We may calculate de Broglie wavelength as
λ=hp=hmv.\lambda = \dfrac{h}{p} = \dfrac{h}{mv}.λ=ph=mvh. The kinetic energy is, on the one hand, Ek=mv22E_k = \dfrac{mv^2}{2}Ek=2mv2 and, on the other hand, Ek=qeΔφ.E_k = q_e \Delta\varphi.Ek=qeΔφ.
Therefore, λ=hmv=h2qemΔφ=6.6⋅10−34 J⋅s2⋅9.1⋅10−31 kg⋅1.6⋅10−19 C⋅150 V=10−10 m=1\lambda =\dfrac{h}{mv} = \dfrac{h}{\sqrt{2q_em\Delta\varphi}} = \dfrac{6.6\cdot10^{-34}\,\mathrm{J\cdot s}}{\sqrt{2\cdot 9.1\cdot10^{-31}\,\mathrm{kg}\cdot1.6\cdot10^{-19}\,\mathrm{C}\cdot 150\,\mathrm{V}}} = 10^{-10}\,\mathrm{m} = 1λ=mvh=2qemΔφh=2⋅9.1⋅10−31kg⋅1.6⋅10−19C⋅150V6.6⋅10−34J⋅s=10−10m=1 Å.
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