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Asked in GSEB Board July 2021 · de Broglie relation and its graphs
Given: m = 0.040 kg, v = 1 km s⁻¹ = 1000 m s⁻¹, h = 6.63×10⁻³⁴ J s.
Idea: de Broglie wavelength λ = h/p = h/(mv).
Momentum p = 0.040×1000 = 40 kg m s⁻¹.
λ = (6.63×10⁻³⁴)/(40) = 1.66×10⁻³⁵ m ≈ 1.7×10⁻³⁵ m
This is far too small to detect, which is why a bullet shows no wave behaviour.
So λ ≈ 1.7×10⁻³⁵ m.
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