**Unformatted text preview:**

Example Problem Photoelectric E ect Physics 1251 TA Brian Clark 11 30 2015 1 A piece of metal has a cuto wavelength of cutof f 450 nm Consider illuminating this piece of metal with two di erent wavelengths of light a 1 500 nm beam and a 2 400 nm beam For each of the two beams nd a The maximum kinetic energy of ejected electrons We can only eject an electron if the illuminating beams are energetic enough Because E hf hc we can only eject an electron if illuminating cutof f So for 1 which is greater than cutof f we eject no electrons However for 2 which is less than cutof f we do eject electrons The energy of these electrons is given by E hf hc That is we take the incoming energy of the photon hc subtract o the energy required to bind the electron the work function and we are left with the kinetic energy of the electron hc cutof f 1 cutof f E hf hc 2 hc cid 17 6 626 10 34 J s 3 108 m s cid 16 1 2 1 400 10 9 E2 5 21 10 20 J 0 344 eV cid 16 1 450 10 9 cid 17 b What is their speed Because 1 never ejects electrons it does not make sense to speak of their speed For 2 we will simply solve using our kinetic energy formula from 1250 E mv2 v 1 2 cid 114 cid 115 2E m 2 5 21 10 20 J 9 11 10 31 kg v2 3 48 105 m s c What is their de Broglie wavelength Again because 1 never ejects electrons it does not make sense to speak of their de Broglie wavelength For 2 we can apply the de Broglie wavelength formula h p 6 626 10 34 J s 9 11 10 31 kg 3 48 105 m s 2 dB 2 09 10 9 m 2 09 nm 2 An electron a proton and a photon each have a wavelength of 0 24 nm For each one nd the momentum the energy and where relevant the accelerating voltage needed to achieve that wavelength The momentum calculation is the same for all three particles We employ the de Broglie relation p pp pe h 6 626 10 34 J s 0 24 10 9 m pp pe p 2 76 10 24 kg m s Now let s specialize to the massive particles rst For a massive particle we can apply that p mv and E 1 p2 m So 2 Ee 2mv2 to nd E 1 p2 e me p2 p mp 1 2 1 2 1 2 1 2 Ep 2 76 10 24 kg m s 2 9 11 10 31 kg 2 76 10 24 kg m s 2 1 67 10 27 kg Ee 4 044 10 18 J 25 24 eV Ep 2 206 10 21 J 0 014 eV Because these are massive particles they can be brought to this energy by an accelerating potential given by E q V Ve Vp Ee qe Ep qp 4 044 10 18 J 1 602 10 19 C 2 206 10 21 J 1 602 10 19 C Ve 25 24 V Ve 0 013 V For the photon the massless particle the calculation is more straightforward E hf hc 6 626 10 34 J s 3 108 m s 0 24 10 9 m E 8 28 10 16 J 5 68 keV Which is termed a soft x ray in the astrophysics community by the way 3 What is the de Broglie wavelength of an electron that has 2 0 keV of kinetic energy What about an electron with 200 keV of kinetic energy The second one requires relativity why For the 2 keV electron we can apply h p h m v h cid 113 2E m m h 2Em 6 626 10 34 J s cid 112 2 3 2 10 16 J 9 11 10 31 kg 2 keV 2 74 10 11 m We must be more careful for the 200 keV electron because its velocity is to rst order cid 114 v cid 115 2E m 2 3 204 10 14 J 9 11 10 31 kg 2 65 108 m s which is very near the speed of light So to be relativistically correct we must apply the relativistic kinetic energy formula to get p E cid 112 mc2 2 p2c2 mc2 p cid 112 E mc2 2 mc2 2 1 c p cid 112 3 204 10 14 J 9 11 10 31 kg 3 108 m s 2 2 9 11 10 31 kg 3 108 m s 2 2 1 c p 2 64 10 22 kg m s Now we can plug this correct momenta into the de Broglie equation h p 6 626 10 34 J s 2 64 10 22 kg m s 200 keV 2 51 10 12 m

View Full Document