Solving the Uncertainty Principle Homework

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SUMMARY

The discussion focuses on solving a homework problem related to the Uncertainty Principle in quantum mechanics. The participant initially calculated momentum (P) using a wavelength of 0.1 femtometer and derived an energy (E) of 1.5e14 eV. However, the expected answer was 12 GeV. The resolution involved recognizing the necessity of using the relativistic energy equation, E² = m₀²c⁴ + p²c², instead of the non-relativistic equation E = p²/2m, due to the high energy involved.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically the Uncertainty Principle.
  • Familiarity with momentum and energy equations in physics.
  • Knowledge of relativistic physics and its implications on energy calculations.
  • Ability to perform calculations involving femtometer-scale measurements.
NEXT STEPS
  • Study the derivation and applications of the relativistic energy equation E² = m₀²c⁴ + p²c².
  • Learn about the implications of relativistic effects in high-energy physics.
  • Explore the concept of momentum in quantum mechanics and its relation to wavelength.
  • Review examples of energy calculations in particle physics, particularly at femtometer scales.
USEFUL FOR

Students of physics, particularly those studying quantum mechanics and high-energy particle physics, as well as educators seeking to clarify the application of relativistic equations in problem-solving.

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Homework Statement


http://img340.imageshack.us/img340/3249/question.jpg

Homework Equations


P=h/lambda
E=p^2/2m


The Attempt at a Solution



Well, I calculated P using a wavelength of .1 femtometer. Then I plugged it into the energy equation. However, I got 1.5e14 eV, while the answer sheet says it's supposed to be 12 GeV. Have I done something wrong wrong? On the answer sheet my professor calculated P, and I got the same answer he got, but after that he just says therefore E=12GeV, not showing how he came to that conclusion.

Thanks for any help!
 
Last edited by a moderator:
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You cannot neglect relativistic effects at such high energies, thus the equation
[tex]E = \frac{p^{2}}{2m}[/tex] does not hold.
Instead you should use the relativistic equation
[tex]E^{2} = m_{0}^{2} c^{4} + p^{2} c^{2}[/tex]
 
Ahh that makes sense and gives e the right answer. Thanks!
 

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