The de Broglie wavelength is \\lambda \\; =\\; {h}/{p}, but the momentum of a re

Question

The de Broglie wavelength is lambda ; =; {h}/{p}, but the momentum of a relativistic particle is not {mv}. Using the relativistic expressions for kinetic energy and momentum, what is the electron's de Broglie wavelength?

Relativistic: total energy is: E=mc^2+eV, and p=sqrt[(E/c)^2-(mc)^2]

Express your answer using two significant figures.

Explanation / Answer

Answer --> 2.08E+5 m/s How and why Given values: ? = 3.5E-9 m h = 6.62606896E-34 J-s m = 9.10938215E-31 kg Where ? = Wavelength h = Planck's constant m = Mass of the electron Original de Broglie equation ? = h / [mv] Rearrange to solve for v v = h / [m?] Solve: v = (6.62606896E-34 J-s) / [ (9.10938215E-31 kg) * (3.5E-9 m) ] v = (6.63E-34) / [ 3.19E-39 m-kg ] v = 207,826 m/s So 2.08E+5 m/s

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