GATE-2008
Q36.
An X-ray benm of wavelength(λ)
=1.541 A0
is incident on a
cubic crystal having lattice spacing of 4 A0. What will be its 2θ
value (where θ is the glancing angle) on X- ray diffractogram?
cubic crystal having lattice spacing of 4 A0. What will be its 2θ
value (where θ is the glancing angle) on X- ray diffractogram?
A)
11.10 A0
B) 20.10 A0
C) 22.20 A0
D) 22.20 A0
B) 20.10 A0
C) 22.20 A0
D) 22.20 A0
Solution:
Formula:
Bragg’s
law===> 2*d* sinθ
= n*λ-------------------(1)
wavelength
(λ)= 1.541 A0
lattice
spacing (d) = 4 A0
Angle
between the incident angle and the surface of the crystal (θ) = ?
==>
2*d* sinθ= n*λ
==>
sinθ = (n*λ) / (2*d)
==>
sinθ = (1*1.541) / (2* 4)
==>
sinθ = 0.1926
==>
θ= sin-1
(0.1926)
==>
θ= 11.1045
Then,
what is the value of 2θ (given problem)
==>
2θ= 2* 11.1045
==>
2θ=22.209
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