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Published byBraedon Barge Modified about 1 year ago

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About these slides These slides are used as part of my lessons and shouldn’t be considered comprehensive –There’s no excuse for not turning up to lessons! These slides use material from elsewhere on the assumption of fair/educational use –If you own the copyright to any of this material and want it credited/removed please contact me These slides may contain errors –Use at your own risk! This work is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike 3.0 Unported License.Creative Commons Attribution-NonCommercial- ShareAlike 3.0 Unported License

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Recap Wave-particle duality You should be familiar with the idea that light can be thought of as a wave or a particle The particle is called a __________ The energy of this particle is given by the equation E=hf Where h is ________ constant and f is the frequency of the light Rewrite the equation in terms of wavelength,

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Einstein

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Evidence for wave-particle duality One piece of evidence that light can behave as a wave or a particle is the photoelectric effect This was discovered by Hertz as a side effect of his work on radio waves He noticed that metals emit electrons when electromagnetic radiation of a certain frequency is directed at a metal

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The photoelectric effect in action Incident Radiation electrode gold leaf zinc plate

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Photoelectric effect The photoelectric effect gives rise to the following observations: –Emission of electrons only takes place above a certain threshold frequency of incident electromagnetic radiation. This minimum frequency is dependent on the metal. If light was a wave then electrons would be emitted at any frequency, as electrons would gradually gather energy from the waves regardless of their frequency

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Photoelectric effect continued The photoelectric effect gives rise to the following observations: –Electrons are emitted as soon as the source of electromagnetic radiation is switched on, no matter what the intensity of the radiation. If light was a wave then electrons would gather energy more quickly, and be emitted sooner, by high intensity light sources.

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Photoelectric effect continued The photoelectric effect gives rise to the following observations: –The number of electrons emitted per second is proportional to the intensity of the incident radiation, unless its frequency is less than the threshold frequency, in which case emission isn’t observed

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Particles as waves

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Electron diffraction

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If electrons were acting as particles we would see them either pass straight through the crystal, or bounce off Instead we see an interference pattern –Exactly the same as if electrons are acting as waves It is possible to do the Young’s slits experiment and see an interference pattern even if the electrons are fired one at a time

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de Broglie In 1923 de Broglie was the first to suggest that matter could be a wave He said that the de Broglie wavelength,, of a particle is related to its momentum, p, by the equation We know that momentum = mass x velocity so

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? ? Question Calculate the de Broglie wavelength of –An electron moving at 2.0 x 10 7 ms -1 –A proton moving at the same speed I weigh 85kg and can run at 6 ms -1 what would my de Broglie wavelength be?

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? ? Question Calculate the momentum and speed of –An electron that has a de Broglie wavelength of 500nm –A proton that has the same de Broglie wavelength

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The photoelectric effect We saw this last lesson As long as light is above a certain frequency it can eject electrons from a metal surface This can be explained as the photon having a certain energy, E = hf It therefore follows that there is a minimum energy needed to eject an electron from a metal surface at zero potential –This is the work function,

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Explaining the photoelectric effect In order to explain photoelectricity Einstein said that: –Electrons must be absorbing a single photon in order to escape from the metal –This photon transfers energy given by E=hf –If the photon has less energy than the work function, , of the metal the electron cannot escape –The work function is the minimum energy required for an electron to escape the metal surface

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Photoelectric effect We can calculate the maximum kinetic energy of an electron ejected by a photon with energy hf E kmax = hf – So emission will take place as long as hf > Which means there is a threshold frequency above which an electron is emitted:

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Plotting E kmax against frequency Since E Kmax = hf – we can plot a graph of the form y = mx + c So if we plot E Kmax against f we get a straight line What is the gradient? What is the intercept on the x axis? What would the intercept on the y axis be? What is the gradient? What is the intercept on the x axis? What would the intercept on the y axis be? E Kmax /J f / Hz

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? ? Questions Why does photoelectric emission only take place above a certain frequency of incident radiation? Calculate the frequency and energy of a photon of wavelength –450nm –1500nm

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? ? Questions The work function of a metal plate is 1.1 x 10 -19 J Calculate: –The threshold frequency –The maximum kinetic energy of electrons emitted from this plate when light of wavelength 520nm is directed at the metal surface

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Electron volts We’ve already encountered the electron volt It is defined as the unit of energy equivalent to the work done when an electron is moved through a potential difference of 1 volt and is equal to 1.6 x 10 -19 J

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Ionisation An ion is a charged atom Ionisation is the process of making an ion by adding or removing electrons The energy to do this can be provided by electricity, heat, light or by collisions between alpha, beta or gamma radiation and an atom

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Energy levels in atoms Electrons surrounding atoms are like standing waves They have certain allowed energy levels It is the movement between these levels that happens during excitation The lowest energy state of an atom is its ground state If an electron moves into a higher energy level the atom is in an excited state

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Excitation In excitation an electron can be moved from an inner shell to an outer shell This only happens at specific excitation energies which are dependent on the atom and correspond to its excited states This is excitation by collision – one electron hits another and promotes it to a higher energy level

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Excitation continued The excitation energy will always be less than the ionisation energy There are a number of excitation energies for each atom –These correspond to the difference between different shells –Electrons can also ‘jump’ more than one shell at a time

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Energy Levels Example: Mercury Ionised 0.00 eV -1.59 ev -1.60 eV -2.51 eV -2.71 eV -3.74 eV -4.98 eV -5.55 eV -5.77 eV -10.44 eV This is excitation by absorption of a photon. If it has exactly the right frequency it promotes the electron to a higher energy level

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De-excitation The excited states of an atom are unstable They have a gap in a low energy electron shell Electrons from a higher shell can drop into this lower level, emitting a photon in the process

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De-excitation Example: Mercury This is fluorescence An atom in an excited state can de-excite by emitting photons of the same or lesser energy than the one that excited it In this case the photons have energies of 0.79 eV and 4.67 eV Ionised 0.00 eV -1.59 ev -1.60 eV -2.51 eV -2.71 eV -3.74 eV -4.98 eV -5.55 eV -5.77 eV -10.44 eV

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Calculating the frequency of photons Since the photon emitted comes from the transition between two energy levels it will have an energy, E given by: E = E 1 – E 2 Where E 1 is the energy of the higher level and E 2 the energy of the lower We also know that E = hf, so we can calculate the frequency of the emitted photon

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Fluorescent Tubes Fluorescent tubes contain mercury vapour and have a fluorescent coating on the inner surface Mercury Vapour Filament Electrodes

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Fluorescent Tubes When a tube is turned on: –Ionisation and excitation of mercury atoms occurs as they collide with each other and with electrons –Photons are emitted at UV and visible frequencies –UV photons are absorbed by atoms in the fluorescent coating –Coating atoms de-excite and emit visible photons –A range of coatings is used to give a ‘white’ light

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? ? Questions How much energy is needed, in Joules, to excite the atom to its highest excitation level? How many different energies could photons released by a mercury atom in the 5.46 eV excited state have? Ionised 0.00 eV -1.59 ev -1.60 eV -2.51 eV -2.71 eV -3.74 eV -4.98 eV -5.55 eV -5.77 eV -10.44 eV

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? ? Question An atom absorbs a photon of energy 3.8 eV and subsequently emits photons of energy 0.6 eV and 3.2 eV –Sketch an energy level diagram to represent these changes –Describe what is happening to the electrons in the atom during this process –What is the frequency of the photons emitted?

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? ? Answer

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Spectral Lines Since each element has different energy levels, and therefore transitions between levels, each element will emit different frequencies (wavelengths) of electromagnetic radiation This means the photons emitted are characteristic of that element

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Mercury

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Hydrogen

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? ? Question A mercury atom de-excites from its 4.9 eV energy level to the ground state Calculate the wavelength of the photon released c = 3.0 x 10 8 ms -1, e = 1.6 x 10 -19 C, h = 6.63 x 10 -34 Js

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