We wish to see inside an atom. Asuming the atom to have a diameter of 100 pm [1 picometer (pm)= $10^{-12}m$], this means that one must be able to resolve a width of, say 10 pm. If an electron microscope is used, the minimum electron energy required is about:

15 keV

As, $λ=\frac{h}{\sqrt{2mK}}$

or $λ^2=\frac{h^2}{2mk}$

$∴K=\frac{h^2}{2mλ^2}=\frac{(6.6×10^{-34})^2}{2×9.1×10^{-31}×(10×10^{-12})^2}J=15.1keV$