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The wave fucnction of an electron of a hydrogen atom in the ground state takes the from Psi(r )=Ae^(-r//r_(1)), where A is a certain constant, r_(1) is the first Bohr radius. Find: (a) the most probable distance between the electron and the nucleus. (b) the mean value of modulus of the Coulomb force acting on the electron (c )the mean value of the potential energy of the electron in the field of the nucleus. |
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Answer» Solution :we first find `A` by normalization `1=int_(0)^(oo)4piA^(2)e^(-2r//r_(1))r^(2)dr=(piA^(2))/(2)r_(1)^(3)int_(0)^(oo)e^(-X)DX=piA^(2)r_(1)^(3)` since the intergal has the value 2, Thus `A^(2)=(1)/(pir_(1)^(3)) or A=(1)/(sqrt(r1^(3)pi))` (a) The most PROBABLE distacne `r_(pr)` is that value of `r` for which `P(r )=4pir^(2)|Psi(r )|^(2)=(4)/(r_(1)^(3))r^(2)e^(-2r//r_(1))` is maximum. Thus requires `(p')(r )=(4)/(r_(1)^(3))[2r-(2r^(2))/(r_(1))]e^(-2r//r_(1))=0` or `r=r_(1)=r_(pr)` (b) The coulomb firce being given by `-e^(2)//r^(2)`, the mean value of its modules is `LT F ge int_(0)^(oo)4pir^(2)(1)/(pir_(1)^(3))e^(-2r//r_(1))(e^(2))/(r^(2))dr` `=int_(0)^(oo)(4e^(2))/(r_(1)^(3))e^(-2r//r_(1))dr=(2e^(2))/(r_(1)^(2))int_(0)^(oo)xe^(-x)dx= -(e^(2))/(r_(1))` In `MKS` units we should read `(e^(2)//4PI epsilon_(0))` for `e^(2)` (c ) `lt U geint_(0)^(oo)4pir^(2)(1)/(pir_(1)^(3))e^(-2r//r_(3))(-e^(2))/(r )dr= -(e^(2))/(r_(1))int_(0)^(oo)xe^(-x)dx= -(e^(2))/(r_(1))` In MKS units we should read `(e^(2)//4piepsilon_(0))` for `e^(2)`. |
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