# 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

I got

Well, assuming that the

#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 theper-particlemass 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

#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

isthe 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.