Practice portal › Dual Nature of Matter and Radiation › Comparing de Broglie Wavelengths of Particles
Asked in GUJCET 2011 · Same or changed energy
Given: Kₑ = Kₚ = K.
Idea: p = √2mK, so λ = h/(√2mK) ∝ 1/(√m) at fixed K.
(λₑ)/(λₚ) = √(2mₚK)/(2mₑK) = √(mₚ)/(mₑ)
With mₚ ≈ 1836 mₑ this is about 43: the electron's wavelength is much longer.
So the ratio is √(mₚ)/(mₑ).
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