In the case of hydrogen atom the potential energy is given by the attraction between the central molecules and the revolving negatively charged electron.
The electrostatic attraction between the electron and nucleus is given by
V = - Ze²
r
Here,
V = potential energy
Z = atomic number
e = charge
r = radius distance between the nucleus and electron.
For the H atom, since the atomic number = 1 ( z = 1)
V = − e²
r
In this problem, since there are no neutrons in the nucleus, the reduced mass is calculated as follows :
μ = me mp
me + mp
The fundamental schrodinger equation is,
∆²ψ + 8π²μ ( E - V ) ψ = 0
h²
V = e² m = μ
r
∆² ψ + 8π²μ ( E + e² ) ψ = 0
h² r
The laplacian operator should be expressed in terms of polar coordinates.
∆² = ∂² + 2 ∂ + 1 [ sinθ ∂ψ ] +
∂r² r ∂r r²sinθ ∂θ
1 ∂²
r² sin²θ ∂φ²
∴ ∂²ψ + 2 ( ∂ψ) + 1 ( sinθ ∂ψ) +
∂r² r ∂r r²sin θ ∂θ
1 ∂²ψ + 8π²μ ( E + e² ) ψ = 0
r²sin²θ ∂φ² h² r²
Above the equation is Schrodinger equation for hydrogen atom.
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