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[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 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:

[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^2-m_{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[2eV-m_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

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 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:

[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^2-m_{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[2eV-m_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

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