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Determine the expected diffraction angle for the first-order reflection from the (113) set of planes for FCC platinum when monochromatic radiation of wavelength 0.1542 nm is used.

Determine the expected diffraction angle for the first-order reflection from the (113) set of planes for FCC platinum when monochromatic radiation of wavelength 0.1542 nm is used.

Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter7: Analysis Of Stress And Strain
Section: Chapter Questions
Problem 7.7.14P: Solve the preceding problem for the following data: x=1120106,y=430106,xy=780106,and=45.
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how do u know to use molybdenum?

Determine the expected diffraction angle for the first-order reflection from the (113) set of planes for FCC platinum when monochromatic radiation of wavelength 0.1542 nm is used. Step-by-step solution Step 1 of 3 Given: Plane: (113) Material: FCC Platinum 2=0.1542 nm n = 1 According to 's law, 2=2d sin 0 --(1) In case of Cubic unit cell, inter planar spacing is given by a d = --(2) Step 2 of 3 From table 3.1, the atomic radius, R for Molybdenum is R = 0.1387 nm For FCC, a=2R/2 =2x0.1387× /2 =0.3923 nm From equation (2), the inter planar spacing for (111) set of planes is 0.3923 d =- Vi? +1² +3? 0.3923 V11 =0.1183 nm
Transcribed Image Text:Determine the expected diffraction angle for the first-order reflection from the (113) set of planes for FCC platinum when monochromatic radiation of wavelength 0.1542 nm is used. Step-by-step solution Step 1 of 3 Given: Plane: (113) Material: FCC Platinum 2=0.1542 nm n = 1 According to 's law, 2=2d sin 0 --(1) In case of Cubic unit cell, inter planar spacing is given by a d = --(2) Step 2 of 3 From table 3.1, the atomic radius, R for Molybdenum is R = 0.1387 nm For FCC, a=2R/2 =2x0.1387× /2 =0.3923 nm From equation (2), the inter planar spacing for (111) set of planes is 0.3923 d =- Vi? +1² +3? 0.3923 V11 =0.1183 nm
From equation (1), 0.1542=2x0.1183sin 0 sin 0=0.6517 0= 40.69° .: 20 =81.38°
Transcribed Image Text:From equation (1), 0.1542=2x0.1183sin 0 sin 0=0.6517 0= 40.69° .: 20 =81.38°
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