Answer
$1.2\times10^{-10}m$.
Work Step by Step
The spacing between planes, d, for the first-order peaks is given in equation 25–10, $\lambda=2d sin\theta$.
Calculate the wavelength of the electrons from the kinetic energy. The KE is much less than the rest energy of an electron. We may use nonrelativistic relationships. Use equation 27–8.
$$KE=\frac{p^2}{2m}$$
$$p=\sqrt{2m(KE)}$$
$$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$
$$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(72eV)(1.60\times10^{-19}J/eV)}}$$
$$=1.446\times10^{-10}m $$
Now find the spacing.
$$d=\frac{\lambda}{2 sin \theta}=\frac{1.446\times10^{-10}m }{2sin38^{\circ}}=1.2\times10^{-10}m $$