What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K?Assume that the molecule is moving with the root-mean square speed of molecules at this this temperature.(Atomic mass of nitrogen-14.0076 u)

Explanation

Mass of nitrogen atom,
`m=2xx14xx1.66xx10^(-27)=46.48xx10^(-27)kg`
T=300 K
`K_(B)`=Boltzmaan.s constant
`=1.38xx10^(-23)Jmol^(-1)K^(-1)`
`h=6.63xx10^(-34)Js`
`implies` Average kinetic energy
`lt(1)/(2)mv^(2)gt=(3)/(2)k_(B)T`
`therefore v=sqrt((3K_(B)T)/(m))`
`therefore` Momentum p=mv=`sqrt(3mk_(g)T)`
`therefore` de-Broglie wavelength,
`lambda=(h)/(p)`
`therefore lambda=(h)/(sqrt(3mk_(B)T))`
`=(6.63xx10^(-34))/(sqrt(3xx46.49xx10^(-27)xx1.38xx10^(-23)xx300))`
`=(6.63xx10^(-34))/(sqrt(57728.16xx10^(-50)))`
`=(6.63xx10^(-34))/(240.266xx10^(-25))`
`=0.02759xx10^(-9)m`
`~~0.028nm`