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Example Problem: Photoelectric EffectPhysics 1251TA: Brian Clark11/30/20151. A piece of metal has a cutoff wavelength of λcutoff= 450 nm. Consider illuminatingthis piece of metal with two different wavelengths of light: a λ1= 500 nm beam and aλ2= 400 nm beam. For each of the two beams, find:(a) The maximum kinetic energy of ejected electrons.We can only eject an electron if the illuminating beams are energetic enough. BecauseE = hf = hc/λ, we can only eject an electron if λilluminating≤ λcutoff. So, for λ1,whichis greater than λcutoff, we eject no electrons.However, for λ2, which is less than λcutoff, we do eject electrons. The energy of these electronsis given by:E = hf − φ = hc/λ −φThat is, we take the incoming energy of the photon (hc/λ), subtract off the energy required tobind the electron (φ, the work function) and we are left with the kinetic energy of the electron.E = hf − φ =hcλ2−hcλcutoff= hc1λ2−1λcutoff= (6.626 × 10−34J · s)(3 × 108m/s)1400 × 10−9−1450 × 10−9=⇒ E2= 5.21 × 10−20J = 0.344 eV(b) What is their speed?Because λ1never ejects electrons, it does not make sense to speak of their speed. For λ2,we will simply solve using our kinetic energy formula from 1250:E =12mv2=⇒ v =r2Em=s2 · 5.21 × 10−20J9.11 × 10−31kg=⇒ v2= 3.48 × 105m/s(c) What is their de Broglie wavelength?Again, because λ1never ejects electrons, it does not make sense to speak of their de Brogliewavelength. For λ2, we can apply the de Broglie wavelength formulaλ =hp=6.626 × 10−34J · s9.11 × 10−31kg · 3.48 ×105m/s=⇒ λ2,dB= 2.09 × 10−9m = 2.09 nm2. An electron, a proton, and a photon each have a wavelength of 0.24 nm. For each one,find the momentum, the energy, and, where relevant, the accelerating voltage neededto achieve that wavelength:The momentum calculation is the same for all three particles. We employ the de Broglie relation:p =hλ=⇒ pp= pe=6.626 × 10−34J · s0.24 × 10−9m=⇒ pp= pe= pγ= 2.76 × 10−24kg m/sNow, let’s specialize to the massive particles first. For a massive particle, we can apply that p = mvand E =12mv2, to find: E =12p2m. So:Ee=12p2eme=12·(2.76 × 10−24kg m/s)29.11 × 10−31kg=⇒ Ee= 4.044 × 10−18J = 25.24 eVEp=12p2pmp=12·(2.76 × 10−24kg m/s)21.67 × 10−27kg=⇒ Ep= 2.206 × 10−21J = 0.014 eVBecause these are massive particles, they can be brought to this energy by an accelerating potential,given by E = q∆V :|∆Ve| =Eeqe=4.044 × 10−18J1.602 × 10−19C=⇒ Ve= 25.24 V|∆Vp| =Epqp=2.206 × 10−21J1.602 × 10−19C=⇒ Ve= 0.013 VFor the photon, the massless particle, the calculation is more straightforward:Eγ= hf =hcλ=(6.626 × 10−34J · s)(3 × 108m/s)0.24 × 10−9m=⇒ Eγ= 8.28 × 10−16J = 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 requiresrelativity–why?For the 2 keV electron, we can apply:λ =hp=hm · v=hm ·q2Em=h√2Em=6.626 × 10−34J · sp2 · (3.2 × 10−16J) · (9.11 × 10−31kg)=⇒ λ2 keV= 2.74 × 10−11mWe must be more careful for the 200 keV electron, because its velocity is, to first order:v =r2Em=s2 · 3.204 × 10−14J9.11 × 10−31kg≈ 2.65 × 108m/swhich is very near the speed of light.So, to be relativistically correct, we must apply the relativistic kinetic energy formula to get p:E =p(mc2)2+ p2c2− mc2=⇒ p =1cp(E + mc2)2− (mc2)2p =1cp[3.204 × 10−14J + (9.11 × 10−31kg) · (3 ×108m/s)2]2− [(9.11 × 10−31kg)(3 × 108m/s)2]2=⇒ p = 2.64 × 10−22kg m/sNow, we can plug this correct momenta into the de Broglie equationλ =hp=6.626 × 10−34J · s2.64 × 10−22kg m/s=⇒λ200 keV= 2.51 ×

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