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The Debroglie wavelength for helium atom travelling at 1000 m/sec is equal to

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$199.4 \times 10^{- 18} m$

$99 \times 10^{- 6} m$

$199.4 \times 10^{- 12} m$

$99.7 \times 10^{- 12} m$

answer is A.

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Detailed Solution

$λ = \frac{h}{mv}; Mass of helium atom = 4 \times 1.66 \times 10^{- 27} Kg; λ = \frac{6.625 \times 10^{- 34} Kg m^{2} \sec^{- 2} . \sec}{4 \times 1.66 \times 10^{- 27} kg \times 10^{3} m {Sec}^{- 1}}$

$λ = 99.7 \times 10^{- 12} m$

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