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Unformatted text preview: 11What is the frequency (per second) and energy per quantum (in joules) of xray beams of wavelength 0.71Å (Mo Kα) and 1.54Å (Cu Kα)?SolutionConverting wavelength to frequency requires the simple relation given in the text, Eq. (12) on page 2, and accurate unit conversion (1Å = 1010 m). Photon frequency is related to energy through Planck’s constant (h= 6.63x1034J⋅sec). Substituting the wavelength values from above yields the answers.E(Mo Kα) = hν= [6.63 ×1034J⋅sec] [4.23 ×1018sec1] = 2.8 ×1015JE(Cu Kα) = hν= [6.63 ×1034J⋅sec] [1.95 ×1018sec1] = 1.29 ×1015Jν(Mo Kα)=cλ=3×108m/sec.71×1010m=4.23×1018sec1ν(Cu Kα)=cλ=3×108m/sec1.54×1010m=1.95×1018sec1PROBLEM 11B.D. Cullity and S.R. Stock, Elements of XRay Diffraction, 3rdEd., Prentice Hall, (2001)MSE 104 Materials CharacterizationProfessor R. Gronskypage 1 of 116Graphically verify Eq. (113) for a lead absorber and Mo Kα, Rh Kαand Ag Kαradiation. (The mass absorption coefficients of lead for these radiations are 122.8, 84.13 and 66.14 cm2/g, respectively). From the curve, determine the mass absorption coefficient of lead for the shortest wavelength radiation from a tube operated at 30,000 volts.SolutionAn element with atomic number Zand density ρabsorbs radiation with wavelength λaccording to the expression for its mass absorption coefficient, Eq. (113) on page 12, Using the data given above, and the wavelength values given in Appendix 7, the following plot is obtained....
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This note was uploaded on 04/07/2008 for the course MSE 104 taught by Professor Gronsky during the Spring '08 term at Berkeley.
 Spring '08
 GRONSKY

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