Electrons and Matter Waves — Question 1

Question 1

Accelerating voltage sets an electron wavelength with a controlled approximation

An electron starts with negligible kinetic energy and gains 150 eV through an accelerating potential. The exact relation is $pc=\sqrt{K(K+2m_ec^2)}$. Use $hc=1240\,\mathrm{eV\,nm}$ and $m_ec^2=511000\,\mathrm{eV}$.

1. Find the exact de Broglie wavelength.

2. Find its nonrelativistic estimate and the fractional overestimate of that estimate.

Question 1 - Solution

Accelerating voltage sets an electron wavelength with a controlled approximation

(a) Method and calculation

$$\lambda=\frac hp=\frac{hc}{\sqrt{K(K+2m_ec^2)}}.$$$$\boxed{\lambda=\frac{1240}{\sqrt{150(150+1022000)}}\,\mathrm{nm}=.100143\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$\lambda_{\mathrm{nr}}=\frac{hc}{\sqrt{2m_ec^2K}}=.100150\,\mathrm{nm}.$$$$\boxed{\frac{\lambda_{\mathrm{nr}}}\lambda-1=\sqrt{1+\frac K{2m_ec^2}}-1=7.33828\times10^{-5}}.$$

Physical check

The nonrelativistic expression slightly overestimates the wavelength because it underestimates momentum at fixed kinetic energy.