Wave-Particle Duality

  • Thread starter bennyq
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  • #1
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Homework Statement


Calculate the de Broglie wavelengths of an electron with energy 120 eV ...


Homework Equations


lambda = h\p where p = sqrt(2*Me*E)

The Attempt at a Solution


E=1.6E-19*120ev..
Then sub into equation and I get 1.1E-10m for the wavelength, which is the answer quoted.

The question that concerns me is why can you not use E=hf, where rearranged gives lambda=(hc/E) which gives
a different answer...
Thanks
 

Answers and Replies

  • #2
ehild
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E=hf is valid if E is the total energy, (E=mc2). The formula p = sqrt(2*Me*E) is valid for speeds much less than c, and E means the kinetic energy.
 
  • #3
Orodruin
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E=hf is valid if E is the total energy, (E=mc2).

Note that you can only use this for the de Broglie wavelength the way done in the OP if the particle is relativistic and thus has momentum essentially equal to its energy. In other words, when the velocity is close to c - otherwise the relation between wavelength and frequency changes.
 

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