The de-broglie wavelength associated with particle of mass 10^-6 kg moving with a velocity of 10 m/sec is:

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$6.63 \times 10^{- 29} m$

$6.63 \times 10^{- 21} m$

$6.63 \times 10^{- 7} m$

$6.63 \times 10^{- 16} m$

answer is D.

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Detailed Solution

According to de-Broglie equation
$λ = \frac{h}{mv} = \frac{6 .63 \times 10^{- 34}}{10^{- 6} \times 10} \Rightarrow \frac{6 .63 \times 10^{- 34}}{10^{- 5}} = 6 .63 \times 10^{- 29} m$

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