Find the typical de-Broglie wavelength associated with a H-atom in helium gas at room temperature `(27^(@)C)` and 1atm pressure, and compare it with the mean separation between two atom under these conditions.

Explanation

De Broglie wavelength associated with He atom =`0.7268xx10^(-10) m`
Room temperature , `T=27^@ C`=27+273=300 K
Atmospheric pressure , P=1 atm =`1.01xx10^5` Pa
Atomic weight of a He atom =4
Avogadro's number , `N_A=6.023xx10^(23)`
Boltzmann constant, `k=1.38xx10^(-23) J mol^(-1) K^(-1)`
Average energy of a gas at temperature T, is given as : `E=3/2 kT`
De Broglie wavelength is given by the relation :
`lambda=h/sqrt(2mE)`
Where, m=Mass of a He atom
`=("Atomic weight")/(N_A)`
`=4/(6.023xx10^21)`
`=6.64xx10^(-24) = 6.64xx10^(-27) Kg`
`therefore lambda=h/sqrt(3mkT)`
`=(6.6xx10^(-34))/sqrt(3xx6.64xx10^(-27)xx1.38xx10^(-23)xx300)`
`=0.7268xx10^(-10) m`
We have the ideal gas formula :
PV=RT
PV=kNT
`V/N=(kT)/P`
Where,
V=Volume of the gas
N=Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation :
`r=(V/N)^(1/3) =((kT)/P)^(1/3)`
`=[(1.38xx10^(-23)xx300)/(1.01xx10^5)]^(1/3)`
`=3.35xx10^(-9) m`
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.