Find the de Broglie wavelength of hydrogen molecules, which corresponds to their most probable velocity at room temperature.

Explanation

In thermodynamics equilibrium, Maxwell's velocity distribution law holds :
`dN=(v)=Phi (v)dv=AV^(2)e^(-mv^(2)//2kT)dv`
`Phi(v)` is maximum when
`Phi(v)=Phi(v)[(2)/(v)-(mv)/(kT)]=0`
The difines the most prabable velocity.
`V_(pr)=sqrt((2kT)/(m))`
The de Broglie wavelength of `H` molecule with the most probable velocity is
`lambda=(2pi ħ)/(mv_(pr))=(2pi ħ)/(sqrt(2mkT))`
Substituting the appropriate value especiaaly
`m=m_(H_(2)=2m_(H)),T=300K`, we get
`lambda=126p m`