What is the de-Broglie wavelength of nitrogen molecule in air at 300K ? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen `=14.0076U`) Plank's constant=`6.63xx10^(-34)Js` , Boltzmann constant `=1.38xx10^(-23)JK^(-1)`

Explanation

Temperature of the nitrogen molecule, T=300 K
Atomic mass of nitrogen = 14.0076 u
Hence , mass of the nitrogen molecule, m=2 x 14.0076 = 28.0152 u
But 1 u =`1.66xx10^(-27)` Kg
`therefore m=28.0152xx1.66xx10^(-27)` kg
Planck's constant, `h=6.63xx10^(-34) Js`
Boltzmann constant, `k=1.38xx10^(-23) J K^(-1)`
We have the expression that relates mean kinetic energy `(3/2kT)` of the nitrogen molecule with the root mean square speed `(v_(rms))` as :
`1/2mv_(rms)^2 =3/2 kT`
`V_(rms)=sqrt((3kT)/m)` Hence, the de broglie wavelength of the nitrogen molecule is given as :
`lambda=h/(mv_(rms))=h/sqrt(3mkT)`
`=(6.63xx10^(-34))/sqrt(3xx28.0152xx1.66xx10^(-27)xx1.38xx10^(-23)xx300)`
`=0.028xx10^(-9) m`
`=0.028 nm`
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.