Question

X- rays with a wavelgnth of 0.0700 nm diffract from a crystal. Two adjacne angels of x-ray diffraciton are 45.6 and 21.0 degrees. What is the distance in nm between the atomic planes responsible for the diffraction? (Use 2dcos(θ)=m(λ))

Answer 1

Answer:

0.15 nm

Explanation:

d = Distance between the atomic planes

m = Order

$\theta_{m}$ = First angle = $45.6^{\circ}$

$\theta_{m+1}$ = Adjacent angle = $21^{\circ}$

$\lambda$ = Wavelength = 0.07 nm

From Bragg's relation we know

$2d\cos\theta_{m}=m\lambda$

$2d\cos45.6^{\circ}=m0.07$

$2d\cos\theta_{m+1}=(m+1)\lambda$

$2d\cos21^{\circ}=(m+1)0.07\\\Rightarrow 2d\cos21^{\circ}=m(0.07)+0.07$

So

$2d\cos21^{\circ}=2d\cos45.6^{\circ}+0.07\\\Rightarrow 2d(\cos21^{\circ}-\cos45.6^{\circ})=0.07\\\Rightarrow d=\dfrac{0.07}{2(\cos21^{\circ}-\cos45.6^{\circ})}\\\Rightarrow d=0.14962\ \text{nm}$

The distance between the atomic planes is 0.15 nm.