Quantum Physics: Electron Microscope

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SUMMARY

The discussion focuses on calculating the smallest object size that can be studied using an electron microscope operating at an accelerating voltage of 15 kV, with a lens diameter of 0.5 cm and a focal length of 50 cm. The resolution requirement is set to one-tenth the size of the object. The energy of electrons is derived from the equation E = qV, while the energy of photons with the same wavelength is calculated using E = hc/λ. The momentum of photons is determined using p = h/λ. The formula for the smallest object size is given by 1.22 * (f * λ / D).

PREREQUISITES
  • Understanding of electron microscopy principles
  • Familiarity with quantum physics equations, specifically E = hc/λ and p = h/λ
  • Knowledge of the relationship between energy, voltage, and wavelength in electron beams
  • Basic grasp of momentum concepts in quantum mechanics
NEXT STEPS
  • Calculate the wavelength of electrons using the equation λ = h/p
  • Explore the implications of accelerating voltage on electron energy in electron microscopy
  • Investigate the relationship between photon and electron energies for different wavelengths
  • Study the effects of lens diameter and focal length on resolution in electron microscopy
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Students and professionals in physics, particularly those studying quantum mechanics and electron microscopy, as well as researchers focused on materials science and nanotechnology.

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Quantum Physics: Electron Microscope

Homework Statement


Consider electrons in the accelerating voltage of 15 kV used in an electron microscope with a 'lens' diameter D = 0.5 cm and focal length f= 50 cm.
a) We require that the resolution is a tenth of the size of the object we want to study. What is the size of the smallest object that we can study with this electron microscope ?

b) Electrons in this microscope have wavelength λe. If we had photons with the same wavelength, λγ = λe, how would their energy compare to the energy of the electrons in the microscope ?

c) What is the momentum of photons with λγ = λe ?

Homework Equations



E = hc/λ
p = h/λ
There are definitely others that are needed but I don't know which ones.

3. Attempt
I don't even know how to begin with part a, so if anyone can give me some advice that would be great. You don't have to do the work for me, just tell me what to do and why.
For part b, I think the energy of the photon is hc/λγ, while the energy of the electron can be found through the equation E = h2/(2mλe2) so the result will depend on the wavelength, which I don't know.
Finally, for part c, I know that the momentum is p = h/λγ.
 
Last edited:
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The smallest object you can study focal distance away is given by 1.22 \frac{f \lambda}{D} where D is the diameter of the lens

the energy of the electron is given by qV and you should be able to work out the wavelength from that and consequently the rest of the question
 

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