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Relativistic Wavelength of Electron in Transmission Electron Microscop 
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#1
Feb614, 12:33 PM

P: 27

[I could not fit the last "e" into the title]
Hi! I am writing a report on TEM, and was asked to consider 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 waveparticle duality of matter that de Broglie found. So, I found a relativistic expression of the electron wavelength online, and it goes like this: [itex] \lambda = \sqrt{\frac{h^2 c^2}{eV \cdot [2m_0c^2 + eV]}} [/itex] 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: [itex] E = \sqrt{p^2 c^2 + m_{0}^{2} c^4} [/itex] I then substituted [itex] p = \frac{h}{\lambda} [/itex] into the equation, and obtained: [itex] E = \sqrt{\frac{h^2}{\lambda^2}c^2 + m_{0}^{2} c^4} [/itex] Solving for [itex]\lambda[/itex], I got: [itex] \lambda = \sqrt{\frac{h^2 c^2}{E^2m_{0}^{2} c^4}}, equation 1 [/itex] Next, I use that the kinetic energy of an electron with charge e in a potential V is expressed as: [itex] eV = \frac{1}{2}m_0 v^2 [/itex] Multiplying both sides with 2m and taking the root, I obtain [itex] mv = \sqrt{2m_oeV} = p [/itex] Substituting the above for [itex] E = pc [/itex] and squaring yields [itex] E^2 = c^2 2m_o eV [/itex] Now I plug this into equation 1, which ultimately gives me [itex] \lambda = \sqrt{\frac{h^2}{m_0[2eVm_0 c^2]}} [/itex] 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. Am I even going in the correct direction with this derivation? Kind regards, Anders 


#2
Feb614, 01:16 PM

Mentor
P: 11,777

When you accelerate an electron with charge q, from rest through a potential difference V, it ends up with kinetic energy equal to qV. This should give you enough information to complete the derivation. Also, your first equation (the one you found somewhere) is incorrect. The units are inconsistent. In the denominator, m_{0}c^{2} has units of energy (joules), and V has units of volts. You can't add them. Nevertheless, that equation is close to the correct one. If you finish your derivation correctly, you'll see what's missing. There's a clue in my statement about accelerating the electron, three paragraphs above this one. 


#3
Feb614, 01:28 PM

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P: 11,777

Ah, you were working while I was typing...
You shouldn't need to use the velocity v at all in this derivation. 


#4
Feb614, 01:32 PM

P: 27

Relativistic Wavelength of Electron in Transmission Electron Microscop
Thank you! I updated the post just before you answered. But I not sure I have made it. My expression leaves me with a complex wavelength.
edit: Man, it happened again! 


#5
Feb614, 01:33 PM

P: 27

Ah, yes, E = pc is a special case for the einstein relativistic energy. I will give it another try.



#6
Feb614, 01:53 PM

P: 27

Yes, I did it! Starting from scratch with equation 1, which was
[itex] \lambda = \sqrt{\frac{h^2c^2}{E^2  m_{0}^2c^4}} [/itex] You said E represents the total energy, the sum of the kinetic energy and the rest energy. The rest energy is given as the equation already stated at the beginning (relativistic energy of particle). The kinetic energy of a charged particle in a potential is then given as [itex] E_K = \frac{1}{2}mv^2 = eV, equation 2 [/itex] I then say the total energy is [itex] E_{total} =eV + \sqrt{p^2c^2 + m_{0}^2 c^4} [/itex] When I plug this into equation 1, the two terms [itex] \pm m_0^2c^4 [/itex] cancels. This leaves: [itex] \lambda = \sqrt{\frac{h^2c^2}{(eV)^2 + p^2c^2}}, equation 3 [/itex] Again, solving equation 2 for [itex] (mv)^2 = p^2 [/itex] gives [itex] p^2 = 2m_0eV [/itex] which I plug into equation 3. This gives, after some cleaning [itex] \underline{ \lambda = \sqrt{\frac{h^2c^2}{eV[2m_0c^2 + eV]}} } [/itex] Thank you jtbell for your help! It always feels great to derive something when you have spent an hour or more trying! Thanks again! 


#7
Feb614, 02:27 PM

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P: 11,777

##E_{kinetic} = eV##. Now, what is ##E_{rest}##? You did in fact end up with the correct equation, so there's another mistake somewhere along the line that "undid" your first mistake. 


#8
Feb814, 07:26 AM

P: 27

Oh. I thought in the lines of that since the relativistic energy is expressed in terms of the rest mass, that equaled the rest energy, i.e. the energy of the electron at rest.
If I just substitute [itex]E^2[/itex] with [itex]E = \sqrt{p^2c^2+m_0^2c^4}[/itex], I end up with: [itex] \lambda = \sqrt{\frac{h^2c^2}{p^2c^2}} [/itex] The [itex]c[/itex]'s cancel, and I can plug in [itex] p^2 = m_0^2v^2 = 2m_0eV [/itex] to obtain [itex] \lambda = \sqrt{\frac{h^2}{2m_0eV}} [/itex] which is far from the correct expression. But then again, [itex]E = \sqrt{p^2c^2+m_0^2c^4}[/itex] includes the momentum, and so I agree that would be the total energy. I am not sure what to do now, honestly. 


#9
Feb814, 10:15 AM

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P: 11,777

OK... your starting point is
$$\lambda = \sqrt{\frac{h^2c^2}{E^2  m_{0}^2c^4}}$$ E is the total energy, namely the sum of the rest and kinetic energies, so $$\lambda = \sqrt{\frac{h^2c^2}{(E_{rest} + E_{kinetic})^2  m_{0}^2c^4}}$$ You know the kinetic energy is eV. What is the restenergy? Substitute them and simplify... 


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