HW01Solution - 1-1What is the frequency (per second) and...

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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 0.71 (Mo K) and 1.54 (Cu K)?SolutionConverting wavelength to frequency requires the simple relation given in the text, Eq. (1-2) on page 2, and accurate unit conversion (1 = 10-10 m). Photon frequency is related to energy through Plancks constant (h= 6.63x10-34Jsec). Substituting the wavelength values from above yields the answers.E(Mo K) = h= [6.63 10-34Jsec] [4.23 1018sec-1] = 2.8 10-15JE(Cu K) = h= [6.63 10-34Jsec] [1.95 1018sec-1] = 1.29 10-15J(Mo K)=c=3108m/sec.7110-10m=4.231018sec-1(Cu K)=c=3108m/sec1.5410-10m=1.951018sec-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 K, Rh Kand Ag Kradiation. (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 absorbs radiation with wavelength according 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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HW01Solution - 1-1What is the frequency (per second) and...

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