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Unformatted text preview: 1-1What is the frequency (per second) and energy per quantum (in joules) of x-ray 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. (1-2) on page 2, and accurate unit conversion (1Å = 10-10 m). Photon frequency is related to energy through Planck’s constant (h= 6.63x10-34J⋅sec). Substituting the wavelength values from above yields the answers.E(Mo Kα) = hν= [6.63 ×10-34J⋅sec] [4.23 ×1018sec-1] = 2.8 ×10-15JE(Cu Kα) = hν= [6.63 ×10-34J⋅sec] [1.95 ×1018sec-1] = 1.29 ×10-15Jν(Mo Kα)=cλ=3×108m/sec.71×10-10m=4.23×1018sec-1ν(Cu Kα)=cλ=3×108m/sec1.54×10-10m=1.95×1018sec-1PROBLEM 1-1B.D. Cullity and S.R. Stock, Elements of X-Ray Diffraction, 3rdEd., Prentice Hall, (2001)MSE 104 Materials CharacterizationProfessor R. Gronskypage 1 of 11-6Graphically verify Eq. (1-13) 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. (1-13) 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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- Spring '08
- Electron, Professor R. Gronsky, Materials Characterization Professor, Characterization Professor R.