Q.1. What is the De-Broglie wavelength of He atom in a container at room temperature? Use #u_(avg) = sqrt((8k_BT)/(pi m))#

1 Answer
Jul 18, 2017

I got #"0.0794 nm"#, or #"79.4 pm"#, over twice its atomic radius.


Well, assuming that the #"He"# atoms are all moving in the same direction, and follow a Maxwell-Boltzmann distribution:

#f(u) = 4pi (m/(2pik_BT))^(3//2) u^2 e^(-m u^2//2k_BT)#

they thus have the average speed

#u_(avg) = int_(0)^(oo) uf(u)du = [ . . . ] = sqrt((8k_B T)/(pim))#,

where:

  • #k_B = 1.38065 xx 10^(-23)# is the Boltzmann constant in #"J/K"#, or #"kg"cdot"m"^2"/s"^2cdot"K"#.
  • #T# is the temperature in #"K"#.
  • #m# is the per-particle mass in #"kg"#, i.e. #M//N_A = m#, where #M# is the molar mass in #"kg/mol"# and #N_A = 6.0221413 xx 10^(23) "mol"^(-1)#.

We then use this average speed as the velocity #v# in the de Broglie relation

#lambda = h/(mv)#,

where #h = 6.626 xx 10^(-34) "J"cdot"s"#, or #"kg"cdot"m"^2"/s"#, is Planck's constant and #m# is as defined before,

since the particles are assumed to all be moving in the same direction so that the velocity is the speed.

Therefore, the wavelength is:

#color(blue)(lambda) = h/m xx sqrt((pim)/(8k_B T))#

#= sqrt((pih^2)/(8mk_B T))#

#= sqrt((pi(6.626 xx 10^(-34) cancel("kg")cdot"m"^cancel(2)"/"cancel("s"))^2)/(8(0.0040026 cancel("kg")"/"cancel"mol" xx cancel"1 mol"/(6.0221413 xx 10^(23)))(1.38065 xx 10^(-23) cancel("kg")cdotcancel("m"^2)"/"cancel("s"^2)cdotcancel"K")(298.15 cancel("K"))))#

#= 7.938 xx 10^(-11)# #"m"#

#=# #color(blue)("0.0794 nm")#,

which is in the gamma-ray region of the EM spectrum.