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Electron Microscopes

01Electron Microscopes

Resolving power

This section covers why the short de Broglie wavelength of accelerated electrons gives electron microscopes high resolving power and how the required accelerating voltage can be estimated.

Electron wavelength and resolution

Resolving power describes how effectively a microscope can show two nearby details as separate rather than merging them into one feature.

The wavelength used to form an image limits the size of detail that can be resolved. Accelerated electrons can have wavelengths far shorter than those used by an optical microscope, so electron microscopes can distinguish much finer structures.

Increasing accelerating voltage

For an electron accelerated through a potential difference VV:

λ=h2meV\lambda=\frac{h}{\sqrt{2meV}}

Therefore:

  • VV increases
  • electron momentum increases
  • λ\lambda decreases
  • the potential resolving power increases

Atomic-scale wavelengths

Atomic dimensions are of order 10−10 m10^{-10}\ \mathrm m. An accelerating voltage can be estimated by setting the electron wavelength equal to the required atomic-scale distance.

Rearranging:

V=h22meλ2V=\frac{h^2}{2me\lambda^2}

Worked example: Estimate the accelerating potential difference required to give an electron a wavelength of 1.5×10−10 m1.5\times10^{-10}\ \mathrm m.

Use:

V=h22meλ2V=\frac{h^2}{2me\lambda^2}

with h=6.63×10−34 J sh=6.63\times10^{-34}\ \mathrm{J\,s}, m=9.11×10−31 kgm=9.11\times10^{-31}\ \mathrm{kg}, e=1.60×10−19 Ce=1.60\times10^{-19}\ \mathrm C and λ=1.5×10−10 m\lambda=1.5\times10^{-10}\ \mathrm m:

V=(6.63×10−34)22(9.11×10−31)(1.60×10−19)(1.5×10−10)2V= \frac{(6.63\times10^{-34})^2} {2(9.11\times10^{-31})(1.60\times10^{-19})(1.5\times10^{-10})^2}

V≈67.0 VV\approx67.0\ \mathrm V

Exam Tip: If asked to estimate an anode voltage for an atomic-scale wavelength, start from λ=h/2meV\lambda=h/\sqrt{2meV} and rearrange for VV. Remember that λ\lambda is squared in the denominator.

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