Answer
a. 6 nm.
b. Diffraction will occur.
Work Step by Step
a. Calculate the kinetic energy of the electrons using equation13–8. At 300K, the electrons are nonrelativistic, so p = mv. Relate the KE to momentum by $KE=(p^2)/(2m)$.
$$KE=\frac{3}{2}kT=\frac{p^2}{2m}$$
$$p=\sqrt{3mkT}$$
Now calculate the wavelength.
$$\lambda=\frac{h}{p}=\frac{h}{\sqrt{3mkT}}$$
$$=\frac{6.626\times10^{-34}J \cdot s}{\sqrt{3(9\times10^{-31}kg)(1.38\times 10^{-23}J/K)(300K)}}$$
$$=6.27\times10^{-9}m\approx 6\;nm$$
b. The wavelength we just calculated is about 20 times the size of the atomic spacing in a semiconductor. At room temperature, electrons will diffract when traveling through the semiconductor lattice.
