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

Mass of one atom of helium is
`m=("atomic weight")/("Avogadro's number")=4/(6xx10^(23))g`
`=(4xx10^-3)/(6xx10^(23))kg =2/3xx10^(-26) kg`
T=27+273=300K
Average K.E. of a helium atom at absolute
temperature T is
`1/2mv^2=3/2 kT or mv^2=3kT`
or `mv =sqrt(3mkT)`
de-Broglie wavelength, `lambda=h/(mv)=h/(sqrt(3mkT))`
`=(6.63xx10^(-34))/((3xx((2)/3xx10^(-26))xx1.38xx10^(-23)xx300)^(1//2)`
`=0.73xx10^(-10)m=0.73Å`
According to gas equation for 1 mol of a gas
`PV=RT` or `PV=k NT` `( :' k=(R )/(N))`
or `(V)/(N) = (kT)/(P)`
Mean separation between two atoms of a gas is
`r=((V)/(N))^(1//3)=((kT)/P)^(1//3)`
`=[((1.38xx10^(-23))xx300)/(1.01xx10^5)]^(1//3)=34xx10^(-10)m`
`:. lambda/r=0.73/34 lt 1` or `lambda lt r`.
This shows that the wave packets associated with helium atoms do not overlap. Therefore, the helium atom can be seen separately.