Simple De Broglie Wavelength Problem

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SUMMARY

The discussion centers on calculating the de Broglie wavelength of a proton with a kinetic energy of 10 eV using the formula λ = hc / pc. The user initially misapplied the momentum formula pc = √(KE² + 2KEm₀c²), leading to an incorrect wavelength calculation of 12.2 nm. The confusion arose from a misunderstanding of the mass-energy equivalence, specifically mistaking 0.511 MeV for 511 eV. The correct application of these formulas is crucial for accurate results in quantum mechanics.

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vsage
I have the formula for the de Broglie wavelength but for some reason I'm not using it properly. Here's the question:

What is the De Broglie wavelength (in nm) of a proton with a kinetic energy of 10 eV? (Take the mass of a proton to be 938 MeV/c2.)

[tex]\lambda = \frac{hc}{pc}[/tex]

hc = 1239.8 nm*eV
[tex]pc = \sqrt{{KE}^2 + 2KEm_0c^2}[/tex] which should be around 101.5 eV, so the answer would be 12.2nm which is wrong! What did I do wrong? It should have been a simple plug and chug problem after deriving pc in terms of KE and the rest mass.
 
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Nevermind! Apparently in my crazy mind 0.511MeV = 511eV
 
What would you need that number?

Daniel.
 

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