* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download CHAPTER 17 LEARNING OBJECTIVES - crypt
Survey
Document related concepts
Internal energy wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Standard Model wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Conservation of energy wikipedia , lookup
Photon polarization wikipedia , lookup
Nuclear drip line wikipedia , lookup
Density of states wikipedia , lookup
Hydrogen atom wikipedia , lookup
Nuclear structure wikipedia , lookup
History of subatomic physics wikipedia , lookup
Elementary particle wikipedia , lookup
Atomic nucleus wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Transcript
17 PROBING DEEP INTO MATTER Creation and Annihilation • Describe differences between matter and antimatter • Apply conservation laws to annihilation and materialisation events • Explain how antimatter is used in medical imaging (PET) Starter: Can you give reasons why imaging the brain using x rays may have limitations or drawbacks. • Conventional x ray imaging does not show soft tissue well • Conventional x ray imaging is a “2 dimensional” technique • Computerised axial tomography can give 3D information but x ray dose is higher than conventional x ray imaging • X ray techniques cannot map brain activity, only show structures Read pages 189-191 Answer the following: 1) What does PET stand for? 2) Try and summarise in you own words how it works. 3) What is the main difference between matter and antimatter? What is the one thing that is the same? Ext: 4) What is meant by creation, annihilation and pair production? Energy is conserved Energy, momentum Conserved quantities and Electric Charge conserved es total energy before = kinetic energy of particles + rest energy of particles minimum value of energy before e– is rest energy: Simplify: assume head-on collision with equal speeds e– energy after is energy of gamma photons = 2 0.511 MeV e+ = 2 mc 2 = 2 0.511 MeV e Momentum is conserved + total energy before total linear momentum before Energy is conserved total energy after = e– = kinetic energy of particles + rest energy of particles = total linear momentum after e+ = same mass; equal and opposite velocities total energy afterenergy E, momentum p = E/c photons identical, momentums opposite energy after is total momentum before =0 total momentum = 0 minimum value of energy total energy after = energy of gamma before is rest energy: photons = 2 mc 2 = 2 0.511 MeV Electric charge is conserved = 2 0.511 MeV total charge before energy after is charge energy of gamma Momentum is conserved (–e) + (+e) = 0 photons = total charge after charge 0 +0= 0 Making PET scans A pair of gamma rays are emitted in opposite directions as a result of electron/positron annihilation inside the patient Scintillator – captures gamma ray photon and emits lower energy photons into photomultiplier tubes signal processing Photomultiplier – incoming photon creates a cascade of electrons, giving an electrical pulse output One pair of detectors will respond almost simultaneously. This near coincidence shows that the two gamma rays came from a common source. The tiny time difference between the two signals is then used to work out where they came from along the line between the detectors. Scintillators are arranged in a grid on the inside surface of the scanner. In any short period of time many detectors will respond to gamma rays from many different annihilations inside the body. A computer produces a slice-by-slice map of activity in the brain. Annihilation and Creation Pair annihilation and creation Annihilation gamma energy = 2 0.511 MeV plus kinetic energy of electrons e– e+ Creation? e– e+ extremely rare (cannot bring two identical photons together) Pair Production Pair creation e– e+ gamma energy = 2 0.511 MeV (minimum) nucleus carries away momentum, to conserve momentum and energy close to nucleus Starter… An electron and a positron with negligible kinetic energy annihilate and produce two identical gamma ray photons. (Rest mass of electron =9.11x10-31kg, h=6.63x10-34Js, c = 3x108 ms-1) Calculate a) the energy released (in J and MeV) b) the frequency of the gamma-photons (in Hz). Particle interactions • Describe how charged particles interact via virtual photon exchange • Explain how to describe these processes using Feynman diagrams and the “try all paths” approach • Discuss consequences of the differences between fermions and bosons Starter: What are the maximum and minimum amplitudes that can result from adding two phasors, each of length 1 unit, and what are the phase differences in each case? Starter: Explain the point this diagram is making…. “Try all paths” is a quantum rule obeyed by allSource photons and electrons. Detector The same idea is applied to interactions of particles. Richard Feynman invented a type of diagram to help physicists keep track of all the possible ways that particles can interact. The rule “try all paths” changes to “try everything allowed” or more technically “everything that is not forbidden is compulsory”. Quantum fields create and destroy particles cannot happen cannot happen Electromagnetic quantum field creates a photon charge e photon carries away energy and momentum Electromagnetic quantum field destroys a photon photon delivers energy and momentum charge e energy and momentum not both conserved combined process can happen virtual (unobservable) photon charge e charge e photon exchanged: charges exchange energy and momentum energy and momentum conserved overall Forces between charges arise from the exchange of momentum through exchange of virtual photons Feynman diagrams show possibilities to be combined Propagation and interaction of a pair of electrons Electrons at A and B arrive at C and D no photon exchange C D A A B electrons just travel A to C, and B to D Electrons are identical, so there is no way to tell these apart: D C add the phasor arrows for both diagrams B electrons just travel A to D, and B to C one photon exchange D D C C B A electrons exchange a photon A B No way to tell these apart: add the phasor arrows for both diagrams electrons exchange a photon .... plus diagrams with more photons.... For each diagram, add phasor arrows for all possible space-time locations A, B, C, D. Add total phasor arrows for each type of diagram Quantum rule: Try all possible ways to interact Ways for an electron to scatter a photon In each diagram one electron and one photon come in, and one electron and one photon go out. All diagrams represent the same process. interaction with virtual pair creation and annihilation Simple electron–photon interactions e– e– B B e– A e+ A e– Electron: absorbs photon at A travels to B emits photon at B A e– B e– Electron: emits photon at A travels to B absorbs photon at B photon creates e–, e+ pair at B e+ goes to A e– and e+ annihilate at A, emitting photon Phasor arrows for all diagrams are added to find total amplitude and phase for scattering plus other more complex diagrams To find out.......... What is a fermion? Give an example of a particle that is a fermion. What is a boson? Give an example of a particle that is a boson. What happens if you try to squeeze two fermions into the same region of space? What happens if two identical bosons encounter each other? Can you describe one important consequence of this? Can you give one important practical application of this? Electron spin Photon spin Fermions and Bosons Fermions Bosons ½ integer spin (1/2, 3/2,....) Electrons, Protons etc. Integer spin (0,1,...) Photons Never occupy same quantum state (Avoid each other always) Can occupy same quantum state (Can “flock” together in step) Consequently..... Consequently..... • Two electrons in same orbital of • In a laser, lots of photons join an atom must have opposite spin together to produce beam of (Pauli exclusion principle) photons all of identical phase and polarisation • Matter is “hard”: atoms are difficult to squash! Identical particles in the same state Identical particles in the same state Electrons (fermions) Photons (bosons) electrons are indistinguishable these possibilities cannot be distinguished add the phasors for the two diagrams D C X C Y A photons are indistinguishable these possibilities cannot be distinguished add the phasors for the two diagrams D X Y X electron X goes to D electron Y goes to C electron X goes to C electron Y goes to D C Y D X Y B A B D C A B A B photon X goes to D photon Y goes to C photon X goes to C photon Y goes to D bring points C and D together bring points C and D together E E Exchanging particles reverses phase. Adding phasors gives zero amplitude A B + Phasors for the two diagrams become the same. Adding phasors gives 2 amplitude 4 intensity =0 A B + = There is zero probability for two electrons to be at the same space-time point (be in same state) This is the Pauli exclusion principle Photons in a given state increase the chance of others joining them in the same state Fermions and Bosons Fermions Bosons ½ integer spin (1/2, 3/2,....) Electrons, Protons etc. Integer spin (0,1,...) Photons Never occupy same quantum state (Avoid each other always) Can occupy same quantum state (Can “flock” together in step) Consequently..... Consequently..... • Two electrons in same orbital of an atom must have opposite spin (Pauli exclusion principle) • Matter is “hard”: atoms are difficult to squash! • In a laser, lots of photons join together to produce beam of photons all of identical phase and polarisation Light Amplification by Stimulated Emission of Radiation Starter For each of the following statements, say whether it applies to FERMIONS or BOSONS: Q1. A photon is an example of this class of particle. Q2. These particles have integer spin values. Q3. Particles which cannot occupy the same quantum state. Q4. “Matter” particles, like protons, neutrons and electrons belong to this class. Q5. Virtual particles which are exchanged between interacting matter particles belong to this class. Q6. These particles can have spin values of 1/2, 3/2, 5/2 etc. Q7. (For chemistry students): Two electrons cannot occupy the same quantum state. How is it then possible to get two electrons into the same orbital? Hint: in what way are the two electrons distinguishable when in the same orbital? Conservation in nuclear processes • Explain why a new particle was needed to account for the energy spectrum of beta particles • Balance nuclear equations: charge, mass-energy, baryon number, lepton number The Baryon Family Baryons contain three quarks (we come to them later). Protons and neutrons are baryons! Baryon number must be conserved. Baryon number is +1 for all protons and -1 for antiprotons. Note: Protons and neutrons are also described as nucleons. Leptons – fundamental particles Leptons are fundamental particles. As far as we are aware they are not made up of anything smaller. Examples are electrons and neutrinos. Leptons are given a property called lepton number. Electrons and neutrinos are given lepton number 1. Where as the antiparticles are given lepton number -1. All hadrons (non-leptons, which we learn more about later) have lepton number 0. ‘Missing’ energy in beta decay Beta decay of strontium–90 rest energy strontium–90 – 0.546 MeV yttrium–90 38 protons 52 neutrons 39 protons 51 neutrons Energy spectrum of beta decay of strontium–90 fraction of beta particles per unit energy range 0 energy/MeV 0.546 Beta decay of strontium–90, including antineutrino emission rest energy strontium–90 – energy E 0.546 MeV energy 0.546 MeV – E yttrium–90 Rutherford’s experiment • Describe how alpha scattering changed our view of atomic structure • Explore effects of changing alpha particle energy and nuclear charge on scattering • Estimate upper limit on nuclear size Starter: Q1. I have a charge of +1 and a lepton number of -1. What could I be? Q2. I have a charge of zero and a lepton number of +1. What could I be? Q3. I have a baryon number of +1, a lepton number of zero and a charge of +1. What could I be? Q4. I have a baryon number of +1 and a charge of zero. What could I be? Q5. I have a baryon number of -1 and a charge of -1. What could I be? Starter: Q1. I have a charge of +1 and a lepton number of -1. What could I be? ANTIELECTRON (POSITRON) Q2. I have a charge of zero and a lepton number of +1. What could I be? NEUTRINO Q3. I have a baryon number of +1, a lepton number of zero and a charge of +1. What could I be? PROTON Q4. I have a baryon number of +1 and a charge of zero. What could I be? NEUTRON Q5. I have a baryon number of -1 and a charge of -1. What could I be? ANTIPROTON Gravitational and electric fields compared Q1. (a) Write down the expression for the force F between two masses, M and m, separated by a distance R. Q1. (b) Write down the corresponding expression for the force F between two charges Q and q, separated by a distance R. (The constant in the equation is known as the electric force constant, and is denoted by k.) Q2. (a) Write down the expression for the gravitational potential energy for an object of mass m at a distance R from the centre of a planet of mass M. Q2. (b) Write down the corresponding expression for the gravitational potential energy of a charge +q at a distance R from another charge +Q. Q3. (a) How much kinetic energy would you need to give the object in Q2. (a) for it to be able to “climb out” of the potential well of the planet? Q3. (b) How much kinetic energy would the charge +q in Q3. (a) have if it was released and allowed to coast far away from +Q? Rutherford’s experiment lead block to select narrow beam of alpha particles radium source of alpha particles thin gold foil microscope to view zinc sulphide screen and count alpha particles vary angle of scattering observed Note: an alpha particle Is a helium nucleus with the electrons removed. So it is positively charged! scattered alpha particles alpha particle beam zinc sulphide screen; tiny dots of light where struck by alpha particle Copy and complete the table using the statements provided Rutherford’s observation Explanation Model of the atom A. Most alpha particles passed straight through the foil, undeflected B. Some were deflected off course as they passed through nearly all the mass of the atom is in the nucleus C. Some bounced right back from the foil there must be centres of + charge in the atom the centres of + charge must be much heavier than the alpha particles the nucleus is positively charged, while the electrons are outside it, far away the atom is mostly empty space the nucleus is tiny compared to the overall size of the atom Rutherford’s observation Explanation Model of the atom A. Most alpha The atom is mostly particles passed empty space. straight through the foil, undeflected The nucleus is tiny compared to the overall size of the atom B. Some were There must be deflected off course centres of + charge as they passed in the atom. through The nucleus is positively charged, while the electrons are outside it, far away The centres of Nearly all of the mass of the C. Some bounced +charge must be atom is in the nucleus right back from the much heavier than foil the alpha particles 10 0 0 30 60 90 120 150 180 s c attering angle/degree F o r calculatio n s fo rce F = a lpha particle s c attered 2Z e 2 4 0d 2 charge +2e s cattering angle d aiming error b As sump tion s: a lpha pa rtic le is the He nuc leus , cha rge + 2e gold nucleus h as charge + Z e, and is muc h m ore m as siv e than alpha particles s c attering forc e is inv erse sq uare el ectrical repuls ion gold nuc leus c harge + Ze e qual forc e F b ut nuc leus is mas siv e, s o little r ecoil TEST: Are slowed -down alp ha particles scat tered m ore? Z TEST: Does using nuclei of smaller charge scatter alpha particles less? TEST: Are slowed -down alp ha particles scat tered m ore? Z reduce alpha energy with absorber less energetic alpha particle turned around further from the nucleus lower speed Z alpha d er from TEST: Does using nuclei of smaller charge scatter alpha particles less? Z r eplace foil by metal of smaller atomic number alpha particle gets closer to nucleus of smaller charge and is deflected less smaller nucleus with less charge, e.g. aluminium Measuring the size of nuclei • Explain how electron diffraction can be used to measure nuclear size accurately and precisely • Determine nuclear diameter from scattering curves • Describe and explain the relationship between nuclear size and nucleon number Starter: From your AS physics waves knowledge, explain the appearance of all features of the single-slit diffraction curve shown below: JJ and GP Thomson: father and son Nobel physics prize winners Listen sonny, I got the Nobel prize for showing that the electron is a particle... Whatever, Dad. I got it for showing that the electron is a wave! Wave-particle duality Evidence for wave-like character Light Electrons Evidence for particle-like character Electron diffraction This eerie green glow is caused by low energy electrons in a cathode ray tube striking the phosphorescent coating on the inside of the glass bulb just behind the ruler. In this case the diffraction is caused by the electrons passing through a thin layer of polycrystalline graphite (pencil "lead"). The regular array of carbon atoms in the crystals is responsible for the diffraction effects. How does the electron energy affect what is seen in an electron diffraction experiment? Diffraction of low energy electrons Diffraction by planes of atoms of low energy electrons gives a diffraction pattern that reveals the inter atomic spacing. Here, the de Broglie wavelength of the electrons is quite large, as their momentum is small. The wavelength is comparable to the inter atomic spacing, so we get a lot of diffraction by the planes of atoms in the manner of a diffraction grating diffracting light. Diffraction of high energy electrons With very high energy electrons, the de Broglie wavelength is comparable to the size of a nucleus, and the diffraction effects seen are essentially the same as single slit diffraction. The atomic nucleus behaves as an obstacle for the electrons to diffract around, and the diffraction patterns seen is essentially the same as we get when light passes through a single slit. Electron diffraction Q1. Explain why there is a minimum in the curve. Q2. What would be the effect on the curve of using higher energy electrons? Q3. What would be the effect on the curve of using a sample of argon-40 in place of neon-20? Density of nuclear matter Volume of nucleus increases linearly with number of nucleons Electron scattering measures radius r of nucleus calculate volume = 43 r3 1500 197 1000 122 59 500 28 16 O 0 C He 4 1 H 88 Sb Sr Si 12 0 Co Au 50 150 100 number of nucleons 200 Estimate from graph: 100 nucleons in volume 700 10–45 m3 Data: mass per nucleon u = 1.7 10–27 kg volume per nucleon = 7 10–45 m3 Calculate density: density = density = mass volume 1.7 10–27 kg 7 10–45 m3 = 2.4 1017 kg m–3 Density of ‘nuclear matter’ is roughly 2 1017 kg m–3 A matchbox-full of nuclear matter would have a mass of five billion tonnes Density of nuclear m atter Volum e of nucleus increases linearly with number of nucleons Electron scattering measures radius r of nucleus calculate v olume = 4 3 KEY POINT Volume of nucleus is proportional to number of particles it contains. r 3 1500 1 97 1000 122 59 500 28 16 Co 88 Au Sb Sr Si O 12 0 0 1 C 4 He H 50 100 150 number of nucleons 200 Nuclear density questions... Q1. If nuclear volume is proportional to the number of nucleons (A) in the nucleus, explain why nuclear radius r is given by: r = r0 A1/3 where r0 is a constant. Q2. The nucleon number of a gold nucleus is 197. a) The radius of the nucleon is 1.2x10-15m. Calculate the radius r of the gold nucleus. b) Calculate volume of the gold nucleus. c) The mass of a nucleon is 1.67x10-27kg, calculate the nuclear density. Q3. Silver has a mass number of 108. What is its nuclear density? The structure of nucleons • Explain why electrons are well suited to be a probe of nucleon structure • Deduce the quark composition of a range of hadrons (baryons and mesons) Starter: Explain why alpha particles and protons are not well suited to probing the size of nuclei Deep inelastic scattering Medium energy: elastic scattering electron proton Quarks move rapidly inside proton. The interaction time is long enough for the proton to behave like a blur of charge. Deep inelastic scattering High energy: deep inelastic scattering electron scattered at large angle u electron d Electron can hit one quark, and be scattered. Exchange of high energy photons leads to the creation of a jet of particles and antiparticles. u ‘jet’ of particles, mainly mesons Particle Family Tree All particles Fundamental particles (no Non-fundamental Particles Hadrons internal structure) (made of quarks) Leptons (electron, muon, tau, neutrinos) Baryons Exchange Particles (contains 3 quarks) Gluon, W, Z, photon e.g. Proton, neutron As we currently understand!! Mesons (quark + antiquark) e.g. Pion, kaon Quarks The building blocks of protons and neutrons, and other fundamental particles. Two flavours of quark… The up quark (+ 2/3 e) and the down quark (– 1/3 e). The first direct evidence for quarks was obtained when very high-energy electrons (approx. 20Gev) in a beam were scattered from a stationary target as if there were point-like scattering centres in each proton or neutron. Quarks do not exist in isolation. They are bound together by the exchange of gluons (since they are the glue that hold the quarks together). Quark–gluon interaction red quark blue quark red–blue gluon red quark blue quark Quarks interact by exchanging gluons, which change the quark colours. Here a red quark and a blue quark exchange a red–blue gluon. The red quark becomes blue and the blue quark becomes red. The quarks exchange energy and momentum. Colour charge Quarks have a new type of charge called colour charge. The charge comes in 3 types. Red, Green and Blue. They aren't actually coloured. Reason for colour charge is you need one of each colour to make a proton or neutron just like you need all colours to make white. Quarks pulled apart make m ore quarks quark quark two quarks held together by the gluon field... gluon field ...pull the quarks apart. The gluon field increases in energy... quark quark quark quark antiquark quark ...a quark–antiquark pair materialises from the gluon field You cant really separate quarks!!! Starter: Copy the table and fill in each box with example(s) of each phenomenon WAVES Light Electrons PARTICLES Chemistry questions that only physics can answer..... DISCUSS IN PAIRS o Why are atoms mostly empty space? o Why don’t the electrons just fall into the nucleus? o Why are only some energy levels allowed for electrons in an atom or a molecule? o Why does delocalisation of electrons stabilise a molecule? o Why are carrots orange? The classical atom dethroned • Explain the flaws in the “solar system” model of the atom • Investigate an electron-wave atomic model • Use an electron-wave model to explain both the stability of aromatics and the origin of colour in chemistry What is the problem?!? Maxwell’s theory which states that an electron being accelerated in an electric field will emit radiation. Hint: Think about the Diamond visit! AS-RECAP Wave-Particle Duality (for electrons) What is the evidence?!? de Broglie λ = h/p = h/mv -31 m=9.1x10 kg -34 h=6.6x10 Js AS-RECAP Standing Waves… at rest Allowed discrete values of wavelengths only…. fundamental n = 1 L = λ/2 L= λ L= 3λ/2 Ie L = n λ /2 harmonic n = 2 L-length of string n – harmonic harmonic n = 3 Electrons in atoms… (QM) Atoms can be thought of as a box in which electrons are trapped. Results of model: Electrons can only occupy certain discrete energy levels in atoms. Each level has a different associated standing wave of a particular λ. A guitar–string atom Sim plify: energy C hange the 1/r potential well of the nucleus into a pair of fixed high walls energy d trapped electron nucleus + trapped electron 1/ r potential well Putting de Broglie and standing waves together…. Beta carotene….. ….gives carrots and pumpkins their orange colour ….in humans, is converted into retinal, which is essential for the vision process • A conjugated system with alternating single and double bonds • Average carbon-carbon bond length = 140 pm (1 pm = 10-12 m) • 2 delocalisable electrons per double bond Challenge Can you use the equation for electron delocalisation in one dimension (particle in a box) to : (a) determine the pattern of energy levels in beta carotene, (b) calculate the wavelength at which it absorbs light. Beta-carotene: a chromophore modelled as a particle in a box • A conjugated system with alternating single and double bonds • Average carbon-carbon bond length = 140 pm • Calculate the length over which electrons are delocalised • Identify the quantum numbers of the highest-occupied and lowest unoccupied energy levels • Use the formula E=n2h2/8mL2 to calculate the energies of these levels • Calculate the energy and wavelength of the photon that would be absorbed if an electron was excited from the highest-occupied to lowest-unoccupied level Chemistry questions that only physics can answer..... o Why are atoms mostly empty space? o Why don’t the electrons just fall into the nucleus? o Why are only some energy levels allowed for electrons in an atom or a molecule? EXPLAINED WITH BASIC MODEL o Why does delocalisation of electrons stabilise a molecule? EXPLAINED WITH BASIC MODEL o Why are carrots orange? EXPLAINED WITH BASIC MODEL Starter In the last lesson, we modelled the electron in an atom by treating the electron as a standing wave confined to a potential well. Q1. Which of the following feature(s) of the atom did this model reproduce? - Quantised electron energies (electrons can only have certain energies within the atom)? - Energy levels becoming closer together with increasing energy, as is observed in the spectra of atoms? Q2. The model could be used to describe delocalised electrons in molecules such as benzene and carotene, and thereby explain why delocalisation (resonance) stabilises these molecules, and how colour arises in them. (a) Using the energy level expression derived from the model, explain why allowing electrons to spread out in a molecule leads to stabilisation. (b) Explain how to calculate the energy of a photon absorbed by a molecule with delocalised electrons. Finally, a model for the atom that fits experimental data! • Develop an electron-wave model that incorporates potential energy • Use the model to investigate atom stability • Calculate energy levels and atomic radii with the model, and see if they match experiment Considering Potential energy properly… How small c ould a hydrogen atom be? imaginary box d = 2r = 4r 1/r potential + nucleus Replace 1/r potential by a box of width d = 2r Calculate kinetic energy for waves = 2d = 4r Calculate potential energy at r potential energy E p = – e2 4 0r standing wave /2 = d momentum p = h/ 2 kinetic energy = p /2m h2 kinetic energy E k = 2m 2 Find the m in im um rad iu s of an atom , fo r total energy < 0 s hort wavelength kinetic unstable energy 120 2 potential energy E p = – nucleus kinetic energy E k = 4 0r 2m 2 Considering Potential energy properly… Find the m in im um rad iu s of an atom , fo r total energy < 0 s hort wavelength kinetic unstable energy 120 kinetic energy = 100 + small radius Ek + Ep > 0 potential energy medium wavelength just stable kinetic ener gy h2 2m minimum radius of bound atom 2 80 total energy > 0 unstable 60 40 total energy = 0 just stable 20 + potential energy medium radius Ek + Ep = 0 0 0.02 long wavelength kinetic energy stab le –20 0.04 0.06 radius r /nm total energy < 0 bound –40 potential energy = – potential energy large radius Ek + Ep < 0 + –60 0.08 e2 40 r KEY POINT: If size is too small, the kinetic energy is too large for electrical potential to bind the electron. The Bohr model of the atom Total Energy = Kinetic Energy + Potential Energy Q1. Which of the terms in the equation above is always positive, and which is always negative? Q2. For a stable atom, one where the electron is bound, to the nucleus, what must be true about the sum of KE+PE? Kinetic Energy KE = h2 / 2mλ2 where λ is the wavelength of the standing wave describing the electron What will happen to the KE if the electron is confined in a smaller space closer to the nucleus? Potential energy PE = -Zke2 / r where r is the nucleus-electron distance, e is the charge on the electron and k is a constant (=1/4πε0) What will happen to the PE if the electron is confined in a smaller space closer to the nucleus? KE + PE for hydrogen atom, with electron modelled as a "particle on a ring", 2-dimensional standing wave 4E-18 At very small radii, the KE is very large as the electron is confined to a very small orbit. The result is that KE + 3E-18 Total energy / J 2E-18 1E-18 0 0.00 At r = 0.059 nm, the total energy is a minimum. This is the distance at which the electron settles, and determines the radius of the atom. The value of r agrees exactly with experiment in this case, showing that the model is a valid one. 0.05 0.10 0.15 At large radii, the KE term is small. The atom is bound because of the negative PE term, but not as stable as it could be if the electron approached the nucleus more closely. 0.20 0.25 -1E-18 -2E-18 -3E-18 r / nm 0.30 0.35 0.40 0.45 Chemistry questions that only physics can answer..... o Why are atoms mostly empty space? EXPLAINED WITH BOHR MODEL (electron wave + PE) o Why don’t the electrons just fall into the nucleus? EXPLAINED WITH BOHR MODEL (electron wave + PE) o Why are only some energy levels allowed for electrons in an atom or a molecule? EXPLAINED CORRECTLY WITH BOHR MODEL (electron wave + PE) o Why does delocalisation of electrons stabilise a molecule? EXPLAINED WITH BASIC MODEL (particle in a box) o Why are carrots orange? EXPLAINED WITH BASIC MODEL (particle in a box) Starter : Explaining atomic structure Explain the following statements carefully, using the key words in italics provided if need be. • An electron in orbit around a nucleus does not radiate energy and fall into the nucleus as predicted by classical physics. • Matter is mostly empty space, with the atomic nucleus occupying a tiny fraction of the volume of the atom. • The energy levels occupied by an electron in atom can have only certain values described by a quantum number n, where n = 1, 2, 3 etc. • Atoms absorb and emit light only at certain specific wavelengths. wave-particle duality standing wave destructive interference de Broglie wavelength kinetic energy potential energy stability photon energy Energy levels and spectra • Use atomic energy level data to calculate emission and absorption spectra • Interpret results of the Franck-Hertz experiment in terms of atomic energy levels Energy levels in hydrogen En = 2 -13.6/n Questions…. Q1. What is the wavelength of the Red and blue Balmer lines? Q2. If the wavelength of an emitted photon was 102 nm, which transition from the Lyman series caused it? The Franck-Hertz experiment The Franck-Hertz experiment • Why does the current suddenly drop sharply? • Why does the drop occur at different beam energies for different elements? A hydrogen atom has energy levels at -13.6 eV (ground state, containing one electron), -3.4 eV, -1.5 eV, -0.9 eV. The energy levels are measured with respect to the ionisation limit, 0.0 eV. (a) Calculate the wavelength of the lowest-energy photon that could be absorbed when an electron in the ground state of a hydrogen atom is excited to an upper energy level. (Data: h = 6.6 x 10-34 Js, c = 3 x 108 ms-1, e = 1.6 x 10-19 C) [3] (b) Electrons of energy 9 eV are fired through hydrogen gas in its ground state. The electrons are scattered without loss of energy. When the experiment is repeated with electrons of 11 eV energy, the electrons are scattered inelastically, emerging with energies of about 1 eV. Explain these observations. [3]