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Asked in GUJCET 2025 · de Broglie relation and its graphs
Given: m = 0.033 kg, v = 1 km/s = 1000 m s⁻¹, h = 6.6×10⁻³⁴ J s.
Idea: de Broglie's relation λ = h/p, with p = mv for the bullet.
mv = 0.033×1000 = 33 kg m s⁻¹
λ = (6.6×10⁻³⁴)/(33) = 2×10⁻³⁵ m
The division is exact: 6.6÷33 = 0.2, so λ = 0.2×10⁻³⁴ m with no rounding.
So the bullet's de-Broglie wavelength is 2×10⁻³⁵ m.
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