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The de-Broglie wave length associated with a particle of mass 1 mg moving with a velocity of 1 m/sec is

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$6.626 \times 10^{- 29} m$

$6.626 \times 10^{- 22} m$

$6.626 \times 10^{- 28} m$

$6.626 \times 10^{- 31} m$

answer is A.

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Detailed Solution

$λ = \frac{h}{mv} = \frac{6.626 \times 10^{- 34} kg . m^{2} . s^{- 1}}{1 mg \times \frac{10^{- 6} kg}{1 mg} \times 10 m / s} = 6.626 \times 10^{- 29} m$

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