The de Broglie wavelength of 1 mg grain of sand blown by a 20 m/s wind (approxiamately) is:
A
4.4×10−29m
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B
2.2×10−29m
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C
1.1×10−29m
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D
3.3×10−29m
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Solution
The correct option is D3.3×10−29m Given : Planck's constant(h)=6.63×10−34Jsmass of sand grain(m)=10−6kgv=20m/sec de Broglie wavelength(λ)=hmv=6.63×10−3410−6×20=3.315×10−29m