De broglie wavelength, [tex]\lambda = \frac{h}{mv}[/tex], where h is the Planck's constant, m is the mass and v is the velocity.
[tex]h = 6.63*10^{-34}[/tex]
Mass of hydrogen atom, [tex]m = 1.67*10^{-27}kg[/tex]
v = 440 m/s
Substituting
Wavelength [tex]\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m[/tex]
[tex]1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm[/tex]
So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm
