PROBLEM 43 &amp; 44 May 19 2010

PROBLEM 43 &amp; 44 May 19 2010 - wavelength of the...

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Solving Problems 43 and 44. Physics 002 C May 19, 2010 Problem 43. (Like Problem 28.22) Given a beam of 75 eV electrons normally incident on a Ni(1,0,0) surface where the line spacing is 0.249 nm, what is the minimum diffraction angle? (a) 14 o ; (b) 75 o ; (c) 65 o ; (d) 22 o ; (e) 34 o . What do you know? nm 1.221 / p h for a 1 eV electron so a 75 eV electron has a wavelength equal to nm 141 . 0 75 nm/ 1.221 . With a normally incident beam, the path length difference between neighboring diffracting rays is sin d . In lowest order sin d for constructive interference. Thus d / sin =0.141 nm/0.249 = 0.566 and = 34.5 o . Problem 44. (Like problem 28.28) The spatial resolution of a microscope is approximately the
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Unformatted text preview: wavelength of the illumination times the f-number of the objective ( f-number = focal length divided by the objective diameter). What is the theoretical resolution of an electron microscope with f-number = 32 that uses 100 keV electrons? Use the result nm 1.221 / p h for a 1 eV electron and forget the relativistic correction. (a) 0.12 nm; (b) 0.34 nm; (c) 0.56 nm; (d) 0.78 nm; (e) 0.90 nm. What do you know? nm 1.221 / p h for a 1 eV electron so a 100,000 eV electron has a wavelength equal to nm 00386 . 100000 nm/ 1.221 . Multiply by the f-number to get a resolution of 0.123 nm....
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This note was uploaded on 09/23/2010 for the course PH 02c taught by Professor Mile during the Spring '04 term at Riverside Community College.

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