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Understanding the Hydrogen Atom: Structure and Energy Levels

Explore the structure of the hydrogen atom and its energy levels through an analysis of its angular and radial nodes. Discover the orbital shapes and corresponding energy values.

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Understanding the Hydrogen Atom: Structure and Energy Levels

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  1. What does an atom look like? • http://www.biologie.uni-hamburg.de/b-online/e16/hydrogen.htm

  2. ˆ L2 Hydrogen Atom(Variable separation in radial coords) -(ħ22/2m)F = -ħ2/2m[1/r.∂2(rF)/∂r2] + +1/2mr2[-ħ2{1/sinq.∂/∂q(sinq∂F/∂q) + 1/sin2q.∂2F/∂j2}]

  3. What does L look like? L = r x p Uncertainty prevents complete specification of Lx, Ly, Lz [x, px] = iħ and so on… (easy to prove from definition of p) This gives us [Lx, Ly] = iħLz and so on… Can choose only one (say, Lz) and also it turns out, L2

  4. So what’s the best we can do? So work only with L2= -ħ2{1/sinq.∂/∂q(sinq∂/∂q) + 1/sin2q.∂2/∂j2} and Lz= -iħ(x∂/∂y – y∂/∂x) = -iħ∂/∂j (like linear momentum in azimuthal coords)

  5. The components of L and their eigenstates Ylm(q,j): eigenstates of L2 and Lz with eigen-indices l, m

  6. Eigenstates of Lz Lz = -iħ∂/∂j LzY = -iħ∂/∂jY = CħY Solution: Y = f(q)eiCj Boundary condition: Y(j=0) = Y(j=2p) • C = integer m • Ylm = Plm(q)eimj

  7. Range of values for m? L2 = Lx2 + Ly2 + Lz2 Max Lz = l, minimum = -l, with ‘graininess’ ħ ie, Lz = mħ, m = -l, -(l-1), -(l-2), …, (l-2), (l-1), l (2l+1) possibilities

  8. Range of values for L2? L2 = Lx2 + Ly2 + Lz2 Naively, we expect L2 = ħ2l2 But the commutator (ie, ordering need) for Lx and Ly in terms of ħLz makes things a little more complex, so that we get an extra ħl L2 = l(l+1)ħ2

  9. Angular momentum quantization ˆ ˆ L2 L2 Lz Lz z-component Quantized mħ m = -l, -(l-1),... (l-1), l Angular Mom Quantized l(l+1)ħ2 l = 0, 1, 2,.. (n-1)

  10. Hydrogen Atom F = Rnl(r)/r x Ylm(q,j) n: Principal Quantum Number (Size, r) l: Azimuthal Quantum Number (Angle q) m: Magnetic Quantum Number (Angles q, j) s: Spin Quantum Number (up, down)

  11. Hydrogen Atom F = Rnl(r)/r x Ylm(q,j) For given n, l = 0, 1, 2, …, n-1 (called s, p, d, f, …) m = -l, -(l-1), -(l-2), … ,l-2, l-1, l (px, py, pz, dxy, dyz, dxz… “Orbitals”) 2l+1 of multiplets

  12. Angular space (Orbitals) Check  they satisfy L2Ylm = l(l+1)ħ2Ylm Y00(q,j) Indep. of angle (l=0) L2 = -ħ2[1/sinq.∂/∂q(sinq.∂/dq)+ 1/sin2q.∂2/∂j2]

  13. Angular space (Orbitals) Px = sinqcosj Py = sinqsinj Pz = cosq Y11(q,j) + Y1,-1(q,j) Y11(q,j) - Y1,-1(q,j) Y10(q,j) Check  they satisfy L2Ylm = l(l+1)ħ2Ylm LzYlm = mħYlm E.g. L2cosq = 2ħ2cosq 1st order polynomials L2 = -ħ2[1/sinq.∂/∂q(sinq.∂/dq)+ 1/sin2q.∂2/∂j2], Lz = -iħ∂/∂j l = 1 (“p” Orbitals), m=-1, 0, 1

  14. d orbitals (l = 2) z z z z z d3z2-r2 = 3cos2q-1 dyz = sinjsin2q dx2-y2 = cos2jsin2q dxz = cosjsin2q dxy = sin2jsin2q y y y y y x x x x x Y22(q,j) + Y22(q,j) - Y21(q,j) + Y21(q,j) - Y2,-2(q,j) Y20(q,j) Y2,-1(q,j) Y2,-1(q,j) Y2,-2(q,j) 2nd order polynomials

  15. Like the modes of a drum 1s 2s 2px 3dxy

  16. And like particle in a 2D (q,j) box Note that particle is ‘free’ in angular direction, as U is independent of angle f11 (“1s”) f12 (“2px”) f22 (“3dxy”) f21 (“2py”) 16

  17. What about radial part? F = Rnl(r)/r x Ylm(q,j) n: Principal Quantum Number (Size, r) l: Azimuthal Quantum Number (Angle q) m: Magnetic Quantum Number (Angles q, j) s: Spin Quantum Number (up, down)

  18. Normalizing the wavefunction dV F = Rnl(r)/r x Ylm(q,j) ∫F2.r2drdW = 1 dW = sinqdqdj ∫Y2dW = 1 ∫(R/r)2.r2dr = 1

  19. What about radial part? Ueff(r) Centrifugal Barrier Uc i.e, we use F = R(r)Ylm(q,j)/r Then -ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R= ER Looks like 1D. Then why R/r? L = mvr (constant) Fc = mv2/r = L2/mr3 Uc = -∫Fdr = L2/2mr2 L2 L2  l(l+1)ħ2 Normalization over r2dr ∫(R/r)2.r2dr = 1 • ∫R2.dr = 1 This looks normalized in 1D ! ˆ

  20. Hydrogen Atom Ueff(r) Uc Centrifugal Barrier Uc Ueff U r -ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R= ER Effective 1-D problem with centrifugal barrier (we know how to solve this numerically with Finite difference!)

  21. Hydrogen Atom Uc Ueff U r [-(ħ2d2/dr2)/2m + Ueff(r)]R= 0 Ueff(r) = –Zq2/4pe0r + l(l+1)ħ2/2mr2 l = 0, 1, 2, ...

  22. Hydrogen Atom P orbitals (l=1) Particle stays away from nucleus (n-2) nodes D orbitals (l=2) n-3 nodes R(r)/r 2p -3.51 eV 3p -1.434 eV 4p 0.32 eV 3d -1.509 eV 4d -0.09 eV What do finite difference 1D radial solutions look like? Depends on l F = Rnl(r)/r x Ylm(q,j) S orbitals (l=0) Particle falls in (No centrif. barrier) n-1 nodes 1s -14.05 eV 2s -3.5 eV 3s -1.38 eV Angular orthogonality between s, p, d due to angular nodes in Ylm Radial orthogonality within s, p, d set due to (n-l-1) radial nodes in R

  23. Hydrogen Atom R(r)/r n-1 Rnl/r ~ rl.e-Zr/na0.L (r/a0) l-1 F = Rnl(r)/r x Ylm(q,j) n gives ‘size’, l gives ‘angular momentum’, m gives z component

  24. n-l-1 nodes  Aufbau principle wikipedia

  25. The shell structure of the H-atom Shows angular and radial nodes for various (nlm) combos Blue: negative, Red: positive

  26. Hydrogen Energy Levels Numerical results 1s -14.05 eV 2s, 2p -3.5 eV 3s, 3p, 3d -1.38 eV Bohr’s results En = (Z2/n2)E0 E0 = -mq4/8ħ2e0 = -13.6 eV E1 = -3.4 eV E2 = -1.51 eV (Grid issues – fine, but enough # points) Accidentally, energy of 2s, 2p same Similarly for 3s, 3p, 3d (ie, E independent of l, dep. only on n) True only for Hydrogen !

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