What is the Angle of Refraction for Electrons in a Metal Surface?

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SUMMARY

The discussion focuses on calculating the angle of refraction for electrons incident on a metal surface, applying Snell's law. The incident energy of the electron beam is 1.6E-18 J, and the energy drop upon entering the metal is 1.9E-18 J. The calculations reveal that if the energy drop exceeds the initial energy, the resulting frequency becomes negative, indicating a fundamental issue in the approach. The participants emphasize the need to correctly interpret the energy drop as potential energy to resolve the calculation discrepancies.

PREREQUISITES
  • Understanding of Snell's law in the context of wave refraction
  • Familiarity with electron wave behavior and potential energy concepts
  • Basic knowledge of quantum mechanics, particularly energy and frequency relationships
  • Ability to perform calculations involving energy, mass, and velocity of particles
NEXT STEPS
  • Study the implications of potential energy on electron behavior in metals
  • Learn about wave-particle duality and its effects on electron waves
  • Explore advanced topics in quantum mechanics, such as the Schrödinger equation
  • Investigate experimental methods for measuring electron refraction angles
USEFUL FOR

Students and researchers in physics, particularly those focusing on quantum mechanics and electron behavior in materials, will benefit from this discussion.

kasse
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Homework Statement



When an electron beam is incident on a metal surface at an angle tetha 1 to the normal, electron waves are refracted into the metal at an angle tetha 2, following Snell's law. The effect on an electron can be represented approximately by a drop in its potential energy.

For a particular metal surface, the size of this drop is 1.9E-18 J. If the energy of the incident electron beam is 1.6E-18 J, and its angle of incidence on the surface is 45 degrees, calculate the wavelengths of the electron waves inside and outside the metal and hence the angle of refraction of the electrons in the metal.

2. The attempt at a solution

E = 1.6E-18 = hf1 --> f1 = 2.42E15 Hz.

E = (1/2)mv1^2 = hf1 --> v = 1.87E6 m/s

lambda 1 = 1.87E6/2.41E15 = 7.8E-10 m

If the energy drop is bigger than the initial energy, the energy in the metal will be negative, and hence the frequency will be negative (impossible!), and I won't be able to calculate lambda 2. What is wrong here?
 
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I think the energy drop refers to potential energy. Perhaps classical conservation-of-energy methods can be used.
 
"If the energy drop is bigger than the initial energy, the energy in the metal will be negative, and hence the frequency will be negative (impossible!),..."
I agree. Are you sure the incident energy is 1.6E-18 J?
 

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