Question

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.

Answer 1

Given `T = 27 + 273 = 300 K, K = 1.38 xx10^(-23) `J/mol/K P =1 atm `= 1.01 xx 10^(5)` Pa
Mass of atom` =("Atomic weight")/("Avagadro number") = 4/(6xx 10^(23)) g = 4/(6xx10^(26))kg`
` lambda - h/(sqrt(3"mKT"))= (6.63 xx10^(-34))/(sqrt(3 xx 4/(6xx10^26)xx 1.38 xx10^(-23) xx300) )= 0.73 xx10^(-10)` m
Now , PV = Rt = KNT, `V/N = (KT)/P`
Mean seperation r = `[(V)/N]^(1/3)= [(KT)/P]^(1//3) = [ (1.38xx10^(-23)xx300)/(1.01 xx10^(5))]^(1//3)`
` r = 3.4 xx 10^(-9) m , lambda /r = (0.73xx10^(-10))/(3.4 xx10^(-9))= 0.021 `
We can see that the wavelength `lambda` with mean seperation r, it can be observed `(r gt gt lambda)` that seperation is larger than wavelength .