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If the Planck's constant h = 6.6 × 10^–34 Js, the de-Broglie’s wave length of a particle having momentum of 3.3 × 10^–24 kg.ms^–1 will be

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$1 \times 10^{- 15} m$

$2 \times 10^{- 10} m$

$10^{- 5} m$

$4 \times 10^{- 10} m$

answer is A.

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Detailed Solution

$λ = \frac{h}{p} = \frac{6.626 \times 10^{- 34}}{3.3 \times 10^{- 24}} = 2 \times 10^{- 10} m$

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