The binding energies of K-shell and L-shell electrons in copper are 8.979 and 0.951 keV, respectively. If a K_\alpha x ray from copper is incident on a sodium chloride crystal and gives a first-order Bragg reflection at an angle of 74.1° measured relative to parallel planes of sodium atoms, what is the spacing between these parallel planes?
We use Eq. 36-31, Eq. 39-6, and hc = 1240 eV·nm = 1240 keV·pm. Letting 2d sinθ = mλ = mhc / ΔE , where θ = 74.1°, we solve for d:
R=\frac{\lambda_{ avg }}{\Delta \lambda} \quad \text { (resolving power defined). } (36-31)
h f=\frac{h c}{\lambda}=\Delta E=E_{ high }-E_{ low } (39-6)
d=\frac{m h c}{2 \Delta E \sin \theta}=\frac{(1)(1240 \,keV \cdot nm )}{2(8.979\,keV -0.951 \,keV )\left(\sin 74.1^{\circ}\right)}=80.3 \,pm