The momentum of a particle having de Broglie wavelength of 0.1 nm is

A
$6.626 \times 10^{- 21} kg m s^{- 1}$
B
$6.626 \times 10^{- 22} kg m s^{- 1}$
C
$6.626 \times 10^{- 23} kg m s^{- 1}$
D
$6.626 \times 10^{- 24} kg m s^{- 1}$

Solution:

$p = \frac{h}{λ} = \frac{6 .626 \times 10^{- 34} Js}{0.1 \times 10^{- 9} m} = 6 .626 \times 10^{- 24} {Jsm}^{- 1} = 6 .626 \times 10^{- 24} kg m s^{- 1}$

Solution:

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