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Physics Quantum
Physics Spring Y12
| Question | Answer |
|---|---|
| What is wave energy normally based on, and what is it for EM waves? | Amplitude, but EM waves have energy based on frequency |
| What is a photon | A quantum (discreet packet) of EM radiation |
| Energy of photon depends on what | Frequency E = hf Energy of photon = planck's constant [J] = [kgm^2s^-1] [Hz] so E = hc/lambda |
| Electronvolts | Used often for energy of a photon 1 eV = 1.6 x 10^-19 J e.g. Electron accelerated by p.d. of 12 kV Gains 12000 eV |
| The photoelectric effect | A one to one interaction: an electron can absorb a photon to gain enough energy to leave a metal surface. If photons dont have enough energy, no electrons will be able to leave no matter how many photons are incident on the surface |
| The work function | Φ. The amount of energy needed to leave the metal surface - depends on the metal itself It's the minimum energy required to release a photoelecton (aka an electron) when light is incident on metal |
| What happens if the photons dont have enough energy | They just reflect |
| Charge on a photon | They are chargeless |
| Energy of photoelectron equation | hf (energy of photon) = Φ [if electron emitted with no KE] hf = Φ + Ek [if electron has KE) Phi here just shows the energy needed to release the electron |
| What happens to the photon after giving energy to the electron? | It's gone - its just a packet of energy |
| Threshold frequency | The minimum frequency to release a photoelectron |
| What does increasing the frequency of the photons do? | Increases the max energy the electrons have once released (because photons have more energy) Inner shell electrons spend more energy escaping so still have low KE |
| What is the equation for the energy into an LED? | IVt (current x potential difference = power) = QV |
| What is the equation for the energy out of an LED? | hf |
| Equation manipulation for LED energy for the Planck's constant experiment | hf = QV Q = e f = c/lambda eV = hc/lambda eV = (hc)(1/lambda) hc = gradient When you do 1/lambda, will leave you with e.g. 735 x 10^9 |
| Planck's constant experiment | Can adjust p.d. and wavelength of LEDs. Find minimum p.d. for a given LED to barely light up. This means the energy required to release a single electron, so J = V. Then convert from J to eV for the table. |
| If asked to explain eVmin = hc/wavelength | The energy eV of the electron going in is equal to the energy hc/lambda of the photon coming out (For LEDs you dont have a work function because the electrons arent trapped) |
| Other way of explaining the eVstop equation | hc/lambda - phi = eVstop hc/lambda - phi = the kinetic energy of the photon when released eVstop is the energy lost by the photon due to attraction to surface released from |
| Vstop | The electrons aren't stopping in place they still go back to the anode, theyre just no longer hitting the cathode The energy removed from each electron is e.g. 5 eV, so in hf = phi + Ek, Ek = 5 eV |
| On a graph of Vstop on the y axis and frequency on the x axis | x intercept is threshold frequency (hf = phi + eVstop. when Vstop = 0, then hf = phi, so hf is minimum energy to beat the threshold) y intercept is -phi/e (REMEMBER THE MINUS) gradient = h/e = CONSTANT |
| How to work out work function from just threshold freq | At threshold frequency, hf = phi |
| How EM waves act as waves | Young's double slits they diffract. Waves do, particles dont |
| How EM waves act as particles | Photoelectric effect - EM waves behave as particles (photons) carrying quantised energy 'packets' |
| How electrons act as particles | Electrical circuits - carry quantised charge and energy in 'energy levels' |
| How electrons act as waves | Electron diffraction - electrons accelerated at a crystalline target will exhibit wave-like properties |
| Know that for all waves | They sometimes act as particles and vice versa |
| What is the wavelength of a particle called | The deBroglie wavelength. ALWAYS say it 'is of a particle' if you ever say its name. Also shown by lambda. lambda = h/momentum of particle momentum of particle can be found from only p.d. of electron gun |
| Key points when talking about the planck constant experiment | Connect flying lead to one LED at a time. Adjust potential divider to zero first - easier to see when it lights up. Dark room and shield with opaque tube eVmin = hc/l Repeat each and between Draw lobf through origin Calc values of 1/wl |
| Very careful if talking about electrons | (or other particles) - can't use the normal equations for them which assume they travel at c, can only use the bottom 2 |
| What to say for why intensity doesnt affect EM energy | Say E proportional to F so E doesnt depend on intensity |
| How function of blue LED relates to shape of curve | LED turns on at 2.6V. Electron passes through pd of 2.6V and loses energy to produce a photoelecton of blue light is not directly proportional as is not a straight line |
| Rough wavelengths of EM | Radio ~1m Micro cm Infra -5 UV -9 X -11 Gamma <-12 |
| Why gold leaf does that | (Instantaneous) discharge due to emission of e. Define photons UV photon energy > rho while light < because E = hf. Energy independent to intensity Energy conserved in interaction Mention Einstein's equation |
| Why wave theory wrong for gold leaf | Suggests lead would fall with light because based on amplitude. |
| Why gold leaf discharges faster when UV close | *Rate* of incident photons is more at smaller distances so quicker fall for UV (not visible) |
| Einstein's equation | hf = rho + eVstop |
| What to mention for analysis in 6 markers | Mention if line is straight line graph and if through origin Draw lobf and determine gradient using a 'large triangle' Define accurate + precise when using them Do any calcs you can - for accuracy turn all uncertainties into % |
| Why electrons diffracted by crystalline target | Not by holes in metal. Not by electrons repelling each other. Instead just due to wavelength of electrons |
| Wavelength of electrons to be diffracted | 10^-10m - size of gap in atoms |
| How to work out efficiency of LED based on light emitted per second when at 20mA Part 1 - working out electron energy | At point when photons first start being emitted (where graph curves up), voltage is 2.6V, so energy is 2.6e on single electron (when photons first emitted. one-to-one), so total energy flow at 20mA is the number of electrons flowing x 2.6e |
| How to work out efficiency of LED based on light emitted per second when at 20mA Part 2 - working out efficiency | Photon energy = electron energy. Work out efficiency with electron energy/basic energy of circuit |
| How to work out efficiency of LED based on light emitted per second when at 20mA Part 3 - conclusions to be drawn | Increases in p.d. of LED increase intensity (as more can overcome) over time but reduce efficiency as the photons still have to leave with the same energy, so each electron having higher energy doesnt matter on the higher levels |
| What happens when you double the intensity on a photocell? | Double amount of photons. IF A QUESTION GIVES ANYTHING NUMERICAL E.G. DOUBLED, SAY 'the rate of photons incident on M is doubled so the rate of emission of photoelectrons/current is doubled' |
| What happens when electrons shot at crystalline target | Electrons diffracted passing through thin sheet of graphite, diffracting producing rings/pattern (on a fluorescent screen) |
| Explain how the KEmax against frequency graph can't be explained with the wave model | Shouldn't have threshold frequency as should be intensity based. Shouldn't have KE of electrons dependent on frequency. |
| Just read the axis more carefully daniel | If there are 10 gaps between 5 and 10 notice it, like come on |
Created by:
Pyrogearos2