Potential Difference in and Electron/de Broglie Equation

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SUMMARY

The discussion centers on calculating the potential difference required to accelerate an electron from rest so that its de Broglie wavelength is 0.1 nm. The initial calculations used the equations E = hf and E = hλ/c, leading to an incorrect energy value of 2.21e-58 J. After recalculating and converting the energy into electron volts, the correct potential difference determined is 150V. This highlights the importance of unit conversion in physics calculations.

PREREQUISITES
  • Understanding of the de Broglie wavelength concept
  • Familiarity with the equations E = hf and E = hλ/c
  • Knowledge of electron volts as a unit of energy
  • Basic principles of electron acceleration in electric fields
NEXT STEPS
  • Study the relationship between energy and potential difference in electric fields
  • Learn about the de Broglie wavelength and its implications in quantum mechanics
  • Explore the concept of electron volts and their application in particle physics
  • Review common mistakes in unit conversions in physics calculations
USEFUL FOR

Students studying physics, particularly those focusing on quantum mechanics and electromagnetism, as well as educators looking for practical examples of energy calculations involving electrons.

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Homework Statement



Through what potential difference should you accelerate an electron from rest, such that it's de Broglie wavelength will be 0.1 nm.

Homework Equations



E = hf
λf = c

The Attempt at a Solution



My Thinking was;

E = hf
E = hλ/c
E= 6.626e-34 * 1e-10/3e8

which turned out 2.21e-58 J, which I'm pretty sure is wrong.

Anyone with some ideas or a worked solution would be fantastic :D

Regards all,

Will.
 
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You're right so far, though you should re-check your final calculation, since you're off by a couple orders of 10. So now you need to figure out what voltage difference is going to supply the energy you calculated. I would recommend converting the energy you calculate into electron volts, since those will make the voltage calculation easier.
 
Solved, 150V, thanks jd :)

Will.
 

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