Question

Calculate the most probable de Broglie wavelength of hydrogen molecules being in thermodynamics equilibrium at room temperature.

Answer 1

To find the most proabable de Broglie wavelength of a gas in the thermodynamics equilibrium we determine the distribution is `lambda` corresponding to Maxellian velocity distribution.
It is given by
`Psi(lambda)d lambda=-Phi(v)dV`
(where-sign takes account of the fact that `lambda` decreases as `v` increases). Now
`lambda=(2pi ħ)/(mv) or v=(2pi ħ)/(m lambda)`
`dv=(2pi ħ)/(m lambda^(2))d lambda`
Thus `Psi(lambda)= +AV^(2)e^(-mv^(2)//2kT)(-(dv)/(d lambda))`
`A((2pi ħ)/(m lambda))^(2)((2pi ħ)/(m lambda^(2)))e^(-(m)/(2kT).((2pi ħ)/(m lambda))^(2))`
`=Const. lambda^(-4)e^(-a//lambda^(2))`
where `a=(2pi^(2)ħ^(2))/(mkT)`
This is maximum when
`Psi(lambda)=0 =Psi(lambda)[(-4)/(lambda)+(2a)/(lambda^(3))]`
or `lambda_(pr)=(126)/(sqrt(2))p m= 89.1p m`