17.-Quantum-Physics-master.pptx physics IGCSE and IAL
- 1. Lesson 35 : Light Energy
- 2. 22. Quantum Physics 22.1 Energy and momentum of a photon Learning outcomes 1 understand that electromagnetic radiation has a particulate nature 2 understand that a photon is a quantum of electromagnetic energy 3 recall and use E = hf 4 use the electronvolt (eV) as a unit of energy 5 understand that a photon has momentum and that the momentum is given by p = E / c
- 3. Light: Particle or Wave?!
- 4. The first ever photograph of light as both a particle and wave
- 5. Photons • Einstein postulated that light is composed of wave packets called photons. Where the energy of each photon could be found using: Energy of photon = hf Planck constant= 6.63x10-34 Js frequency
- 6. From this equation we can see that: E f Thus, high frequency means high energy. Since c=f We can say that: 𝐸= h𝑐
- 7. Recall: Electron volts An Electron volt (eV) is a unit of energy equal to the work done when 1 electron is moved through a p.d. of 1V. Work Done = Charge x Voltage E = QV or Where Q is the charge of an electron (1.6x10-19 c) ∴ therefore 1eV = 1.6x10-19 C x 1V 1eV = 1.6x10-19 J
- 9. Kinetic energy You can find the speed of the electron from the anode to cathode voltage V using: m= eV
- 10. Practical: Estimating the value for Planck constant.
- 11. Momentum of photons Photons are massless particles! Is it possible for something to have zero mass and have momentum? Recall the formula for momentum! ( p = mv) Einstein postulated that the steady stream of photons in space were responsible for exerting force (pressure) on the objects in space (eg the tail on the comet pointing away from the sun). Thus, it was said that photons must possess some form of momentum. Momentum of PHOTON: 𝑝= 𝐸 𝑐
- 13. Extension: Past paper question
- 14. Lesson 36: Photoelectric effect
- 15. 22. Quantum Physics 22.2 Photoelectric effect Learning outcomes 1 understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation 2 understand and use the terms threshold frequency and threshold wavelength 3 explain photoelectric emission in terms of photon energy and work function energy 4 recall and use hf = Φ + 1/2mvmax 5 explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity
- 16. Photoelectric effect. te: Each photoelectron only absorbs 1 photon. No. of photoelectrons emitted ∴ second (ie photoelectric current) is proportional to the intensity of the incident l suming the frequency is greater than the threshold frequency)
- 19. Problems with the classical Wave model of light 1) Photoelectric emission does not occur if the frequency is below a certain value. This is called the threshold frequency. No matter how much the intensity of the incident radiation, electron emission will not occur if below this threshold frequency. 2) The photoelectric effect occurs without delay (provided the frequency is greater than the threshold frequency) 3) Number of electrons emitted is proportional to the intensity of radiation provided frequency above threshold frequency
- 21. Continued…
- 22. hf = Ek(max) + 𝛟 The minimum energy needed to free the electron
- 24. Question: • Lithium and Berylium have work functions of 2.3eV and 3.9eV respectively. a) Determine the threshold frequency for each material b) If a 400nm light is incident on each of these materials determine which of them will exhibit the photoelectric effect. c) The maximum K,E for photo electrons in each case.
- 25. Lesson 37: Wave-particle duality
- 26. 22. Quantum Physics 22.3 Wave-Particle duality Learning outcomes 1 understand that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature 2 describe and interpret qualitatively the evidence provided by electron diffraction for the wave nature of particles 3 understand the de Broglie wavelength as the wavelength associated with a moving particle 4 recall and use λ = h / p
- 28. de Broglie’s wavelength In 1924 Louis De Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass – such as electrons – do as well. The wavelength of these ‘material waves’- also known as the de Broglie wavelength – can be calculated from using: or
- 29. Electron Diffraction A beam of electrons directed towards a graphite target diffracts like a wave to produce diffraction rings thus proving that matter has wave like properties too. Braggs equation: The wavelength of an electron can be found using: Where d the lattice spacing of material (graphite) λ=2⋅d sin( ⋅ θ) d
- 30. People waves From the de Broglie equation it is apparent that any object with mass and speed has a wavelength. Thus, humans behave like waves too. Calculate/estimate your own wavelength walking through a door. If humans have wavelengths, then why do we not randomly diffract when walking through a door?
- 31. Fig. 28.25. diffraction pattern for an alloy of titanium or Nickel Fig. 28.26. Structure of the giant molecule DNA, deduced from electron diffraction
- 32. Lesson 38: Line Spectra
- 33. 22. Quantum Physics 22.4 Line spectra Learning outcomes 1 understand that there are discrete electron energy levels in isolated atoms (e.g. atomic hydrogen) 2 understand the appearance and formation of emission and absorption line spectra 3 recall and use hf = E1 – E2
- 34. Dispersion Light can be broken up into its constituent parts by shining it through a prism as shown to produce a spectrum.
- 35. Emission & absorption Spectra Note: the emission and absorption Line spectra are for the same gas.
- 36. • Different gases have unique emission spectras and thus we can use this to determine the composition of distant stars.
- 37. Structure of atoms • An atom has many different levels (orbital shells)in which electrons can exist. Each shell can accommodate a fixed number of electrons. (1st shell:2, 2nd shell: 8, 3rd shell:8 and so on...). • An electron in a shell near the nucleus will possess less energy than one further away from it. • All electrons strive to be in the lowest possible state at all times (ie. Ground state). Diagram: examples of structures of atoms and different orbits. So for example Sodium has 11 electrons in 3 different shells all of which are all in their ground state. However there are other higher orbital shells for sodium that an electron could jump to, but would need extra energy to excite it to that level The red electron represents the valence electron.
- 38. Energy levels Let us consider the energy shells in a mercury atom.
- 39. Excitation We can see that an electron would need to absorb 4.9eV to jump to the next orbital. However, it does not necessarily jump only 1 level at a time. It may jump up as many levels as long as it has absorbed the energy totaling the difference between the two shells. 4.9- 0- Ground state
- 40. hf = E2 – E1 As the electron is in a higher shell it becomes unstable and soon drops down to lower energy level (as seen on the right). The electron will thus lose the energy equal to the difference between the two levels (in this case it will lose 4.9eV). This is called De-excitation. When it de-excites, a photon is released. The energy of the photon emitted in this case is equal to the energy lost Ie. 4.9eV (see below) -10.4 0- Ground state 4.9eV De-excitation Energy of photon
- 41. An electron may either jump multiple levels to its ground state or may proceed to the ground state via intermediate levels as long as the energy lost is equal to the difference between the energy levels. 4.9- 0- Ground state
Editor's Notes
- #3: Intro to light https://www.youtube.com/watch?v=IXxZRZxafEQ Introduces Wave –particle issues. https://www.youtube.com/watch?v=J1yIApZtLos
- #4: Give a brief intro into Einstein and how he came up with this idea
- #6: Interesting link on momentum of light. Do not need for CIE but useful extension https://wtamu.edu/~cbaird/sq/2014/04/01/light-has-no-mass-so-it-also-has-no-energy-according-to-einstein-but-how-can-sunlight-warm-the-earth-without-energy/#:~:text=Since%20photons%20(particles%20of%20light,zero%20(E%20%3D%200).
- #10: See page 567 Textbook
- #19: 2. The Photoelectric Effect When light shines on the surface of a metallic substance, electrons in the metal absorb the energy of the light and they can escape from the metal's surface. This is called the photoelectric effect, and it is used to produce the electric current that runs many solar-powered devices. Using the idea that light is a wave with the energy distributed evenly throughout the wave, classical physicists expected that when using very dim light, it would take some time for enough light energy to build up to eject an electron from a metallic surface. WRONG!! Experiments show that if light of a certain frequency can eject electrons from a metal, it makes no difference how dim the light is. There is never a time delay. In 1905, Albert Einstein came up with the solution. If Max Planck's idea that energy comes in clumps (quanta) is correct, then light must consist of a stream of clumps of energy. Each clump of light energy is called a photon, said Einstein, and each photon has an energy equal to hf (Planck's constant times the frequency of the light). Therefore, the energy of light is not evenly distributed along the wave, but is concentrated in the photons. A dimmer light means fewer photons, but simply turning down the light (without changing its frequency) does not alter the energy of an individual photon. So for a specific frequency light, if a single photon has enough energy to eject an electron from a metallic surface, then electrons will always be ejected immediately after the light is turned on and the photons hit the metal
- #22: Emission can only take place if Ek(max) >0
- #24: a) When determining Freq. must change eV to Joules Lithium = 5.55 x 10^14Hz Berylium= 9.41 x 10^14 Hz b Energy of photon = 3.11eV therefore lithium has photoelectric effect whereas beryllium doesn’t c) Ke= Energy of photon –work function = 3.11eV-2.3eV= 0.81eV (lithium only)
- #27: Dr. Quantums double slit experiment: starter to introduce wave particle duality of electrons https://www.youtube.com/watch?v=Q1YqgPAtzho So far we have looked at the dual nature of light waves. We will now look at electrons. Up till now we have treated electrons as purely particles. But is it possible that Electrons also have this dual nature. Wave-particle like nature. Strange as it may sound, interference can only occur when no one is watching. Once an observer begins to watch the particles going through the openings, the picture changes dramatically: if a particle can be seen going through one opening, then it's clear it didn't go through another. In other words, when under observation, electrons are being "forced" to behave like particles and not like waves. Thus the mere act of observation affects the experimental findings.
- #28: The acceleration of electrons in an electron beam gun with the acceleration voltage Va results can be also used to find the de Broglie wavelength by rearranging for and substituting into = h / (me) see https://virtuelle-experimente.de/en/elektronenbeugung/wellenlaenge/de-broglie-wellenlaenge.php
- #29: Simulation if no electron gun to show diffraction https://phet.colorado.edu/en/simulations/davisson-germer Derivation of braggs equation https://www.youtube.com/watch?v=Cjce4QumZNk Or can watch youtube demo Electron Diffraction - F-J's Physics - Video 45 – YouTube
- #30: Wavelengths significantly smaller than the gap therefore no diffraction.
- #31: Video showing neutron diffraction https://www.youtube.com/watch?v=CFFKoLS5b3o Key point from this is that particles can be sped up to particular speeds such that the wavelength of the particle is comparable to the spacings in the atom being studied. Max diffraction occurs when gap between molecules is similar to wavelength.
- #35: https://www.youtube.com/watch?v=oae5fa-f0S0 for emission spectra demos A hot gas will emit a very unique set of wavelengths. However, if white light is to shine through a cold gas cloud, then the gas will absorb distinct wavelengths and will produce an absorption spectra. Any heated body experiences thermal emission - that is, as the atoms and molecules move around they undergo accelerations due to interactions with other atoms and molecules. As the particles approach one another, the electrons interact due to their charges. Now when a charged particle is accelerated, it emits photons - that’s thermal emission. So a heated body composed of many atoms will in general emit a lot of photons due to these interactions (that’s how the heated body cools down in a vacuum, losing energy by emitting photons). These photons will be of many different energies and will be radiated in all directions. So essentially you will get an emission profile similar to a black body spectrum coming off of the body (not exactly black body as the properties of the atoms and molecules affect the emission spectrum somewhat, but let’s ignore that for now). So that’s what the heated body is putting out - a spectrum that, if the body is hot enough, you can observe it with a spectroscope (a prism). Now consider a cold gas between you and the hot body (maybe a galactic cloud?). The gas is so cold, it’s putting out very little energy in terms of thermal emission, especially in the visible part of the spectrum. If the gas cloud is thick or dense, the photons from the hot body passing through it on the way to your spectroscope will interact with atoms as they pass through. Those photons that have exactly the right energy to move an electron in the cold gas up to a higher orbital stand an excellent change of being absorbed and kicking the electron up to that orbital. So, those particular parts of the spectrum that correspond to differences in energy levels in the atoms making up the cold gas are absorbed by the atoms, leaving dark or absorption lines in the spectrum you observe. Those characteristic dark lines allow you to identify the cold gas. But what about those excited electrons, you may ask? Won’t they emit the photon of that wavelength some time later as the electron falls back to its ground state? The answer is yes, of course, but they will emit that photon in a random direction, maybe even back towards the hot body. So, your spectroscope may see a little energy in that dark line, but it is so much less compared to the rest of the spectrum that it looks dark. Now, let’s say the heated body is a tube of rarefied gas such as neon excited by a current, that is, ionized by application of a voltage. Looking at that through a spectroscope you would see emission lines emitted as the electrons rejoin their atoms, while the rest of the spectrum is much dimmer. These bright lines correspond to the dark lines (same wavelength) seen when viewing a black body spectrum through a thick cloud of cold neon gas. The energy levels are those of neon in either case, but in the latter case they are emitted, in the former absorbed and then re-emitted in random directions.
- #37: The ground state does not necessarily mean the first shell. a valence electron is an outer shell electron that is associated with an atom and thus a valence electron in the 3rd shell is also at its ground state.
- #38: The electron represented is the valent electron. The other shells represent the higher orbital shells it could exist in if sufficient enough energy. Note the values on the right are negative , as it shows that the electron is trapped within the atom by the attractive forces from the nucleus. To liberate this electron permanently 10.4eV of energy is required.
- #40: In text book formula is hf= E1 –E2 this is also ok as we are just interested in the change of energy as it is a scalar. However, it is also if we do E2 –E 1 then if we get a negative value, it means a photo is emitted (energy lost) or positive means photon absorbed (energy gained by electron)
- #42: 2nd to 1st state is not visible. Extension question. What part of the EM Spectrum is it? Use E=hf to determine its freq.