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Calculate the most probable de Broglie wavelength of hydrogen molecules being in thermodynamics equilibrium at room temperature. |
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Answer» SOLUTION :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` |
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