Answer
$80.48\;pm$
Work Step by Step
The energy of $\text{Cu}\;K_{\alpha}$ x-ray is
$E=(8.979-0.951)\;keV=8.028\;keV$
Therefore, the wavelengthof $\text{Cu}\;K_{\alpha}$ x-ray is
$\lambda=\frac{ch}{E}$
or, $\lambda=\frac{3\times 10^{8}\times 6.63\times 10^{-34}}{8.028\times10^3\times1.6\times10^{-19}}\;m$
or, $\lambda=1.548\times 10^{-10}\;m$
According to the Bragg’s law for x-ray diffraction i
$2d\sin θ = mλ$
where, $d$ is the spacing between atomic planes.
For first order diffraction $m=1$. Therefore, for first order diffraction, we obtain
$2d\sin θ = λ$
or, $d=\frac{λ}{2\sin θ}$
Substituting $\theta=74.1^{\circ}$ and $\lambda=1.548\times 10^{-10}\;m$, we obtain
$d=\frac{1.548\times 10^{-10}}{2\times\sin 74.1^{\circ}}\;m$
$d\approx8.048\times 10^{-11}\;m$
$d=80.48\;pm$
Therefore, the spacing between the parallel atomic planes is $80.48\;pm$
