Problem Set 1 Solutions

Problem Set 1 Solutions - Chemistry 120A, Spring 2009...

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Chemistry 120A, Spring 2009 Problem Set 1 Solutions Due January 30, 2009 Problems 1. The Davison-Germer experiment of the diffraction of electrons by an ordered metal surface gives a beautiful illustration of the validity of the de Broglie relation connecting the particle- like properties of electrons (their momentum) to their wave-like properties (their wavelength, which causes diffraction if same as interatomic spacing) (a) A 40 eV beam of electrons shows no diffraction from a silver surface, a 54 eV beam exhibits a clear diffraction patter. By 60 eV the diffraction structure is again largely lost. Work out the wavelengths that these three beam energies correspond to. We know from de Broglie that λ = h p , and KE = 1 2 mv 2 = p 2 2 m Therefore, p = 2 mE and λ = h 2 mE For 40 eV λ = 6 . 626 × 10 - 34 Js r 2 · 9 . 109 × 10 - 31 kg · 40 eV · 1 . 602 × 10 - 19 J 1 eV λ = 1 . 9393 × 10 - 10 m = . 19393 nm For 54 eV λ = 6 . 626 × 10 - 34 Js r 2 · 9 . 109 × 10 - 31 kg · 54 eV · 1 . 602 × 10 - 19 J 1 eV λ = 1 . 6691 × 10 - 10 m = . 16691 nm For 60 eV λ = 6 . 626 × 10 - 34 Js r 2 · 9 . 109 × 10 - 31 kg · 60 eV · 1 . 602 × 10 - 19 J 1 eV λ = 1 . 5834 × 10 - 10 m = . 15834 nm 1
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(b) If one were worried about damage to the surface caused by the energy of the electrons necessary to cause diffraction, one could consider using a beam comprised of different particles. Work out the energy of a beam of helium atoms that could exhibit diffraction on the same surface and discuss whether or not this is a gentler probe. In order to get diffraction, we need a wavelength of 1.6691 ˚ A, as that is the wavelength of the 54 eV electron beam. Using λ = h p and E = p 2 2 m we get E = 1 2 m · ± h λ ² 2 E = 1 2 m He · ³ h λ ´ 2 = 1 2 · 6 . 6423 × 10 - 27 kg · 6 . 626 × 10 - 34 Js 1 . 6691 × 10 - 10 m ! 2 = 1 . 1864 × 10 - 21 J · 1 eV 1 . 602 × 10 - 19 J = . 007406 eV The beam of Helium atoms has a lower energy, and therefore will be a gentler probe of the surface. 2. Consider an electron that is initially far away from a proton and has no kinetic energy.
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Problem Set 1 Solutions - Chemistry 120A, Spring 2009...

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