Electrons and Matter Waves — Question 5

Section 38-5 · Physics pages 6340–6341

Question 5 Electron interference can be used to infer accelerating voltage A coherent nonrelativistic electron beam passes through slits separated by d = 1.00 µm. The distant screen is L = 1.00 m away and the fringe spacing is ∆y = .100 mm. Use the small-angle relation ∆y = λL/d; neglect initial kinetic energy. Use hc = 1240 eV nm and mec2 = 511000 eV. P L Schematic; labels, not drawing lengths, specify the data. (a) Infer the wavelength and accelerating potential U. (b) Predict the new fringe spacing if U is increased by a factor of four.
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Question 5 - Solution Electron interference can be used to infer accelerating voltage P L (a) Method and calculation λ = d∆y/L = 1.00 × 10−10 m = .100 nm . eU = (hc)2 2mec2λ2 = 150.450 eV , U = 150.450 V. (b) Interpretation and further calculation λ ∝K−1/2 ∝U−1/2. ∆ynew = ∆y/2 = .0500 mm . Physical check The voltage determines longitudinal momentum; transverse coherence and a stable apparatus are also needed for observable fringes.

Question 5

Electron interference can be used to infer accelerating voltage

A coherent nonrelativistic electron beam passes through slits separated by d=1.00μmd=1.00\,\mu\mathrm m. The distant screen is L=1.00mL=1.00\,\mathrm m away and the fringe spacing is Δy=.100mm\Delta y=.100\,\mathrm{mm}. Use the small-angle relation Δy=λL/d\Delta y=\lambda L/d; neglect initial kinetic energy. Use hc=1240eVnmhc=1240\,\mathrm{eV\,nm} and mec2=511000eVm_ec^2=511000\,\mathrm{eV}.

Diagram shown in the original SVG worksheet.

  1. Infer the wavelength and accelerating potential U.

  2. Predict the new fringe spacing if U is increased by a factor of four.

Question 5 - Solution

Electron interference can be used to infer accelerating voltage

Diagram shown in the original SVG worksheet.

(a) Method and calculation

λ=dΔy/L=1.00×10−10m=.100nm.\boxed{\lambda=d\Delta y/L=1.00\times10^{-10}\,\mathrm m=.100\,\mathrm{nm}}.eU=(hc)22mec2λ2=150.450eV,U=150.450V.\boxed{eU=\frac{(hc)^2}{2m_ec^2\lambda^2}=150.450\,\mathrm{eV}},\quad U=150.450\,\mathrm V.

(b) Interpretation and further calculation

λ∝K−1/2∝U−1/2.\lambda\propto K^{-1/2}\propto U^{-1/2}.Δynew=Δy/2=.0500mm.\boxed{\Delta y_{\mathrm{new}}=\Delta y/2=.0500\,\mathrm{mm}}.

Physical check

The voltage determines longitudinal momentum; transverse coherence and a stable apparatus are also needed for observable fringes.