An electron and a photon each have a wavelength of 1.5 Å. Calculate their momenta and energies. (\(h=6.63\times10^{-34}\) J s, \(m_e=9.1\times10^{-31}\) kg, \(c=3\times10^8\) m/s)

Answer: Both have \(p=4.42\times10^{-24}\) kg m/s. Photon: \(E=1.326\times10^{-15}\) J \(\approx8.29\) keV. Electron: \(E=1.07\times10^{-17}\) J \(\approx67.1\) eV.

Momentum. For any particle or photon, \(p=\dfrac{h}{\lambda}\). With \(\lambda=1.5\times10^{-10}\) m:

\[p=\frac{6.63\times10^{-34}}{1.5\times10^{-10}}=4.42\times10^{-24}\ \text{kg m/s}\]

So the electron and the photon have the same momentum.

Energy of photon: \(E=pc=\dfrac{hc}{\lambda}=4.42\times10^{-24}\times3\times10^8=1.326\times10^{-15}\) J

\(=\dfrac{1.326\times10^{-15}}{1.6\times10^{-19}}=8.29\times10^{3}\) eV \(\approx8.29\) keV.

Energy of electron (non-relativistic, kinetic): \(E=\dfrac{p^2}{2m}\)

\[E=\frac{(4.42\times10^{-24})^2}{2\times9.1\times10^{-31}}=\frac{1.954\times10^{-47}}{1.82\times10^{-30}}=1.07\times10^{-17}\ \text{J}=67.1\ \text{eV}\]

The photon carries about 124 times more energy than the electron of the same wavelength. (The electron's speed is \(v=p/m=4.86\times10^6\) m/s, well below \(c\), so the non-relativistic formula is valid.)