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Find the energy of an electron fo a hydrogenatom in a stationary state for which the wave function takes the from Psi(r )=A(1+ar)e^(-alpha r), where A, a and alpha are constant. |
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Answer» Solution :The Schrodinger equation for the problem in Gaussian units `grad^(2)Psi+(2m)/(ħ^(2))[E+(e^(2))/(r )]Psi=0` In `MKS` units we should read `(e^(2)//4pi epsilon_(0))` for `e^(2)` we put `Psi=(chi(r ))/(r ).` Then `chi''+(2m)/(ħ^(2))[E+(e^(2))/(r )]xho=0` ....(1) We are given that `chi=rPsi=Ar(1+ar)e^(-ar)` so `chi'=A(1+2ar)e^(-ar)-alphaAr(1+ar)e^(-ar)` `chi''=alpha^(2)Ar(1+ar)e^(-ar)-2alphaA(1+2ar)+2aAe^(-ar)` Substituting in (1) GIVES the condition `alpha^(2)(r+ar^(2))=2alpha(1+2ar)+2a+(2m)/(ħ^(2))=0`....(2) `a alpha^(2)+(2m)/(ħ^(2))Ea=0`....(3) `a^(2)-4a alpha+(2m)/(ħ^(2))(E+e^(2)a)=0` (4) From (3) either `a=0`, In the first case `alpha=(me^(2))/(ħ^(2))=-(ħ^(2))/(2m)alpha^(2)=-(me^(4))/(2ħ^(2))` This state is the GROUND state. `a=alpha-(me^(4))/(8ħ^(2))` and `a= -(1)/(2)(me^(2))/(ħ^(2))` This state is one with `n=2(2s)`. |
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