Find the de broglie wavelength of hydrogen atom at a temperature 303 K?

I got #"0.145 nm"#.

Well, we assume that for hydrogen atom, the temperature is high enough that the equipartition theorem applies and the average kinetic energy is given by:

#<< kappa >> = N/2 k_BT#

where #N# is the number of degrees of freedom, #k_B = 1.38065 xx 10^(-23) "J/particle"cdot"K"# is the Boltzmann constant, and #T# is the temperature in #"K"#.

#N = 3# because atoms can only fly forward in three dimensions, and not rotate or vibrate. One hydrogen atom would then have a kinetic energy that is its own average, so:

#<< kappa >> = 3/2 cdot 1.38065 xx 10^(-23) "J/atom"cdotcancel"K" cdot 303 cancel"K"#

#= 6.275 xx 10^(-21) "J/atom"#

The kinetic energy of a single atom is also equal to

#<< kappa >>_"single atom" = K = 1/2mv^2 = p^2/(2m)#

where #p# is the forward linear momentum, and #m# is the mass of the atom.

Therefore, its momentum is:

#p = sqrt(2m<< kappa >>)#

#= sqrt(2cdot"0.0010079 kg"//cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23) cancel"particles") cdot 6.275 xx 10^(-21) "kg"cdot"m"^2"/s"^2cdotcancel"atom")#

#= 4.583 xx 10^(-24) "kg"cdot"m/s"#

And now, by the de Broglie relation:

#lambda = h/p#,

where #h = 6.626 xx 10^(-34) "J"cdot"s"# is Planck's constant and #lambda# is the wavelength in #"m"# of a mass-ive particle.

Therefore:

#color(blue)(lambda) = (6.626 xx 10^(-34) cancel"kg"cdot"m"^cancel(2)"/"cancel"s")/(4.583 xx 10^(-24) cancel("kg"cdot"m/s"))#

#= 1.446 xx 10^(-10) "m"#

#=# #color(blue)("0.145 nm")#