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Find the mean electrostatic potential produced by an electron in the centre of a hydrogen atom if the electron is in the ground state for which the wave function is Psi(r )=Ae^(-r//r_(1)), where A is a certain constant, r_(1) is the first Bohr radius. |
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Answer» Solution :We find `A` by normalization as above. We get `A=(1)/(sqrt(pir_(1)^(3)))` Then the electronic charge density is `rho=e|Psi|^(2)=-ee^(-2r//r_(1))/(pir_(1)^(3))~=rho (VEC(r ))` The potential `Psi(vec(r ))` due to this charge density is `varphi(vec(r ))=(1)/(4piepsilon_(0)) INT(rho(vecr ))/(|vec(r )-vec(r )|)d^(3)r'` so at the origin `varphi(0)=(1)/(4pi epsilon_(0))int_(0)^(oo)(rho(r'))/(r') 4pi r'^(2) dr'=(-e)/(4pi epsilon_(0))int_(0)^(oo)(4r')/(r_(1)^(3))e^(-2r//r_(1))dr` `=-(e)/(4pi epsilon_(0)r_(1))int_(0)^(oo)xe^(-X)DX= -(e )/((4pi epsilon_(0)r_(1)))` |
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