Practice portal › Dual Nature of Matter and Radiation › de Broglie Wavelength and Accelerating Potential
Asked in RS Academy GUJCET booklet · Accelerated through a potential difference
Given: kinetic energy K = 10 keV, as if the electron were accelerated through V = 10⁴ V.
Idea: for an electron, λ = h/(√2meV) = (12.27)/(√V) A.
λ = (12.27)/(√10⁴) A = (12.27)/(100) A
λ ≈ 0.123 A
So the wavelength is about 0.12 A.
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