Q.Calculate the de Broglie wavelength of an electron whose kinetic energy is $150\ \text{eV}$, and state how this wavelength compares with the typical spacing between atoms in a crystal (about $0.1$-$0.3\ \text{nm}$).

Calculation. Since kinetic energy in eV numerically equals the accelerating voltage in volts for an electron starting from rest, $\lambda\approx1.23/\sqrt{V}\ \text{nm}$ applies directly with $V=150$: $\lambda\approx\frac{1.23}{\sqrt{150}}=\frac{1.23}{12.25}\approx0.100\ \text{nm}$ …

Use $\lambda\approx1.23/\sqrt{V}\ \text{nm}$ with the given kinetic energy (in eV) treated as the equivalent accelerating voltage, then c …

• Forgetting that kinetic energy in eV and accelerating voltage in volts are numerically interchangeable for an electron released from rest, and instead trying (unnecessarily) to convert to joules and use $\lambda=h/\sqrt{2mE}$ fro …