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Q.

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

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a

2 × 10^–10m

b

1 × 10^–15 m

c

10^–5m

d

4 × 10^–10 m

answer is A.

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Detailed Solution

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

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