Minumum electron energy and microscopes

[Firecloak]

Homework Statement

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to "see" inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolve a width of, say, 10 pm.

(a) If an electron microscope is used, what minimum electron energy is required?

Homework Equations

E = hc/λ
λ = h/p (DeBroglie wavelength)
p=sqrt(2mK) Momentum
Δλ = (h(1-cos θ))/mc Compton Shift

The Attempt at a Solution

I tried the equation E = hc/λ = (6.63e-34)(3e8)/(1e-12) = 1.24e6 eV, but that isn't the answer.

 

[ideasrule]

E = hc/λ is for photons, not for electrons. What's the equation that gives an object's kinetic energy in terms of its momentum?

 

[Firecloak]

p=sqrt(2mK)

So, do I use that equation in conjunction with DeBroglie's? λ = h/p This one is used for electrons, right?