Q. For a crystal, the angle of diffraction $(2.\theta)$ is $90{^\circ}$and the second order line has a d value of $2.28\ \mathring{\mathrm{A}}.$ The wavelength $(in\ \mathring{\mathrm{A}})\$of X-rays used for Bragg’s diffraction is
A) 1.612
B) 2.00
C) 2.28
D) 4.00
Explanation
(a)
Given, angle of diffraction $(2\theta) = 90{^\circ}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta = 45{^\circ}$
Distance between two planes, $d = 2.28\ \mathring{\mathrm{A}}\$
$\ \ \ \ n = 2$ $\lbrack\because\$ Second order diffraction]
Bragg’s equation is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n\lambda = 2d\sin\theta$
$\ \ \ \ \ \ \ \ \ \ \ \ \ 2 \times \lambda = 2 \times 2.28 \times \sin{45{^\circ}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = 1.612$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.