To find minimum electron energy.. Question of wave optics

Question

We wish to see inside an atom. Assuming the atom to have a diameter of 100pm . 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 

 

1.5keV/ 15keV/ 150keV/ 1.5MeV.

 

which one correct?

Answer 1

λ = h/p ⇒ p = h/λ

 

E = p²/(2m) = (h/λ)²/(2m) = [(4.14 x 10^-15 eV s)/(10 x 10^-12 m)]² / [2(0.511 MeV/c²)]

 

E = (3 x 10⁸ m/s)² (1.714 x 10^-7 eV²s²/m²) / (1.022 x 10⁶ eV) ≅ 15,094 eV =15.094 keV