the cathode ray tube to create an electric field to attempt to change the path of the cathode rays. Magnetic fields were applied to the cathode rays through the glass from outside the tube, and solid objects were placed inside the tube to block the path of the rays.

  12. ACTIVITIES • Prepare a Risk Assessment for the following experiment • Experiment 20B: Properties of Cathode Rays

  13. Sample observations: A fluorescent screen shows that cathode rays can cause fluorescence. This demonstrates that cathode rays have energy. A fluorescent screen can also be used to trace the path of cathode rays being manipulated by other means.

  14. A Maltese cross placed in the path of the cathode rays causes a clearly defined shadow at the end of the tube. This effect is used to infer that cathode raystravel in straight lines and are blocked by solid objects.

  15. Charged plates Pairs of electric plates cause the cathode rays to bend towards the positive plate. This shows that cathode rays have negative charges. Note that Crookes believed that cathode rays could be deflected by an electric field but never succeeded in demonstrating this experimentally.

  16. first A lightweight glass paddle wheel, able to rotate freely, is placed in the path of the cathode rays. The cathode rays cause the wheel to spin and move away from the cathode. This demonstrates that the cathode rays have momentum, and therefore mass, and that they are emitted from the cathode.

  17. A cathode ray in the presence of a magnetic field will deflect (bend). As we already know, moving charges experience a force in a magnetic field (RHPR). This shows that cathode rays carry a negative electrical charge. Use the diagrams to show that the cathode rays are negatively charged

  18. These (and other) experiments showed that cathode rays: • always have the same properties regardless of the metal in the cathode • travel in straight lines • carry energy • have mass • cause a green luminescence when they strike glass • change the colour of some silver salts (i.e. expose photographic film) • pass straight through thin metal foil without damaging it • travel considerably more slowly than light • are deflected by magnetic fields • are deflected by electric fields (we’ll come back to this one!!)

  19. The cathode ray debate – waves or particles? The controversy as to whether cathode rays were streams of charged particles or electromagnetic waves is a great example of the nature of scientific development. This particular debate raged for years.

  20. Some of the properties of cathode rays suggested to physicists that cathode rays were a wavesimilar to light. For instance: • they produced fluorescence • they travelled in straight lines • they pass straight through thin metal foil • they produced similar chemical reactions on photographic film to those produced by light • they were not deflected by electric fields (or so it seemed). • Yet cathode rays were deflected by a magnetic field as if they were negatively charged particles. This apparently inconsistent behaviour of cathode rays led to much controversy over the nature of cathode rays.

  21. The debate was finally resolved in 1897 by J.J. Thompson. The reason the debate took so long to clear up was that initially the cathode rays couldnotbe deflected by an electric field. J.J. Thompson was able to make a cathode ray tube with sufficiently low pressure and using a very high voltage supply he was able to deflect the cathode rays with an electric field. This showed conclusively that cathode rays were beyond doubt streams of charged particles. We now call these charged particles electrons! Before we look at J.J. Thompson’s experiment, we need to look at the effects of charged particles moving in electric and magnetic fields.

  22. ACTIVITIES • Read 10.1 • Complete 1 – CRTs • Complete 2 – Properties of Cathode Rays • Complete Assignment 1

  23. Electric fields An electric field is said to exist in any region in which an electrically charged object experiences a force. An electric field has both strength and direction. • The strength of the electric field due to a point charge diminishes with distance from it. The direction of the field around a point charge: • is directed radially away from a positive point charge; and • is directed radially towards a negative point charge.

  24. The number of lines drawn to represent a field at any point indicates the electric field strength at that point. The stronger the field, the more lines are drawn in a given space. The electric field between two oppositely charged parallel plates is uniform in strength and direction. The field direction is at right angles to the plates and away from the positive plate.

  25. Electric field between two parallel charged plates The electric field due to charged parallel plates It should be noted that the electric field is not uniform at the edges of the parallel plates:

  26. Hence, Note: the syllabus expects you to be able to solve problems using this equation E = V/d The electric field strength, E, between two oppositely charged parallel plates is: • proportional to the potential difference, V, between the plates; • inversely proportional to the separation, d, between the plates; Where: E is electric field strength (V/m) V is the potential difference between the plates (V) d is the distance between the plates (m)

  27. Practice Questions

  28. Example: In a glass tube from which air has been removed, electrons are accelerated by a uniform electric field between a pair of charged parallel plates, separated by 5 cm. If an electrical potential difference of 4 kV is applied across the plates, determine the electric field strength half way between the plates. Solution: This is a trick question. The field is UNIFORM, hence the field strength is the same in all areas between the plates: E = V/d E = 4000 / 0.05 E = 8 x 104 V/m (N/C)

  29. Note: the syllabus expects you to be able to solve problems using this equation F = qE A charged object placed in an electric field will experience a force. The magnitude of the force acting on this charged object is given by: Where: F is the force acting on the charged object (N) q is the magnitude of the charge of the object (C) E is the electric field strength (N/C or V/m)

  30. Example: • A particle carries a charge of -3.2x10-19C. It is placed in a uniform electric field of strength 5000 V/m. • Where might such a uniform field exist? • Determine the electrostatic force on the particle. Solution: • Between two oppositely charged parallel plates • F = Eq • F = 5000 x (-3.2x10-19) • F = -1.6x10-15 N (the negative reminds us that the force is in the opposite direction to the electric field)

  31. X X X X X X X X X X I X X X X X X X X X X F X X X X X B + Moving charges in magnetic field experience a force As we saw in the topic ‘Motors and Generators’, a magnetic field exerts a force on electric currents – that is, on moving charged particles. The direction of the force can be determined using the right hand palm rule. Remember that current is in the opposite direction to the flow of electrons. The magnitude of the force is a maximum when the charge moves perpendicular to the magnetic field:

  32. A constant magnitude force, which is perpendicular to the velocity of a particle, results in the particle travelling in a circular path. Can you explain why? The force, F, on a charge moving through a magnetic field is: • proportional to the size of the charge, q; • proportional to the velocity of the charge, v; • proportional to the magnetic induction, B; and

  33. F = qvB sinq Note: the syllabus expects you to be able to solve problems using this equation • proportional to the sine of the angle, q between the velocity and the magnetic field lines (being a maximum when it is 90o, and zero when the velocity is parallel to the field). F is the force on the charged object (N)q is the charge of the object (C) v is the velocity of the charged object (m s-1)B is the magnetic field flux density (T) q is the angle between the velocity and the magnetic field.

  34. Example 1:A proton travelling at 5.0x103 m s-1 enters a magnetic field of strength 1.0 Tesla at 90o. Determine the magnitude of the force experienced by the proton.Solution:

  35. Example 2: The path of a helium nucleus, travelling at 3.0x103 m s-1, makes an angle of 90oto a magnetic field. The electron experiences a force of 1.2x10-15 N while in the field. Calculate the strength of the field.Solution:

  36. into the page Example 3: A charge of +5.25 mC moving with a velocity of 300 m/s north east, enters a uniform magnetic field of 0.310 T directed from left to right across the page. Calculate the magnetic force acting on the charge. Solution:

  37. ACTIVITIES: • Read 10.2 – 10.3 • Complete Assignment 2 • Complete 4 – Electric Fields • Complete 5 – Charges in Magnetic Fields 1 • Complete 7 – Charges in Magnetic Fields 3

  38. J.J. Thompson’s charge to mass ratio experiment With the knowledge of how charges are influenced by electric and magnetic fields, we can return to the cathode ray controversy. Remember that it was J.J Thomson who finally established the true nature of cathode rays – they were negatively charged particles. In order to do so he carried out an astonishing experiment. He subjected beams of cathode rays to deflection by electric and magnetic fields set at right angles to each other. By doing so, he was able to measure the charge to mass ratioof cathode rays.

  39. In his experiment, cathode rays were passed through two narrow slits to make a thin parallel beam aimed at the centre of a fluorescent screen. Charged parallel plates were placed to create a uniform electric field that exerted a upwards force on the beam. An external electromagnet (called a Helmholtz coil) was placed to produce a uniform magnetic field that exerted a downward force on the beam.

  40. Note: the syllabus does not expect you to be able to solve problems using these equations, but you can use them when you explain Thompson’s experiment. Thomson manipulated the strengths of the two fields until the beam passed through both fields undeflected. This occurred when the two forces on the particles in the beam were equal and opposite. By equating the expressions for these two forces, Thomson calculated the velocity of the particles. Mathematically: FB = FE therefore, qvB = qE hence, v = E/B

  41. Note: the syllabus does not expect you to be able to solve problems using these equations, but you can use them when you explain Thompson’s experiment. Thomson then removed the electric field and calculated the radius of the circular path followed by the particles in the uniform magnetic field alone. By equating the force due to the magnetic field to the centripetal force, he was able to calculate that all cathode ray particles had the same charge to mass ratio of 1.76 x 1011 C kg-1 – this was 1800 times larger than the q/m value for a hydrogen ion (a proton). Mathematically: FB = FC therefore, qvB = mv2/r hence, q/m = v/Br

  42. In other experiments, Thomson showed that the charge on the cathode ray particles was the same magnitude as the charge on a hydrogen ion (a proton). This combined with the fact that the q/m ratio for cathode rays was 1800 times larger than that for the hydrogen ion meant that the mass of the cathode ray particles had to be 1800 times smaller than that of the proton. On the basis of all these results, Thomson suggested that the cathode ray particle was a fundamental constituent of the atom (i.e. a sub-atomic particle). Although he originally referred to the particles as ‘corpuscles’, the name electron slowly became accepted as the official name.

  43. fluorescent screen The cathode ray TUBE • A cathode ray tube (CRT) consists of three main components:  • A fluorescent screen • An electron gun • The deflection system

  44. The Fluorescent Screen: The end of the CRT is coated on the inside surface with some fluorescent material such as zinc sulfide (ZnS). When an electron strikes the end of the tube, the material fluoresces, that is gives off light. This enables a spot of light to appear wherever an electron (from the electron gun) strikes the end of the tube. electron beam deflection plates

  45. electron beam deflection plates The Electron Gun: This produces a narrow beam of electrons. It consists of a filament enclosed in a cylindrical cathode electrode, a ring-shaped electrode called the grid and two cylindrical anode electrodes. The filament is heated by passing electric current through it. Electrons are then emitted due to thermionic emission from the heated filament. These electrons are accelerated towards the anodes by the electric field set up between the cathode and anodes. The dual anode system helps to focus the electron beam into a single beam. The ‘grid’ is negatively charged, and its magnitude can be varied easily. What effect would it have on the brightness of the screen? Why?

  46. The grid is placed between the cathode and anodes and constantly varies its negative charge. This enables the intensity of the electron beam to be controlled – the more negative the grid, the fewer electrons are permitted to leave the electron gun and ultimately the spot on the end of the tube is less bright.

  47. The Deflection System: This allows the electron beam to be deflected from the straight-line trajectory with which it leaves the electron gun. The deflection system usually consists of two sets of parallel plates, one set in the horizontal plane, and the other in the vertical plane. When potential differences are applied between each set of plates, electric fields are set up between the plates. The electrons in the beam then experience forces vertically while passing between the horizontal plates and horizontally while passing through the vertical plates. Thus, by applying appropriate voltages to the deflection plates, the position of the spot on the end of the screen can be controlled.

  48. Both sets of plates are uncharged, and the cathode ray passes through undeflected.

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