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Quantum Physics and Nuclear Physics 13.1 Quantum Physics.

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2 Quantum Physics and Nuclear Physics 13.1 Quantum Physics

3 Quantum Physics Revision When an electron falls to a lower energy level, the change in energy (ΔE) is emitted as a photon of e-m radiation. ΔE = E 2 – E 1 The change in energy is proportional to the frequency of the emitted photon: ( E = Photon energy h = Plank’s constant = 6.6 x m 2 kgs -1 ) E = hf

4 © John Parkinson 3 MAX PLANCK PHOTOELECTRIC EFFECT

5 The photoelectric effect The photoelectric effect is the emission of electrons from the surface of a material due to the exposure of the material to electromagnetic radiation. For example zinc emits electrons when exposed to ultraviolet radiation. If the zinc was initially negatively charged and placed on a gold leaf electroscope, the electroscope’s gold leaf would be initially deflected. However, when exposed to the uv radiation the zinc loses electrons and therefore negative charge. This causes the gold-leaf to fall Gold Leaf Electroscope

6 © John Parkinson 5 At the end of the nineteenth century, Classical Electromagnetic Wave Theory thought of light waves as being like water waves. The wave’s Intensity or energy was directly proportional to the square of the Amplitude, A. A

7 © John Parkinson 6 Potassium metal undergoes photoemission with blue and green light, but not with red light. potassium metal e e Emission! Nothing!! Blue light Green light Red light

8 © John Parkinson 7 THE CLASSICAL THEORY SUGGESTS TRYING MORE INTENSE LIGHT potassium metal Nothing!!Nothing!!

9 © John Parkinson 8 The Classical Theory must be wrong!!!!!

10 Threshold Frequency Electrons will only be emitted from zinc by photoelectric emission if the electromagnetic radiation incident upon its surface has a frequency of 1 x Hz or above. This is called the threshold frequency of zinc. Limitation of Wave Theory of Light Wave theory would suggest that once enough visible light energy had been absorbed by the zinc, the electron would be able to escape. This is not the case. No matter how intense the incident radiation, if its frequency is below the threshold frequency for a particular material, no photoelectric emission will occur.

11 Experimental observations Threshold frequency The photoelectric effect only occurs if the frequency of the electromagnetic radiation is above a certain threshold value, f 0 Variation of threshold frequency The threshold frequency varied with different materials. Affect of radiation intensity The greater the intensity the greater the number of electrons emitted, but only if the radiation was above the threshold frequency. Time of emission Electrons were emitted as soon as the material was exposed. Maximum kinetic energy of photoelectrons This depends only on the frequency of the electromagnetic radiation and the material exposed, not on its intensity. INTENSITY OF RADIATION DOES NOT ATTECT ENERGY OF EMITTED ELECTRONS

12 Problems with the wave theory Up to the time the photoelectric effect was first investigated it was believed that electromagnetic radiation behaved like normal waves. The wave theory could not be used to explain the observations of the photoelectric effect – in particular wave theory predicted: –that there would not be any threshold frequency – all frequencies of radiation should eventually cause electron emission –that increasing intensity would increase the rate of emission at all frequencies – not just those above a certain minimum frequency –that emission would not take place immediately upon exposure – the weaker radiations would take longer to produce electrons.

13 Threshold Frequency Electrons will only be emitted from zinc by photoelectric emission if the electromagnetic radiation incident upon its surface has a frequency of 1 x Hz or above. This is called the threshold frequency of zinc. Limitation of Wave Theory of Light Wave theory would suggest that once enough visible light energy had been absorbed by the zinc, the electron would be able to escape. This is not the case. No matter how intense the incident radiation, if its frequency is below the threshold frequency for a particular material, no photoelectric emission will occur.

14 Photons – the Quantum Model In 1900 Max Planck came up with the idea of energy being ‘quantised’ in some situations. i.e. existing in small ‘packets’. In 1905 Einstein suggested that all e-m radiation is emitted in small quanta called photons rather than in a steady wave. - Intensity of radiation depends on the number of photons being emitted per second (not amplitude as suggested by the wave model). - Energy per photon depends upon its frequency: E = hf

15 Einstein’sEinstein’s explanation Electromagnetic radiation consisted of packets or quanta of energy called photons The energy of these photons: –depended on the frequency of the radiation only –was proportional to this frequency Photons interact one-to-one with electrons in the material If the photon energy was above a certain minimum amount (depending on the material) –the electron was emitted –any excess energy was available for electron kinetic energy Einstein won his only Nobel Prize in 1921 for this explanation. This explanation also began the field of Physics called Quantum Theory, an attempt to explain the behaviour of very small (sub-atomic) particles.

16 © John Parkinson 15 Quantum Theory of the Photoelectric Effect Because of the interaction of this electron with other atoms, it requires a certain minimum energy to escape from the surface. The photons are sufficiently localized, so that the whole quantum of energy [ hf ] can be absorbed by a single electron at one time. The electron can then either share its excess energy with other electrons and the ion lattice or it can use the excess energy to fly out of the metal. The minimum energy required to escape depends on the metal and is called the work function, Φ.

17 © John Parkinson 16 For electron emission, the photon's energy has to be greater than the work function. The maximum kinetic energy the released electron can have is given by: E K = hf - Φ For every metal there is a threshold frequency, f 0, where hf 0 = Φ, that gives the photon enough energy to produce photoemission. It follows that the photo electric current is proportional to the intensity of the radiation provided the frequency of radiation is above threshold frequency. The number of photoelectrons emerging from the metal surface per unit time is proportional to the number of photons striking the surface that in turn depends on the intensity of the incident radiation E K = photon energy – the work function.

18 © John Parkinson 17 Quantum Theory of the Photoelectric Effect In 1905 Einstein developed Planck’s idea, that energy was quantised in quanta or photons, in order to explain the photoelectric effect. Electromagnetic radiation is emitted in bursts of energy – photons. The energy of a photon is given by E = hf, where f is the frequency of the radiation and h is Planck’s constant. [h = 6.6 x Js] But velocity of light = frequency times wavelength Substituting into E = hf

19 © John Parkinson 18 the visible spectrum λ frequency violet light light 400 nm red light light 700 nm uv light < 400 nm Blue photon Red photon Which photon has the most energy ????? BLUE !!!

20 Photon energy (revision) photon energy (E) = h x f where h = the Planck constant = 6.63 x JsPlanck also as f = c / λ; E = hc / λ Calculate the energy of a photon of ultraviolet light (f = 9.0 x Hz) (h = 6.63 x Js) E = h f = (6.63 x Js) x (9.0 x Hz) = 5.37 x J

21 The photoelectric equation hf = φ + E Kmax where: hf = energy of the photons of electromagnetic radiation φ = work function of the exposed material E Kmax = maximum kinetic energy of the photoelectrons Work function, φ This is the minimum energy required for an electron to escape from the surface of a material

22 Threshold frequency f 0 As: hf = φ + E Kmax If the incoming photons are of the threshold frequency f 0, the electrons will have the minimum energy required for emission and E Kmax will be zero therefore: hf 0 = φ and so: f 0 = φ / h

23 Question 1 Calculate the threshold frequency of a metal if the metal’s work function is 1.2 x J. (h = 6.63 x Js) f 0 = φ / h = ( 1.2 x J) / (6.63 x Js) threshold frequency = 1.81 x Hz

24 Question 2 Calculate the maximum kinetic energy of the photoelectrons emitted from a metal of work function 1.5 x J when exposed with photons of frequency 3.0 x Hz. (h = 6.63 x Js) hf = φ + E Kmax (6.63 x Js) x (3.0 x Hz) = (1.5 x J) + E Kmax E Kmax = x x = x J maximum kinetic energy = 4.89 x J

25 The vacuum photocell Light is incident on a metal plate called the photocathode. If the light’s frequency is above the metal’s threshold frequency electrons are emitted. These electrons passing across the vacuum to the anode constitute and electric current which can be measured by the microammeter. The photocell is an application of the photoelectric effect

26 © John Parkinson 25 Radiation mA Anode +ve Cathode -ve electrons The electromagnetic radiation releases electrons from the metal cathode. These electrons are attracted to the anode and complete a circuit allowing a current to flow vacuum

27 © John Parkinson 26 If the polarity is reversed, the pd across the tube can be increased until even the most energetic electrons fail to cross the tube to A. The milliammeter then reads zero. mA A C Radiation electrons The p.d. across the tube measures the maximum kinetic energy of the ejected electrons in electron volts. V

28 Obtaining Planck’s constant By attaching a variable voltage power supply it is possible to measure the maximum kinetic energy of the photoelectrons produced in the photocell. The graph opposite shows how this energy varies with photon frequency. hf = φ + E Kmax becomes: E Kmax = hf – φ which has the form y = mx +c with gradient, m = h Hence Planck’s constant can be found.

29 © John Parkinson 28 Maximum E K emitted electrons / J Frequency f / Hz metal A Work function, Φ Threshold frequency f 0 metal B E K = hf - Φ Gradient of each graph = Planck’s constant, h.

30 Work Function and Photoelectric Emission When u-v light is incident upon a zinc surface, each photon gives its energy to a single electron on the zinc surface. u-v photons have a high frequency. As a result they give enough energy to the electron to escape from the surface. The minimum energy needed to just remove an electron from a metal surface is called the work function, .

31 Low intensity u-v light will still cause electrons to be emitted. Because there are less photons per second there will be less electrons emitted per second. If the incident light has a lower frequency, each photon has less energy and so no electrons are emitted, irrespective of the intensity.

32 Einstein’s Photoelectric Equation If the photon energy (E=hf) is greater than the work function (  ), any remaining energy becomes kinetic energy of the electron (= ½mv 2 ). Thus Einstein stated... This is one version of Einstein's photoelectric equation. If hf is less than  nothing can happen m = mass of an electron v = speed of fastest electrons (ms -1 ) hf =  + ½ mv 2

33 Einstein’s theory was confirmed by Robert Millikan in He realised that if the clean metal emitting surface was given a positive potential, the electron emission could be stopped. Thus, if the p.d. applied was known, the KE ‘removed’ from the fastest electron can be found: We know...V = W / q  W = eV So... Link - PhET simulation V = stopping voltage e = charge on an electron = 1.6 x C  = Work function (Joules) hf =  + eV

34 Testing Einstein's Photoelectric Equation Einstein’s photoelectric equation can be rearranged to give... Thus plotting a graph of V against f enables us to determine Plank’s constant. V = h f -  e e

35 Incident radiation is shone onto a photoelectric cell with a low work function This causes electrons to be emitted from the larger emitting electrode. If they reach the small receiving electrode a current is detected on the ammeter The stopping voltage V is increased until zero current flows through the ammeter. The p.d. has made it impossible for even the fastest electrons to escape from the large electrode. The experiment is repeated with different frequency incident radiation and a set of values for frequency and stopping voltage are collected.

36 Results:

37 Q. Explain why this graph will always have the same gradient, whatever metal is used for the emitting electrode. Q. Explain how you would determine a value for the work function of the metal used to produce the graph above. Q. Wave theory suggests thatSupported or Contradicted by quantum theory? Any frequency can emit electrons from a certain metal surface Current depends on intensity (i.e. the number of photons per unit time) Maximum energy of electrons is independent of frequency Maximum energy of electrons would depend on intensity

38 Photocurrents If the metal surface and frequency of incident radiation are both kept constant, a graph can be plotted showing how the photoelectric current (photocurrent) in a photocell varies with applied p.d. (voltage). Consider these situations and explain the photocurrent that will flow in each case:

39 Photocurrent Applied p.d Stopping potential, V s Saturation current

40 Photocurrent Applied p.d High intensity Low intensity We find the stopping voltage stays the same no matter what the intensity of the light source. The increase in current is due to more electrons being emitted. The intensity of light has no effect on the maximum energy.

41 Photocurrent Applied p.d Low frequency (red) Higher frequency (blue)

42 The electron-volt (revision) The electron-volt (eV) is equal to the kinetic energy gained by an electron when it is accelerated by a potential difference of one volt. 1 eV = 1.6 x J Question: Calculate the energy in electron-volts of a photon of ultraviolet light of frequency 8 x Hz. (h = 6.63 x Js)

43 Ionisation An ion is a charged atom Ions are created by adding or removing electrons from atoms The diagram shows the creation of a positive ion from the collision of an incoming electron. Ionisation can also be caused by: –nuclear radiation – alpha, beta, gamma –heating –passing an electric current through a gas (as in a fluorescent tube) incoming electron

44 Ionisation energy Ionisation energy is the energy required to remove one electron from an atom. Ionisation energy is often expressed in eV. The above defines the FIRST ionisation energy – there are also 2 nd, 3 rd etc ionisation energies.

45 Excitation Excitation is the promotion of electrons from lower to higher energy levels within an atom. In the diagram some of the incoming electron’s kinetic energy has been used to move the electron to a higher energy level. The electron is now said to be in an excited state. Atoms have multiple excitation states and energies. incoming electron

46 Question An electron with 6 x J of kinetic energy can cause (a) ionisation or (b) excitation in an atom. If after each event the electron is left with (a) 4 x J and (b) 5 x J kinetic energy calculate in eV the ionisation and excitation energy of the atom.

47 Electron energy levels in atoms Electrons are bound to the nucleus of an atom by electromagnetic attraction. A particular electron will occupy the nearest possible position to the nucleus. This energy level or shell is called the ground state. It is also the lowest possible energy level for that electron. Only two electrons can exist in the lowest possible energy level at the same time. Further electrons have to occupy higher energy levels.

48 Electron energy levels in atoms Energy levels are measured with respect to the ionisation energy level, which is assigned 0 eV. All other energy levels are therefore negative. The ground state in the diagram opposite is eV. Energy levels above the ground state but below the ionisation level are called excited states. Different types of atom have different energy levels.

49 De-excitation Excited states are usually very unstable. Within about s the electron will fall back to a lower energy level. With each fall in energy level (level E 1 down to level E 2 ) a photon of electromagnetic radiation is emitted. emitted photon energy = hf = E 1 – E 2

50 Energy level question Calculate the frequencies of the photons emitted when an electron falls to the ground state (at 10.4 eV) from excited states (a) 5.4 eV and (b) 1.8 eV. (h = 6.63 x Js)

51 Complete: transitionphoton E 1 / eVE 2 / eVf / PHzλ / nm

52 Excitation using photons An incoming photon may not have enough energy to cause photoelectric emission but it may have enough to cause excitation. However, excitation will only occur if the photon’s energy is exactly equal to the difference in energy of the initial and final energy level. If this is the case the photon will cease to exist once its energy is absorbed.

53 Fluorescence The diagram shows an incoming photon of ultraviolet light of energy 5.7eV causing excitation. This excited electron then de-excites in two steps producing two photons. The first has energy 0.8eV and will be of visible light. The second of energy 4.9eV is of invisible ultraviolet of slightly lower energy and frequency than the original excitating photon. This overall process explains why certain substances fluoresce with visible light when they absorb ultraviolet radiation. Applications include the fluorescent chemicals are added as ‘whiteners’ to toothpaste and washing powder. Electrons can fall back to their ground states in steps.

54 Fluorescent tubes A fluorescent tube consists of a glass tube filled with low pressure mercury vapour and an inner coating of a fluorescent chemical. Ionisation and excitation of the mercury atoms occurs as the collide with each other and with electrons in the tube. The mercury atoms emit ultraviolet photons. The ultraviolet photons are absorbed by the atoms of the fluorescent coating, causing excitation of the atoms. The coating atoms de-excite and emit visible photons. See pages 37 and 38 for further details of the operation of a fluorescent tube

55 Line spectra A line spectrum is produced from the excitation of a low pressure gas. The frequencies of the lines of the spectrum are characteristic of the element in gaseous form. Such spectra can be used to identify elements. Each spectral line corresponds to a particular energy level transition.

56 Question Calculate the energy level transitions (in eV) responsible for (a) a yellow line of frequency 5.0 x Hz and (b) a blue line of wavelength 480 nm.

57 The hydrogen atom

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59 With only one electron, hydrogen has the simplest set of energy levels and corresponding line spectrum. Transitions down to the lowest state, n=1 in the diagram, give rise to a series of ultraviolet lines called the Lyman Series. Transitions down to the n=2 state give rise to a series of visible light lines called the Balmer Series. Transitions down to the n=3, n=4 etc states give rise to sets of infra- red spectral lines

60 The discovery of helium Helium was discovered in the Sun before it was discovered on Earth. Its name comes from the Greek word for the Sun ‘helios’. A pattern of lines was observed in the Sun’s spectrum that did not correspond to any known element of the time. In the Sun helium has been produced as the result of the nuclear fusion of hydrogen. Subsequently helium was discovered on Earth where it has been produced as the result of alpha particle emission from radioactive elements such as uranium.

61 The wave like nature of light Light undergoes diffraction (shown opposite) and displays other wave properties such as polarisation and interference. By the late 19 th century most scientists considered light and other electromagnetic radiations to be like water waves

62 The particle like nature of light Light also produces photoelectric emission which can only be explained by treating light as a stream of particles. These ‘particles’ with wave properties are called photons.

63 The dual nature of electromagnetic radiation Light and other forms of electromagnetic radiation behave like waves and particles. On most occasions one set of properties is the most significant The longer the wavelength of the electromagnetic wave the more significant are the wave properties. Radio waves, the longest wavelength, is the most wavelike. Gamma radiation is the most particle like Light, of intermediate wavelength, is best considered to be equally significant in both

64 Matter waves In 1923 de Broglie proposed that particles such as electrons, protons and atoms also displayed wave like properties. The de Broglie wavelength of such a particle depended on its momentum, p according to the de Broglie relation: λ = h / p As momentum = mass x velocity = mv λ = h / mv This shows that the wavelength of a particle can be altered by changing its velocity.

65 Question 1 Calculate the de Broglie wavelength of an electron moving at 10% of the speed of light. [m e = 9.1 x kg] (h = 6.63 x Js; c = 3.0 x 10 8 ms -1 )

66 Question 2 Calculate the de Broglie wavelength of a person of mass 70 kg moving at 2 ms -1. (h = 6.63 x Js)

67 Question 3 Calculate the effective mass of a photon of red light of wavelength 700 nm. (h = 6.63 x Js)

68 Evidence for de Broglie’s hypothesis A narrow beam of electrons in a vacuum tube is directed at a thin metal foil. On the far side of the foil a circular diffraction pattern is formed on a fluorescent screen. A pattern that is similar to that formed by X-rays with the same metal foil. Electrons forming a diffraction pattern like that formed by X-rays shows that electrons have wave properties. The radii of the circles can be decreased by increasing the speed of the electrons. This is achieved by increasing the potential difference of the tube.

69 Energy levels and electron waves An electron in an atom has a fixed amount of energy that depends on the shell it occupies. Its de Broglie wavelength has to fit the shape and size of the shell.

70 © John Parkinson 69 f / Hz Max E k / eV 1 2 PotassiumMagnesiumAluminium

71 © John Parkinson 70 Summary For any metal there is a minimum threshold frequency, f 0, of the incident radiation, below which no emission of electrons takes place, no matter what the intensity of the incident radiation is or for how long it falls on the surface. Electrons emerge with a range of velocities from zero up to a maximum. The maximum kinetic energy, E k, is found to depend linearly on the frequency of the radiation and to be independent of its intensity. For incident radiation of a given frequency, the number of electrons emitted per second is proportional to the intensity of the radiation. Electron emission takes place immediately after the light shines on the metal with no detectable time delay.

72 Einstein’s Theory  The photoelectric effect is interpreted with photons and the conservation of energy with the equation:  hf =  + ½ mv 2  hf equals the energy of each photon  The photoelectric effect is interpreted with photons and the conservation of energy with the equation:  hf =  + ½ mv 2  hf equals the energy of each photon Source:

73 Kinetic energy of emitted electron vs. Light frequency  Higher-frequency photons have more energy, so they should make the electrons come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If you leave the frequency the same but crank up the intensity, more electrons should come out (because there are more photons to hit them), but they won't come out any faster, because each individual photon still has the same energy. And if the frequency is low enough, then none of the photons will have enough energy to knock an electron out of an atom. So if you use really low-frequency light, you shouldn't get any electrons, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low. Source: content/PhyAPB/lessonnotes/dualnature/ photoelectric.asp

74 Source: Simple Photoelectric Experiment

75 Photoelectric Effect Applications

76  The Photoelectric effect has numerous applications, for example night vision devices take advantage of the effect. Photons entering the device strike a plate which causes electrons to be emitted, these pass through a disk consisting of millions of channels, the current through these are amplified and directed towards a fluorescent screen which glows when electrons hit it. Image converters, image intensifiers, television camera tubes, and image storage tubes also take advantage of the point-by-point emission of the photocathode. In these devices an optical image incident on a semitransparent photocathode is used to transform the light image into an “ electron image. ” The electrons released by each element of the photoemitter are focused by an electron-optical device onto a fluorescent screen, reconverting it in the process again into an optical image

77 Applications: Night Vision Device

78 Photoelectric Effect Applications  Photoelectric Detectors In one type of photoelectric device, smoke can block a light beam. In this case, the reduction in light reaching a photocell sets off the alarm. In the most common type of photoelectric unit, however, light is scattered by smoke particles onto a photocell, initiating an alarm. In this type of detector there is a T-shaped chamber with a light-emitting diode (LED) that shoots a beam of light across the horizontal bar of the T. A photocell, positioned at the bottom of the vertical base of the T, generates a current when it is exposed to light. Under smoke-free conditions, the light beam crosses the top of the T in an uninterrupted straight line, not striking the photocell positioned at a right angle below the beam. When smoke is present, the light is scattered by smoke particles, and some of the light is directed down the vertical part of the T to strike the photocell. When sufficient light hits the cell, the current triggers the alarm. Source:

79 Photoelectric Smoke Detector Source:

80 Applications  Solar panels are nothing more than a series of metallic plates that face the Sun and exploit the photoelectric effect. The light from the Sun will liberate electrons, which can be used to heat your home, run your lights, or, in sufficient enough quantities, power everything in your home. Source: picsolarpannelsmatt.jpg

81 The Wave Nature of Matter We have seen that light behaves both like a wave (it diffracts) and like a particle (in explaining photoelectric emission). It has wave particle duality. Demo 1: Shine a beam of laser light (a wave) through a single diffraction grating (like a very fine gauze) then through many gratings crossed at angles to each other. Demo 2: Fire a beam of electrons (particles of matter) at a thin piece of graphite using a cathode ray tube. Wave - particle duality is the ability of something to exhibit both wave and particle behaviour.

82 VAVA Cathode Control grid Anode Graphite foil Flourescent screen

83 De Broglie’s Equation Considering a photon of light, in 1924 Prince Louis Victor de Broglie equated Einstein’s mass-energy relation and Planck’s equation: E = mc 2 and E = hf thus... mc 2 = hf so... mc 2 = h c λ so...or λ = h p λ = h mc But c = f λ

84 This is the de Broglie equation. ( Where h = Planck’s constant = 6.6 x Js ) By analogy, this equation can be applied to any other particle of matter. Thus de Broglie’s Hypothesis: Q1 A year 13 student runs with joy to his physics lesson. If he runs at 5 ms -1 and has mass 60kg, determine the de Broglie wavelength of his motion. Comment upon your answer. All matter can behave like a wave, with the wavelength given by Plank’s constant divided by the matter’s momentum.

85 Q2 An electron is accelerated in a cathode ray tube through a potential difference of 2kv. i.Determine the velocity of the electron (m e = 9.11 x kg) 2.65 x 10 7 ms -1 ii.Determine the de Broglie wavelength of the electron. 2.7 x m

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87 Energy Levels E.g. E = hf h= Planck constant = 6.6 x m 2 kgs -1 hf = E 2 - E 1

88 Emission Line Spectra Atoms of a gas emit e-m radiation if they become excited. This means the electrons jump to a higher energy level and then fall back, losing potential energy and emitting it as a photon of e-m radiation. This will have a frequency according to ΔE = hf Experiment: Observing emission spectra - Place a slit in front of a hydrogen lamp. - View the light through a diffraction grating or by refracting it through a prism.

89 Results: Conclusion Explain why... a.only certain colours are seen for any particular lamp. b.coloured fringes are produced This is the image seen when light from one lamp is diffracted through a grating:

90 Neon

91 Mercury

92 Argon

93 Xenon Images from this website this website

94 Hydrogen... Helium... Neon...

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96 Absorption spectrum If light with a continuous spectrum of frequencies (a filament light bulb approximately emits this) is shone into a gas its photons will interact with electrons in the gas atoms, boosting them to a higher energy level. Thus the gas will absorb only certain frequencies of the light. These frequencies will be missing from the light that passes through.

97 Absorption spectrum for Hydrogen. Q. Why des the intensity not fall to zero for the absorption lines? Energy is re-emitted as photons in all directions

98 Fraunhoffer Lines This effect is visible in the Fraunhoffer lines seen in spectra of light from the Sun. The Sun emits a virtually continuous spectrum. The absorption lines are due to gases in the Sun’s atmosphere absorbing some frequencies of photon. Hence astronomers can deduce what gases are in a star’s atmosphere as well as their proportions.

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102 The ‘Electron in a Box’ Model of an Atom We have seen how the Bohr model of an atom suggests that the atom has discreet energy levels. Thus the energy of the electrons and of the atom is not continuous. Next we will look at how this idea is compatible with the de Broglie hypothesis. According to de Broglie, the electron must have wave properties. So how does this wave fit into an atom...?

103 We can create a quantum model of an electron trapped in an atom by considering the electron as a standing wave travelling back and forth across a box of side L: Java applet Clearly, if there are nodes at either end, the electron can only have certain fixed wavelengths and frequencies. i.e. λ = 2L(fundamental) λ = L (2 nd harmonic) λ = 2/3 L(3 rd harmonic) So in general λ = 2L n

104 According to de Broglie... The KE of the electron is given by KE = ½ mv 2 Substituting the equations for v and λ gives… The energy of the electron ‘wave’ is clearly quantised (it has discreet values depending upon n), thus fitting scientific observations e.g. work function in the photoelectric effect. λ = h = h  v = h pmv mλ KE = n 2 h 2 8mL 2 n = integer h, m and L are all constants.

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106 Limitations of the Bohr Model of the Atom The Bohr model of the atom worked well with basic ideas of quantum theory (as we have seen). However it has limitations. E.g. It only works well with the Hydrogen atom It does not predict intensities of different spectral lines.

107 Schrödinger's Model of the Atom Schrödinger developed a new model that accounted for all these limitations. He suggested that... - electrons exist in the atom with a position determined by the wavefunction (ψ – psi), a function of position and time. - the position of the electron at any time is undefined - the probability of the finding the electron at any particular position in the atom is determined by the square of the amplitude of the wavefunction i.e. ψ 2 Thus for a given energy there are some places where the electron is more likely to exist. This can be represented by ‘probability clouds’.

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109 Additional note: This illustrates an important feature of quantum mechanics. Its outcomes are probabilities and not certainties; it is a probabilistic model. Classical mechanics (e.g. Newtons laws) are deterministic: the future is determined by the past and is therefore predictable with some certainty!

110 Heisenberg’s Uncertainty Principle Consider one electron amongst a beam of electrons moving with identical momentum towards a narrow slit: If the slit is wide, the electron will pass straight through without diffracting. Hence we can know its direction and thus its momentum beyond the slit. However we cannot be sure where exactly it passed through the slit so the uncertainty in position is large.

111 This is an example of the Heisenberg uncertainty principle which states that... Note: This is a fundamental property of the universe and is nothing to do with our ability to measure accurately. If the slit width is similar to the diameter of the electron, it will diffract as it passes through. So we are not sure about its direction and thus momentum although we can be quite sure of its position. It is not possible to measure the exact position and the exact momentum of a particle at the same time.

112 This imposes a minimum uncertainty on the product of the uncertainties of position and momentum: This can also be written in terms of energy and time: Clearly the value of h is so small that the effects of this uncertainty are not seen in everyday events. Δp Δx ≥ h 4π ΔE Δt ≥ h 4π h = Plank’s constant

113 E.g. Quantum tunnelling As a result of the uncertainty principle energy can be ‘borrowed’ to overcome potential barriers: - Alpha particles should not be able to escape from the nucleus due to the ‘strong force’ between quarks. However they escape by borrowing energy and paying it back within a time limit defined by the equation. - Two protons (hydrogen nuclei) should not be able to fuse together in the Sun at its current density and temperature. They do fuse together by borrowing energy and then paying it back later on. You tube link

114 Q1 Consider a football moving through the air. If the uncertainty in its momentum is 2.25 x kgms -1, Determine the uncertainty in its position (Δx). Δx ≥ +/- 2.33x m so… Δx ≥ h 4π Δp Δx Δp ≥ h 4π Δx ≥ 6.6 x π x (2.25 x )

115 Q2 A proton in the LHC takes 1.0 x s to collide. Assuming a 1% uncertainty in measurement of time, determine the uncertainly in measurements of proton energy (ΔE). 1% uncertainty in time = 1/100 x (1.0 x ) = +/- 1.0 x s so… ΔE ≥ h 4π Δt ΔE Δt ≥ h 4π ΔE ≥ 6.6 x π x (1.0 x ) ΔE ≥ +/- 5.3 x J

116 Q3 Assuming 1% uncertainty in velocity, determine the uncertainty in the position of a proton moving at i.2.5 x 10 7 ms -1 ii.Close to the speed of light (3.0 x 10 8 ms -1 ) ( m p = x kg )

117 Q4 Outline by reference to position and momentum how the Schrödinger model of the hydrogen atom is consistent with the Heisenberg uncertainty principle. - the Schrödinger model assigns a wave function to the electron that is a measure of the probability of finding it somewhere; - therefore the position of the electron is uncertain; - resulting in an uncertainty in its momentum;

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125 Probability Waves From the de Broglie hypothesis we can see that our picture of the electron as a small ‘ball’ is too simplistic. It

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