Interactions of Light and Matter

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SUMMARY

The discussion centers on calculating the kinetic energy of electrons in an electron microscope, specifically those emitted from a tungsten cathode with a work function of 4.5 eV. The de Broglie wavelength of the electron beam is given as 0.040 nm. The kinetic energy equation used is Ek = hf - w, where hf represents the energy of the photon and w is the work function. The calculation reveals confusion regarding the relationship between de Broglie wavelength and the particle's momentum, indicating a need for clarification on these concepts.

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  • Understanding of electron microscopy principles
  • Familiarity with the de Broglie wavelength and its implications
  • Knowledge of the work function in quantum mechanics
  • Basic grasp of kinetic energy calculations in physics
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  • Investigate the role of potential difference in accelerating electrons in electron microscopes
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Homework Statement


The tungsten cathode in an electron microscope has a work function of 4.5 eV and is heated to release electrons that are initially at rest (near the cathode). The electrons are accelerated by a potential difference that creates a beam with a de Broglie wavelength of 0.040 nm.

Calculate the kinetic energy, in Joules, of an electron in the beam.


Homework Equations


Ek = hf - w

F = c / [tex]\lambda[/tex]


The Attempt at a Solution


Ek = hf - w
F = c / [tex]\lambda[/tex]
Therefore, Ek = h x (c/[tex]\lambda[/tex]) - w
Ek = 6.63 x 10^-34 x (3 x 10^8/0.040 x 10^-9) - 4.5
Ek = -4.5?
Feel like I am doing something very wrong here...
 
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How does the de Broglie wavelength relate to the particle's momentum?
 

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