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 helium atom is given by
`m_(He)=("Atomic mass")/("Avogadronumber")`
`=(4xx10^(-3))/(6xx10^(23))kg=6.67xx10^(-27)kg`
The de Broglie wavelength is given by
`implies lamda=(6.62xx10^(-34))/(sqrt(3xx6.67xx10^(-27)xx1.38xx10^(-23)xx3000))`
`=0.73xx10^(-10)m`
Mean separation ,
`r_0=(V/N)^(1/3)`
For T = 300K,
`P=1/3m/Vc^2=2/(3V)xx1/2mc^2`
or `P=2/(3V)xx3/2RT=(RT)/(V)=(kNT)/V`
`V=(kNT)/P`
`:.r_0=((kT)/P)^(1//2)`
`=((1.38xx10^(-25)xx300)/(1.01xx10^5))^(1/3)`
`r_0=3.4xx10^(-9)m`.
We find `r_0 gt gt lamda`
So , the mean separation is much greater than the Broglie wavelength.