Question

The First order reflection $(n=1)$ from a crystal of the $X$-ray from a copper anode tube $(\lambda =1.54\mathring{A})$ occurs at an angle of ${45}^o$. What is the distance between the set of plane causing the diffraction?

• A. $0.1089nm$
• B. $0.1089m$
• C. $10.89\mathring{A}$
• D. $1.089\times {10}^{-9}m$

Answer 1

The correct option is A $0.1089nm$

Given, $n=1$
$\lambda =1.54\mathring{A}$
$Q={45}^o$
$d=$?

From $n\lambda =2d\sin \theta$

$1\times 1.54=2\times d\times \sin {45}^o$

$1.54=2\times d\times 0.7071$

$d=\frac{1.54}{2\times 0.7071}$

$=\frac{1.54}{1.414}$

$d=1.089\mathring{A}$ $(\because 1\mathring{A}={10}^{-10}m)$

$d=1.089\times {10}^{-10}m$

$d=0.1089\times {10}^{-9}m$

$d=0.1089nm$