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

0.15 nm

Step-by-step 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=(0.07)/(2(\cos21^(\circ)-\cos45.6^(\circ)))\\\Rightarrow d=0.14962\ \text{nm}`

The distance between the atomic planes is 0.15 nm.