Question

When to apply relativistic principle to calculate wavelength of electron?

Answer 1

the effect of accelerating voltage on the resolution. Since resolution in the light microscope is limited by the wavelength of visible light, we obtain much higher resolutions in a TEM due to the wave-particle duality of matter that de Broglie found. So, I found a relativistic expression of the electron wavelength online, and it goes like this: λ = √ h 2 c 2 e V ⋅ [ 2 m 0 c 2 + e V ] λ=h2c2eV⋅[2m0c2+eV] I wanted to do a little derivation of this expression. I started from the relativistic energy of a particle as described by Einstein in his special relativity: E = √ p 2 c 2 + m 2 0 c 4 E=p2c2+m02c4 I then substituted p = h λ p=hλ into the equation, and obtained: E = √ h 2 λ 2 c 2 + m 2 0 c 4 E=h2λ2c2+m02c4 Solving for λ λ, I got: λ = √ h 2 c 2 E 2 − m 2 0 c 4 , e q u a t i o n 1 λ=h2c2E2−m02c4,equation1 Next, I use that the kinetic energy of an electron with charge e in a potential V is expressed as: e V = 1 2 m 0 v 2 eV=12m0v2 Multiplying both sides with 2m and taking the root, I obtain m v = √ 2 m o e V = p mv=2moeV=p Substituting the above for E = p c E=pc and squaring yields E 2 = c 2 2 m o e V E2=c22moeV Now I plug this into equation 1, which ultimately gives me λ = √ h 2 m 0 [ 2 e V − m 0 c 2 ] λ=h2m0[2eV−m0c2] Now, this expression is similar, but not identical to the one I stated at the beginning. Where am I going wrong. Or what have I forgotten to do? Could someone help me with this? This derivation is way out of the scope of my TEM exercise, so it is purely out of curiosity. 

Reference https://www.physicsforums.com/threads/relativistic-wavelength-of-electron-in-transmission-electron-microscop.736897/