Deriving ground state electron energy using Boundary Value

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Homework Statement
Derive the ground state electron energy ##E_1## based on the boundary condition of R(r) -> 0, r -> ∞ and write down the corresponding energy function, Psi. (5 points)
Relevant Equations
Schrodinger's Equation (below)
Schrödinger equation for a hydrogen-like atom: -(h̄²/2m)∇²ψ - (1/(4πε₀))(e²/r)ψ = Eψ, written as typeset mathematics.

This is the equation given.
I attempted to use Radial Equation, obtained from separating variables, to solve for ##E_1##.
Handwritten separation-of-variables equations for the hydrogen atom: r^2 R''(r)/R(r) + 2r R'(r)/R(r) - (E r^2 + r/(4 pi epsilon_0) e^2)(-2m/h^2) = Cr; sin^2 theta Theta''(theta)/Theta(theta) + sin^2 theta cot theta Theta'(theta)/Theta(theta) + Cr sin^2 theta = -C phi; phi''(phi)/phi(phi) = C phi.

Handwritten derivation of the radial hydrogen-atom equation R''(r) with 2r R'(r) terms, a substitution R(r) = A e^(λr), the resulting quadratic in λ, and the stated solution R(n,l)(r) = r^l L e^(-r/na0) that satisfies R(r→∞) = 0.
 
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  • Handwritten separation-of-variables equations for the hydrogen atom: (1) r^2 R''(r)/R(r) + 2r R'(r)/R(r) - (Er^2 + r/(4πε0) e^2)(-2m/h^2) = Cr; (2) sin^2θ Θ''(θ)/Θ(θ) + sin^2θ cotθ Θ'(θ)/Θ(θ) + Cr(sin^2θ) = -Cφ; (3) Φ''(φ)/Φ(φ) = Cφ.
    IMAG1361_2.jpg
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currently said:
Here's the problem. the equation was stated in class to be Schrödinger's Equation.
Last problem should say: n, l, and m are arbitrary constants.

View attachment 255295

We are now in part (d) of your exercise. I can hardly read, let alone quote from your jpg.
What constraints are there on ##l##, and what does that mean when in the ground state ##n=0## ?
 
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BvU said:
What constraints are there on ##l##, and what does that mean when in the ground state ##n=0## ?
Okay, will keep in mind. Thanks!

BvU said:
We are now in part (d) of your exercise. I can hardly read, let alone quote from your jpg.
I'm slightly touched you remember which problem this is.. and I apologize for the terrible image quality! I'll be more quotable next post.
 
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