De Broglie Wavelength Calculation for an Electron with 120 eV Energy

AI Thread Summary
The discussion focuses on calculating the de Broglie wavelength of an electron with 120 eV energy, using the formula lambda = h/p, where p is derived from the kinetic energy. The calculated wavelength is approximately 1.1E-10 m, which matches the expected result. A key point raised is the inapplicability of the E=hf formula in this context, as it requires total energy rather than kinetic energy. The distinction between non-relativistic and relativistic conditions is emphasized, noting that the de Broglie wavelength formula is valid only when the particle's speed is significantly less than the speed of light. Understanding these conditions is crucial for accurate calculations in quantum mechanics.
bennyq
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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
 
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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.
 
ehild said:
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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