Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

#### Solution

De Broglie wavelength associated with He atom =`0.7268 xx 10^(-10) m`

Room temperature, *T* = 27°C = 27 + 273 = 300 K

Atmospheric pressure, *P* = 1 atm = 1.01 × 10^{5} Pa

Atomic weight of a He atom = 4

Avogadro’s number, N_{A} = 6.023 × 10^{23}

Boltzmann constant, *k* = 1.38 × 10^{−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/(6.023 xx 10^23)`

`= 6.64 xx 10^(-24) g = 6.64 xx 10^(-27) kg`

`:. lambda = h/sqrt(3mkT)`

`= (6.6 xx 10^(-34))/sqrt(3xx6.64 xx10^(-27)xx 1.38 xx 10^(-23) xx 300)`

`=0.7268 xx 10^(-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)`

`= [(1.38 xx 10^(-23) xx 300)/(1.01 xx 10^5)]^(1/3)`

`= 3.35 xx 10^(-9) m`

Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.