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 whenmonochromatic radiation of wavelength 0.1542 nm is used.Step-by-step solutionStep 1 of 3Given:Plane: (113)Material: FCC Platinum2=0.1542 nmn = 1According to 's law,2=2d sin 0--(1)In case of Cubic unit cell, inter planar spacing is given byad =--(2)Step 2 of 3From table 3.1, the atomic radius, R for Molybdenum isR = 0.1387 nmFor FCC,a=2R/2=2x0.1387× /2=0.3923 nmFrom equation (2), the inter planar spacing for (111) set of planes is0.3923d =-Vi? +1² +3?0.3923V11=0.1183 nm From equation (1),0.1542=2x0.1183sin 0sin 0=0.65170= 40.69°.: 20 =81.38°

In this problem, the Molybdenum is used as an FCC crystal but it has a BCC structure. It is wrongly…

Answered: Determine the expected diffraction…

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.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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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

From equation (1),
0.1542=2x0.1183sin 0
sin 0=0.6517
0= 40.69°
.: 20 =81.38°

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