Question

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Answer 1

De broglie wavelength, $\lambda = \frac{h}{mv}$, where h is the Planck's constant,  m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom,  $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

   Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm