The de Broglie wavelength for your hydrogen atom is #"840 pm"#.
To solve for the de Broglie wavelength of the hydrogen atom, use this equation
#p = h/(lamda)#, where
#p# - the momentum of the atom;
#h# - Planck's constant - #6.626 * 10^(-34)"m"^(2)"kg s"^(-1)#
#lamda# - wavelength;
The momentum of your hydrogen atom can be written as
#p = m * v#, where
#m# - the mass of the hydrogen atom - #1.67 * 10^(-27) "kg"#
#v# - the speed of the atom;
Plug this into the first equation to get
#m * v = h/(lamda) => lamda = h/(m * v)#
#lamda = (6.626 * 10^(-34)"m"^(cancel(2))cancel("kg")cancel("s"^(-1)))/(1.67 * 10^(-27)cancel("kg") * 470cancel("m")cancel("s"^(-1)))#
#lamda = 0.00844 * 10^(-7)"m"#
Expressed in picometers and rounded to two sig figs, the number of sig figs given for 470 m/s, the answer will be
#lamda = color(green)("840 pm")#
