An electron and a photon each have a wavelength of 2A°. What are their momentum and energies?

Given: \(\lambda=2\) Å \(=2\times10^{-10}\) m, \(h=6.63\times10^{-34}\) J s, \(m_e=9.1\times10^{-31}\) kg, \(c=3\times10^8\) m/s.

Momentum (same for both, since \(p=h/\lambda\)):

\[ p=\frac{6.63\times10^{-34}}{2\times10^{-10}}=3.315\times10^{-24}\ \text{kg m/s} \]

Photon energy: \(E=pc=\dfrac{hc}{\lambda}=3.315\times10^{-24}\times3\times10^8=9.945\times10^{-16}\) J \(=6.22\times10^3\) eV ≈ 6.2 keV

Electron energy (non-relativistic):

\[ E=\frac{p^2}{2m}=\frac{(3.315\times10^{-24})^2}{2\times9.1\times10^{-31}}=6.04\times10^{-18}\ \text{J}=37.7\ \text{eV} \]

Both: \(p=3.315\times10^{-24}\) kg m/s. Photon: \(E\approx9.95\times10^{-16}\) J (6.2 keV). Electron: \(E\approx6.04\times10^{-18}\) J (37.7 eV).

So at the same wavelength the photon carries about 165 times more energy than the electron.