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A potential difference V is maintained between the metal target and the collector cup

A potential difference V is maintained between the metal target and the collector cup. Electrons ejected from C travel to A and G detects the flow. i. Apply voltage V between A and C to slow the ejected electrons down. When potential matches the initial

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A potential difference V is maintained between the metal target and the collector cup

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  1. A potential difference V is maintained between the metal target and the collector cup Electrons ejected from C travel to A and G detects the flow i Apply voltage V between A and C to slow the ejected electrons down When potential matches the initial KE of the electrons, the flow stops (most energetic electrons stopped) Kmax= e Vstop

  2. Photoelectric Effect • Vstopdoes not depend on the intensity of the light source for a given frequency f • => classical physics would predict that if we increase the amplitude of the alternating electric field, then a larger kick would be given to the electron? • => if light is composed of photons, then the maximum energy that an electron can pick up is that of a single photon

  3. High intensity Adjust V in negative sense until current vanishes All electrons reach collector Kmax= e Vstop Independent of intensity! Low intensity Measure Vstop as a function of frequency f

  4. Vstop Different metals f f0 Photoelectric Effect • In each case, there is a minimum frequency f0 for the effect to occur • cannot be explained classically

  5. Photoelectric Effect • Classical theory: • oscillating e/m fields in the light cause electrons in the metal to oscillate • average KE ~ amplitude2 ~(electric field)2~Intensity • Quantum theory: • light composed of energy packets called photons E=hf =>an electron absorbs one photon and gains energy hf (this process is independent of the intensity) • not expected classically! => increase intensity or wait longer for electron to absorb enough energy

  6. PE  e Photoelectric Effect Why do electrons stay in metals? Electrical force lowers the potential energy • Electron needs a minimum amount of energy  to escape •  depends on the type of metal => called the work function • if an electron absorbs a photon, then (hf - ) is the amount of energy left over for KE

  7. Photoelectric Effect • Hence we need hf >  to just escape • that is f > /h =f0 • Einstein: Kmax = (hf - ) if no other = e Vstop losses of energy are involved • Vstop =(h/e) f - (/e) Slope =h/e is independent of the metal!

  8. units: volts is a unit of electrical potential • eV = (1.6x10-19) volts is a unit of energy called an electron volt (eV) • eVstop =h f -  = h( f - f0) = Kmax

  9. Problem • A satellite in Earth orbit maintains a panel of solar cells of area 2.60 m2 oriented perpendicular to the direction of the Sun’s rays. Solar energy arrives at the rate of 1.39 kW/m2 (energy/area/time) • (a) at what rate does Solar energy strike the panel? • rate=energy/time = 1.39(2.60) =3.61 kW • (b) at what rate are Solar photons absorbed ? (=550nm) • each photon carries E=hc/ =(6.63x10-34)(3x108)/(550x10-9)=3.61x10-19 J • photons/time = (3.61x103)/(3.61x10-19 ) = 1022 /sec

  10. Problem • (c) how long would it take for a mole of photons to be absorbed? • NA = 6.02 x 1023 • time = NA/(number photons/time) = (6.02 x 1023)/1022 = 60.2 sec

  11. Problem • Light strikes a sodium surface and causes photoelectric emission. If Vstop = 5.0 volts and the work function  is 2.2 eV, what is the wavelength of the light? • Ephoton = hf = hc/ • Kmax = Ephoton -  = hc/ - = e Vstop •  = (hc)/(e Vstop + ) • h = 6.63x10-34 J.s = 6.63x10-34 /1.6x10-19 eV.s = 4.14 x10-15 eV.s •  = (4.14 x10-15 eV.s)(3x108m/s)/[5.0 eV+2.2 eV] =170 nm

  12. Momentum • 1916 Einstein extended the photon idea • when light interacts with matter, not only energy but also linear momentum is transferred via photons • momentum is also transferred in discrete amounts p=hf/c = h/photon momentum • E=hf = hc/ p=hf/c= h/ • => E = pc • recall that E2=p2c2 + m2c4 => m=0 massless • short wavelength photons have more energy and momentum!

  13. `  Compton Effect • 1923 Compton performed an experiment which supported this idea • directed a beam of x-rays of wavelength  onto a carbon target • x-rays are scattered in different directions ` has 2 peaks  = 71.1 pm (10-12 m)

  14. Compton Scattering • Wavelength ` of scattered x-rays has two peaks • these occur at  and  +  •  >0 is the Compton shift • classical physics predicts  =0 • Quantum picture: • a single photon interacts with electrons in the target • light behaves like a ‘particle” of energy E=hf=hc/ and momentum p=h/ => a collision

  15. E`=hf `=hc/`  E=hf=hc/ K=mec2(-1) Compton Scattering • Conservation of energy E = E` + K • => E` < E => f ` < f => ` >  • X-ray momentum p=h/ p`= h/` • electron momentum pe = mev

  16. E`=hf `=hc/`  E=hf=hc/ K=mec2(-1) Compton Scattering • Conservation of energy E = E` + K • => E` < E => f ` < f => ` >  • X-ray momentum p=h/ p`= h/` • electron momentum pe = mev

  17. X-ray scattering • Energy and momentum are conserved • Momentum is a vector! F=dp/dt=0 => p = constant

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