1. Photons SACE Stage 2 Physics

2. Photons Consider a darkened room with the Young’s Double Slit experiment setup. The light source is releasing very low levels of light. Using a photo sensitive screen the pattern observed would look like the following, The image on the screen would indicate that only discrete ‘bits’ of light has reached the screen. If we repeated this experiment a number of times, the same effect would be observed but the positions f the dots would vary.

3. Photons After repeating the experiment a large number of times, and using the same photosensitive screen, the build up of dots on the screen would produce the image seen below. The image is caused by the arrival of localised bundles of energy. As more and more of these bundles arrive at the screen, the image is built up. These localised bundles of energy is the minimum unit or quantum of energy in which light arrives.

4. Photons In 1900, Max Planck developed a theory that light travelled as bundles of energy and behaved like a wave. He also suggested that these bundles of energy only contained certain amounts of energy reflective of the oscillation of the atom that generated it. E = Energy of the light;n = an integer; h = Planck’s constant = 6.63 x 10 -34 J s F = frequency of the light

5. Photons Einstein then postulated that light travelled as quanta (photons). The energy of each Photon is given by: The momentum of a photon is given by the formula: Where p is the momentum of the photon. But, E=mc 2, therefore,

6. Photons Calculate the energy (in joules and electron volts) and momentum of each of the following photons. (1)Orange light of wavelength 600nm (2)An X-ray of wavelength 0.1nm

7. Photons (1)Orange light of wavelength 600nm

8. Photons (2)An X-ray of wavelength 0.1nm

9. Photons A laser emitting monochromatic light of wavelength 780nm, is rated at a power of 0.1mW. How many photons per second is it emitting?

10. Photons A laser emitting monochromatic light of wavelength 780nm, is rated at a power of 0.1mW. How many photons per second is it emitting?

11. The Photo Electric Effect The emission of electrons from materials (usually metals) which have absorbed light. Ordinary metals require Ultra Violet Light unless coated with a special material, visible light can be used.

12. Experiments on the photoelectric effect In summary, the effects result in, 1. Increase frequency,  stopping voltage increases 2. Decrease frequency to find the minimum “threshold” frequency 3. Increase intensity (brightness)  photo-current increases but no change occurs in the stopping voltage. 4. Type of metal affects stopping voltage 5. Gradual decrease in photo-current as potential difference increases.

13. Measuring KE of Emitted Photons

14. As light is incident on the photo sensitive material, a current flows in the circuit. This is measured using an ammeter. The electrons can be prevented from crossing the gap applying a voltage in the opposite direction to the electron flow. As V is increased, some electrons are repelled back. As V is increased, more electrons are repelled until none reach the other plate. It is at this voltage (the stopping voltage, V s ) that the photo-electric current drops to zero. The gradual current decrease is evidence for a range of electron energies inside atoms.

15. Measuring KE of Emitted Photons Results, (a)Time Delay - the time delay between when the light first hits the metal and the time the electrons are first emitted is very small. (b)Kinetic energy (and hence speed) of emitted electrons – depended upon the emitting surface. Maximum KE is depended upon the frequency of light, not intensity. Higher frequencies produced greater kinetic energies. (c)Number of electrons/sec (and hence current) - proportional to the light intensity but not depend on frequency. (d)Threshold frequency - frequency below which no photoelectrons are emitted.

16. Quantum Theory of Light In 1900 Planck had shown that oscillating charges could only change their energy in discrete lumps or quanta. In 1905 Einstein postulated that the energy of each photon was determined by the frequency of the light according to Planck's formula ie. (Note the inconsistency of this theory - it is not a wave but we use frequency, a wave concept, Planck spent many years trying to prove the theory wrong.)

17. Quantum Theory of Light The photoelectric equation of Einstein All metals contain 'free' electrons that move around inside the metal but cannot escape because of the positive charge on the nuclei of the atoms of the metal. If electrons are given extra energy they may escape. Some electrons are more strongly bound than others and so need larger amounts of energy to escape. We define: Work function W, as the minimum energy required to be supplied to the metal to release an electron.(ie has no excess kinetic energy.) The work function varies with the type of metal.

18. Quantum Theory of Light 1:1 interaction of photons and electrons Einstein proposed that the photons of light interact with the electrons on a 1:1 basis, i.e. a photon imparts all its energy to a single electron or none of it. This postulate is fundamental to his explanation.

19. Quantum Theory of Light Threshold frequency The frequency f 0 of the photon which will just bring about the escape of the least strongly bound electron is called the threshold frequency. The photon with this frequency must have just enough energy to give an electron sufficient energy to allow it to escape. ie. hf 0 = W where f 0 = threshold frequency and W = work function of the metal.

20. Quantum Theory of Light Photoelectric equation

21. Quantum Theory of Light If f is less than f 0 no photo-electrons will be emitted because the photon can not supply even the work function to the electrons. If f is larger than f 0 they will have some kinetic energy left over. Where: K max = hf - W ie.K max = hf - hf 0 K max is the maximum amount of energy an electron can have after being emitted. The work function is the minimum energy required to emit an electron. E(photon) = K(electron) + E(lost to metal)  K(electron) = E(photon) - E(lost to metal) 1 / 2 mv 2 = hf - W

22. Quantum Theory of Light Typical Results

23. Quantum Theory of Light If we compare Einstein's equation with the standard equation of a straight line, K max = hf - W with y = m x + c frequency is a straight line with slope equal to Planck's constant, the x intercept equal to the threshold frequency, the negative y intercept equal to the work function of the metal. Note that the lines for all metals should be parallel as they all have the same slope.

24. Quantum Theory of Light Calculate: (1)the threshod frequency of silver if its work function is 4.73eV; (2)the work function of barium, if its threshold frequency is 5.985 x 10 14 Hz.

25. Quantum Theory of Light (1)the threshod frequency of silver if its work function is 4.73eV;

26. Quantum Theory of Light (2)the work function of barium, if its threshold frequency is 5.985 x 10 14 Hz.

27. Quantum Theory of Light The work function of aluminium is 4.08eV. Determine (1)Whether light of frequency 9 x 10 14 Hz will cause electrons to be emitted; (2)The maximum kinetic energy of the electrons if the metal is irradiated with ultra violet light of wavelength 2.0 x 10 -7 m.

28. Quantum Theory of Light (1)Whether light of frequency 9 x 10 14 Hz will cause electrons to be emitted;

29. Quantum Theory of Light (2)The maximum kinetic energy of the electrons if the metal is irradiated with ultra violet light of wavelength 2.0 x 10 -7 m.

30. Quantum Theory of Light The work function W of zinc is 4.31eV. A zinc surface is irradiated with lights of different frequency. Sketch a graph of the maximum kinetic energy of emitted electrons against the frequency of light.

31. Quantum Theory of Light

32. X-Rays PE effect: incident high frequency light produces photo electrons X-rays: incident high energy electrons produce high frequency electromagnetic radiation.

33. X-Rays X-rays can be produced using the following aparatus. (1% of electrons energy produces the X-ray) Filament F produces a current to heat the wire and emit electrons. Hood H used to focus electrons Anode A electrons accelerated away from filament with a large acceleration due to high PD 40kV to 400kV between the Anode and the Cathode. Target T tungsten target on which the electrons impinge.

34. X-Rays Electrons are accelerated through a potential V,the electrons from the Cathode reach the target T with an energy eV. Few electrons penetrate and interact with target and loses energy.

35. X-Rays Energy available to produce photon is, Where hf is the energy of one photon. When photons are produced, most of the energy is converted to heat at the anode and must be removed, usually by a cooling liquid (i.e., silicon oil).

36. X-Rays The diagram shows a bremsstrahlung photon produced by an electron decelerated by interaction with a target atom. There will be a range of final velocities due to the varying degree of interaction (slowing down) so we have a range of X-ray photons produced. This gives us a continuous range of X-ray photons.

37. X-Rays When the incoming electron loses all of its energy to the target atom, we get the maximum energy photon, i.e., E 2 = 0.

38. X-Ray Spectra Intensity of X-rays emitted by a tube is measured as a function of frequency to plot the X-ray spectrum for that tube. Note: As filament current is increased, the intensity of the X-rays is increased.

39. X-Ray Spectra Three characteristics of X-ray spectra. a. Cut off frequency f max depends on the accelerating voltage but does not depend on the target material or the filament current. The higher f max the more penetrating the X- rays. b. "White" background radiation "Bremsstrahlung" radiation: A continuous range of X-ray frequencies present up to f max. The intensity of this background increases with filament current. c. Sharp Peaks of Intensity. The sharp peaks depend on the type of target. eg. tungsten different to gold.

40. Explanation of Cut-off frequency The energy of each X-ray photon produced is limited to the energy of a single electron. Usually the electron loses its energy in stages producing X- ray photons of smaller energy than its own K. eg.

41. Explanation of Cut-off frequency The highest energy photons is emitted when the most energetic electron loses all of its energy to produced the most energetic photon. This is rare so don’t expect to see many photons at f max.

42. Examples X-rays are generated in an X-ray tube with an operating voltage of 50kV. Find(1)The maximum energy of the emitted photons; (2)The maximum frequency of the emitted photons.

43. Examples X-rays are generated in an X-ray tube with an operating voltage of 50kV. Find(1)The maximum energy of the emitted photons; (2)The maximum frequency of the emitted photons. (1)Energy of electrons reaching anode = 50000eV  the maximum energy of the emitted photons = 50000eV (2)

44. Examples An X-ray tube emits radiation with a minimum wavelength of 0.3nm. What is the operating voltage of the tube?

45. Examples Electrons must have 4144eV of energy when arriving at the anode, therefore the voltage of the tube = 4144V.

46. Application: X-Rays in Medicine X-ray photons are of high energy and hence can penetrate human tissues in a manner not possible for lower energy ultra-violet, visible and infra-red electromagnetic radiations. When a beam of X-rays is reduced in intensity we say that it has been ATTENTUATED.

47. Application: X-Rays in Medicine The degree of absorption of an X-ray in the human body depends on, The density of the tissue X-rays will penetrate further into soft tissue than bone due to the density of these materials. That is, the attenuation of X-rays is greater in more denser tissue. The thickness of the tissue A given thickness of soft tissue attenuates a beam less than does the same thickness of bone. The atomic number of the elements that make up that tissue the higher the atomic numbers of the elements present (there are more 'heavy' elements in bone than in muscle) the greater the attenuation.

48. Application: X-Rays in Medicine The penetrating power of X-ray photons depend on; The hardness of the beam depends on the accelerating voltage For a given hardness the intensity of the beam depends on the current of electrons in the tube (tube current) The tube current depends on the filament current.

49. Application: X-Rays in Medicine When using X-rays on a person, it is important to keep exposure time to a minimum. This can be done by using a very sensitive film, increasing the filament current (number of electrons hitting the target) and increasing the Voltage (determines the quality of the X-rays).