What potential difference is required in an electron microscope to give electrons a wavelength of 0.05 nm ?

Answer 1

For that de Broglie wavelength, Electron momentum = p = h/(5*10^-11 m)
where h is Planck's constant.

Momentum p = 6.62*10^-34/5*10^-11
= 1.32*10^-23 kg*m/s

Energy = p^2/(2m) = e*V
m = electron mass
e = electron charge
Solve for potential difference, V.

V = p^2/(2*m*e)