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

Give, mass of nitrogen molecule,
`m=2xx14.0076u`
`=28.0152xx1.67xx10^(-27)kg`
Temperature , T = 300 K
Kinetic energy is given by
`E_k=1/2mc^2=3/2kT`
`impliesc=sqrt((3kT)/m)and `
The de Broglie wavelength is given by
`lamda=h/p=h/(mc)=h/(sqrt(3mkT))`
`implieslamda=(6.62xx10^(-34))/(sqrt(3xx28.0152xx1.67xx10^(-27)xx1.38xx10^(-23)xx300))`
`=(6.62xx10^(-34))/(241xx10^(-25))`
`=0.0275xx10^(-9)=0.028nm`.