Practice portal › Dual Nature of Matter and Radiation › Comparing de Broglie Wavelengths of Particles
Asked in GUJCET 2014 · Same or changed energy
Given: K₂ = 2K₁ for the same electron.
Idea: λ = h/p = h/(√2mK), so λ ∝ 1/(√K).
(λ₂)/(λ₁) = √(K₁)/(K₂) = √1/2 = 1/(√2)
So the new wavelength is 1/(√2) times (about 0.71 times) the initial wavelength.
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