Question

A photon and an electron each have an energy of (6.0 X 10^3 eV). What are their
wavelengths? Which of these would you use to probe atomic structures?

Answer 1

Energy = 6000 eV

for photon

$\lambda=\frac{hc}{E}$ $=\frac{1240}{E}nm=\frac{1240}{6000}nm=0.2067nm$

for electron

$\lambda=\frac{h}{\sqrt{2mE}}=\frac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times6000\times1.6\times10^{-19}}}=\frac{6.63\times10^{-34}}{41.8\times10^{-24}}=1.59\times10^{-9}m=1.59nm$

I would use electron to probe atomic structure