The Hydrogen Atom. Chapter 18. P. J. Grandinetti. Nov 6, Chem P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

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1 The Hydrogen Atom Chapter 18 P. J. Grandinetti Chem Nov 6, 2017 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

2 The Hydrogen Atom Hydrogen atom is simplest atomic system where Schrödinger equation can be solved analytically and compared to experimental measurements. Analytical solution serve as basis for obtaining approximate solutions for multi-electron atoms and molecules, where no analytical solution exists. Warning: Working through analytical solution of H-atom may cause drowsiness. Do not operate heavy machinery during this derivation. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

3 Two Particle Problem z 2 particles with mass m p and m e at r p and r e. Total mass, center of mass, and inter-particle distance vector is y M = m p + m e R = m p r p + m e r e m p + m e and r = r e r p Express r p and r e in terms of M, R and r as x r p = R m e M r and r e = R + m p M r Express individual momenta of 2 particles as p p = m p d r p dt then in terms of M, R and r as ( d R p p = m p dt m ) e d r M dt and p e = m e d r e dt ( d R and p e = m e dt + m ) p d r M dt P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

4 Two Particle Problem Next consider total energy E = p 2 p + p e + V( r) 2m p 2m e 2 V( r) depends only on distance between 2 particles. Write energy in terms of M, R and r and obtain E = 1 ( )2 d 2 M R + 1 ( ) 2 d r dt 2 μ + V( r) dt μ is reduced mass, given by 1 μ = 1 m p + 1 m e Define 2 new momenta associated with center of mass and reduced mass, p R = M d R dt, and p r = μ d r dt P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

5 Two Particle Problem Total energy becomes E = p 2 R 2M + p 2 r 2μ + V( r) 1st term is translational energy of center of mass 2nd term is kinetic energy due to relative motion of 2 particles Translating to quantum mechanics we write time independent Schrödinger equation for 2 particle system as [ ] ħ2 2M 2 R ħ2 2μ 2 r + V( r) f ( R, r) = Ef ( R, r) P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

6 Two Particle Problem [ ] ħ2 2M 2 R ħ2 2μ 2 r + V( r) f ( R, r) = Ef ( R, r) Wave function can be separated into product of two wave functions f ( R, r) = χ( R)ψ( r) χ( R) depending only on center of mass ψ( r) depending only on relative motion of 2 particles Substitute product into Schrödinger Eq above we obtain 2 wave equations where ħ2 2M 2 R χ( R) = E R χ( R) and E = E R + E r [ ħ2 2μ 2 r + V( r) ] ψ( r) = E r ψ( r) On left is wave equation for translational motion of free particle of mass M On right is wave equation for particle with mass μ in potential V( r) For electron bound positively charged nucleus we focus on PDE for ψ( r) P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

7 Schrödinger Equation in Spherical Coordinates Focus on PDE for ψ( r): Electron bound to positively charged nucleus [ ] ħ2 2μ 2 + V(r) ψ( r) = Eψ( r) Coulomb potential is V(r) = Zq2 e 4πε 0 r Central potential with 1 r dependence. V as r 0 V 0 as r P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

8 Schrödinger Equation in Spherical Coordinates Since V(r) only depends on r we adopt spherical coordinates [ ħ2 2 2μ r r r + 1 ( ( 1 sin θ ) + 1 )] 2 ψ+ V(r)ψ = Eψ r 2 sin θ θ θ sin 2 θ φ 2 L 2 ħ 2 2 Take term in parentheses as L 2 ħ 2, rearrange, and simplify to r 2 2 ψ(r, θ, φ) r 2 ψ(r, θ, φ) + 2r + 2μr2 ( E V(r) ) ψ(r, θ, φ) = 1 r ħ 2 ħ L 2 ψ(r, θ, φ) 2 To solve PDE use separation of variables ψ(r, θ, φ) = R(r)Y(θ, φ) P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

9 Schrödinger Equation in Spherical Coordinates Substitute ψ(r, θ, φ) into PDE and dividing both sides by ψ(r, θ, φ) r 2 2 R(r) + 2r R(r) + 2μr2 ( E V(r) ) = R(r) r 2 R(r) r ħ 2 Left side depends only on r Right side depends only on θ and φ. We have turned one PDE into two ODEs. 1 1 Y(θ, φ) ħ L 2 Y(θ, φ) 2 1 We know eigenfunctions and eigenvalues of L 2 on right side, L 2 Y l,m (θ, φ) = l(l + 1)ħ 2 Y l,m (θ, φ) 2 Define l(l + 1) as separation constant and obtain ODE for radial part r 2 d 2 R(r) + 2r dr(r) + 2μr2 ( E V(r) ) = l(l + 1) R(r) dr 2 R(r) dr ħ 2 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

10 Schrödinger Equation in Spherical Coordinates Expanding and rearranging ODE for radial part we obtain [ ( ħ2 d 2 2μ r + 2 ) ] d + ħ2 l(l + 1) + V(r) R(r) = ER(r) 2 r dr 2μ r 2 V eff (r) Looks like 1D time independent Schrödinger Eq with effective potential V eff (r) = ħ2 2μ l(l + 1) r 2 Centrifugal Term Zq2 e 4πε 0 r Total angular momentum, l(l + 1)ħ, creates centrifugal force that pushes electron away from nucleus. Notice 2 terms in V eff (r) always have opposite signs. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

11 V eff (r): Effective Potential of H-Like Atom Switching to Atomic Units: Convenient units for atomic physics Atomic unit of length, also called Bohr radius, is defined as a 0 4πε 0ħ 2 q 2 e m e = pm Atomic unit of energy is 1E h ħ2 m e a 2 0 = q2 e 4πε 0 a 0 = ev Dividing effective potential by E h, set Z = 1, μ m e, and obtain V eff (r) E h = l(l + 1) m e a μ r a 0 2 r l(l + 1) 2ρ 2 1 ρ Define dimensionless radius as ρ = r a 0 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

12 Plot of V eff for l = 0, 1, 2, and Note: potential minimum shifts to higher radii with increasing angular momentum (i.e., centrifugal force). P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

13 Solving the Wave Equation for the Radial Part [ ħ2 2μ ( d 2 r r ) ] d + ħ2 l(l + 1) Zq2 e R(r) = ER(r) dr 2μ r 2 4πε 0 r pour yourself a third cup of coffee... P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

14 Solving the Radial Wave Equation To obtain radial wave function, R(r), we must solve [ ( ħ2 d 2 2μ r ) ] d + ħ2 l(l + 1) r dr 2μ r 2 Zq2 e R(r) = ER(r) 4πε 0 r Begin by dividing and both sides by E and obtain [ ħ2 d 2 2μE r ] ( ) d ħ R(r) + 2 l(l + 1) r dr 2μE r 2 Zq2 e 4πε 0 re 1 R(r) = 0 Define κ 2 2μE ħ 2 or E = ħ2 κ 2 2μ κ is in wave numbers. Rearrange potential energy expression to Zq2 e 4πε 0 re = ρ 0 is dimensionless quantity Zq2 e 4πε 0 r 2μ ħ 2 κ 2 = ρ 0 Zq2 e μ 2πε 0 ħ 2 κ Zq2 e μ 2πε 0 ħ 2 κ 1 κr = 2ρ 0 κr P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

15 Solving the Radial Wave Equation Substitute expression for potential energy into ODE [ 1 d 2 κ 2 r ] d R(r) r dr Define dimensionless variable Rewrite radial part as function of ρ as [ d 2 dρ ] d R(ρ) ρ dρ Further simplify ODE by defining u(ρ) = ρr(ρ), Re-express ODE as du(ρ) dρ d 2 u(ρ) dρ 2 ( l(l + 1) 1 + (κr) 2 + 2ρ 0 κr ρ 2κr ( l(l + 1) 1 + ρ 2 + 2ρ 0 ρ ) R(r) = 0 ) R(ρ) = 0 = ρdr(ρ) dρ + R(ρ), d 2 u(ρ) dρ 2 = ρ d2 R(ρ) dρ dr(ρ) dρ ( 1 4 ρ ) 0 l(l + 1) + 2ρ ρ 2 u(ρ) = 0 ρ varies from 0 to. 1st look at asymptotic solutions: (1) ρ and (2) ρ 0 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

16 Solving the Radial Wave Equation, ρ Starting with d 2 u(ρ) dρ 2 ( 1 4 ρ ) 0 l(l + 1) + 2ρ ρ 2 u(ρ) = 0 (1) ρ. At large values of ρ approximate ODE as d 2 u(ρ) dρ 2 u(ρ) 4 0 Solutions to this ODE have form u(ρ) Ae ρ 2 + Be ρ 2 Reject positive exponent since require that solution be finite everywhere, u(ρ) Ae ρ 2 in limit that ρ P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

17 Solving the Radial Wave Equation, ρ 0 Starting with d 2 u(ρ) dρ 2 ( 1 4 ρ ) 0 l(l + 1) + 2ρ ρ 2 u(ρ) = 0. (2) ρ 0. At small values of ρ approximate ODE as d 2 u(ρ) l(l + 1) dρ 2 ρ 2 u(ρ) 0. Solutions to this ODE have form u(ρ) Aρ l+1 + Bρ l. Reject B term again because require that solution be finite at ρ = 0. u(ρ) Aρ l+1 in limit that ρ 0 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

18 Solving the Radial Wave Equation u(ρ) Ae ρ 2 in limit that ρ u(ρ) Aρ l+1 in limit that ρ 0 Now, back to d 2 u(ρ) dρ 2 ( 1 4 ρ ) 0 l(l + 1) + u(ρ) = 0 2ρ ρ 2 we propose general solution of form u(ρ) = ρ l+1 L(ρ) e ρ 2 This has correct behavior in two limits. Only need to determine L(ρ) to get behavior for all ρ. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

19 Solving the Radial Wave Equation Substitute u(ρ) = ρ l+1 L(ρ) e ρ 2 into d 2 u(ρ) dρ 2 ( 1 4 ρ ) 0 l(l + 1) + u(ρ) = 0 2ρ ρ 2 gives ρ d2 L(ρ) dρ 2 + dl(ρ) dρ (2(l + 1) ρ) + ( ρ0 2 (l + 1) ) L(ρ) = 0 Set j = ρ 0 2 l 1, and k = 2l + 1 and this ODE is recognized as associated Laguerre differential equation d 2 L k j (ρ) dl k j (ρ) ρ + (k + 1 ρ) + jl k dρ 2 dρ j (ρ) = 0 Has nonsingular solutions only if j is non-negative integers, j = 0, 1, 2, These solutions, L k (ρ), are called the associated Laguerre polynomials. j P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

20 Selected associated Laguerre polynomials n l L 2l+1 (ρ) Polynomial Roots n l L 1 0 (ρ) L 1 (ρ) 1 2 ρ L 3 0 (ρ) L 1 2 (ρ) 1 ( ) 2 6 6ρ + ρ L 3 (ρ) 1 4 ρ L 5 0 (ρ) L 1 3 (ρ) 1 ( ρ + 12ρ 2 ρ 3) L 3 2 (ρ) 1 ( ) ρ + ρ L 5 (ρ) 1 6 ρ L 7 0 (ρ) L 1 4 (ρ) L 3 3 (ρ) L 5 2 (ρ) L 7 1 ( ρ + 120ρ 2 20ρ 3 + ρ 4) ( ρ + 18ρ 2 ρ 3) ( 42 14ρ + ρ 2 ) (ρ) 8 ρ L 9 0 (ρ) Associated Laguerre polynomials following definition where L k j (x) = ( 1)k d k dx k L j+k(x). Laguerre polynomial L j (x) is defined by Rodrigues formula: L j (x) = 1 n! ex d j dx j ( x j e x). P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

21 Solving the Radial Wave Equation Since l = 0, 1, 2,, and j = 0, 1, 2,, then ρ 0 2 = j + l + 1 can only take on integer values of ρ 0 2 = 1, 2,. We define this as the principal quantum number which can only take on values of n ρ 0 2 = j + l + 1, n = 1, 2, 3,. Wave functions with same n value form set called a shell. Special letters are sometimes assigned to each n value n = numerical value K L M N O symbol P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

22 Solving the Radial Wave Equation l is called the azimuthal quantum number Rearranging n = j + l + 1 for l gives l = n j 1 and we find that l cannot exceed n 1. Range of l is l = 0,, n 1. Recall that azimuthal quantum number, l, defines total angular momentum of l(l + 1)ħ. m l is called the magnetic quantum number. m l has positive and negative integer values between l and l. When x, y, or z component of the electron s angular momentum is measured only values of m l ħ are observed. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

23 Solving the Radial Wave Equation Wave functions with same value of n and l form set called a sub-shell. Special letters are assigned to sub-shell with given l value, l = numerical value s p d f g h i k l m o q r t symbol and continue afterwards in alphabetical order. 1st four letters arose in pre-quantum atomic spectroscopy for classifying emission lines and stand for sharp, principal, diffuse, and fine. Shorthand nl notation uses principal quantum number with l symbol, Wave function with n = 1, l = 0 is referred to as 1s state n = 2, l = 1 is referred to as 2p state Number of roots of L 2l+1 equal to n l 1. n l 1 is n l 1. Thus, number of radial nodes is Number of angular nodes is l and total number of nodes is n 1. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

24 Energy of the Hydrogen-Like Atom Given the constraint that ρ 0 2 = n we go back to From ρ 0 Zq2 e μ 2πε 0 ħ 2 κ = 2n and rearrange to κ n = Zq2 e μ 4πε 0 ħ 2 n κ 2 2μE ħ 2 rearranges to E n = ħ2 κ 2 n 2μ = Z2 q 4 e μ 32π 2 ε 2 0 ħ2 n 2 = Z2 q4 e μ 8ε 2 0 h2 n 2 E n = Z2 q 4 e μ 8ε 2 0 h2 n 2 Energy of H-like atom e bound to nucleus with charge Z energy only depends on n and Z If Z increases (holding n constant) then E n decreases becomes more negative. Higher Z means nucleus holds e more tightly In limit that n then E 0 and e is unbound P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

25 Energy of the Hydrogen-Like Atom With μ m e, the n = 1 energy is called Rydberg unit of energy (Ry) With μ m e approximation 1Ry = E 1 = q4 e m e = ev 8ε 2 h2 0 E n = 1 Ry n 2 Energy of H atom Can also divide by atomic unit of energy, E h, to obtain E n = 1 E h 2n 2 Energy of H atom Given n, energy is identical for each l. Given l energy is identical for 2l + 1 values of m l. Degeneracy of nth energy level is n 1 g n = (2l + 1) = n + 2 l = n + n(n 1) = n 2 l=0 n 1 l=0 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

26 Energy levels of single electron bound to proton orbitals degeneracy Visible light emission Balmer series UV light emission Lyman series P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

27 Energy of the Hydrogen-Like Atom Remember Balmer series for hydrogen emission spectra obeys relation: ( 1 1 λ = R H 2 1 ) where R 2 n 2 H = /m Followed by Bohr s theory of atom which gave ( ) ( ) ( ) me λ = q 4 e Z πε 0 4πħ 3 c 0 n 2 n 2 f i perfect agreement with R H Taking hν = hν = ΔE = E λ f E i and solving for 1 λ from H-like atom energy gives exact same result as Bohr s theory for R H. At this point Schrödinger knew wave equation approach was working. After determining full wave function for H-like atom next step is solutions for multi-electron atoms where Bohr s theory had failed. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

28 Normalizing the Radial Wave Function Solutions, L k (ρ), are called associated Laguerre polynomials. j From these we construct radial part of wave function. Need to retrace steps from u(ρ) back to R(r) starting with going back through u n,l (ρ n ) = ρ l+1 n L 2l+1 n l 1 (ρ n) e ρ n 2 R n,l (ρ n ) = u n,l(ρ n ) ρ n = ρ l n L2l+1 n l 1 (ρ n) e ρ n 2 and finally returning to R n,l (r) = (2κ n r) l L 2l+1 n l 1 (2κ nr) e κ nr P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

29 Normalizing the Radial Wave Function To normalization radial part we define x = 2κ n r and express radial part as R n,l (x) = A n,l x l L 2l+1 (x) e x 2 n l 1 A n,l is to-be-determined normalization constant. Normalization integral is 0 R n,l (r)r n,l(r)r 2 1 dr = (2κ n ) 3 Substituting R n,l (x) into integral gives A 2 n,l (2κ n ) R n,l (x)r n,l(x)x 2 dx = 1 x 2l+2 [L 2l+1 n l 1 (x)]2 e x dx = 1 Using general integral for associated Laguerre polynomials we find 0 x k+1 [L k j (x)]2 e x (j + k)! dx = (2j + k + 1) j! A 2 n,l 2n(n + l)! (2κ n ) 3 (n l 1)! = 1 or A n,l = (2κ n ) 3 2 (n l 1)! 2n(n + l)! P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

30 Radial Wave Function To further simplify R n,l (r) we introduce quantity analogous to Bohr radius, then we obtain and express ρ n as a μ 4πε 0ħ 2 q 2 e μ κ n = Z na μ ρ n = 2κ n r = 2Zr na μ Finally, we obtain radial part of wave function of hydrogen-like atom ( ) l+3 2 ( ) 2Z (n l 1)! 2Z R n,l (r) = na μ 2n(n + l)! L2l+1 r r l e Zr (na μ ) n l 1 na μ P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

31 Radial part of H-like wave functions for n = 1 to n = 4. ( ) 3 2 Z R 1,0 = 2 e Zr a μ aμ ( ) 3 2 ( ) R 2,0 = 1 2 Z 2 Zr e Zr (2a μ ) 2 a μ a μ ( ) 5 2 R 2,1 = 1 2 Z re Zr (2a μ ) 6 a μ R 3,0 = ( ) Z 81 a μ 6 6 ( 2Zr 3a μ ) ( ) 2 2Zr + 3a μ R 3,1 = 1 ( ) 5 2 ( ( )) 2 Z 2Zr 4 re Zr (3a μ ) 27 3 a μ 3a μ e Zr (3a μ ) ( ) 3 2 ( ) ( ) R 4,0 = 1 2 ( ) 3 Z 1 16 a μ Zr Zr Zr + 12 e Zr (4a μ ) 2a μ 2a μ 2a μ R 4,1 = 1 32 ( ) Z 1 15 aμ 2 ( Zr 2aμ ) ( ) 2 Zr + 2aμ R 4,2 = 1 ( ) 7 2 ( ( )) 1 Z Zr 6 r 2 e Zr (4a μ ) aμ 2aμ R 4,3 = 1 ( ) Z r 3 e Zr (4a μ ) aμ r e Zr (4a μ ) R 3,2 = 2 81 ( ) Z r 2 e Zr (3a μ ) 15 a μ P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

32 R n,l (r) for s orbitals (l = 0) of n = 1, 2, 3, 4, s s s s s s s s s s Wave function extends further out in r, away from the nucleus, as n increases As with harmonic oscillator classically excluded positions are displacements where E < V(r) For hydrogen atom classically excluded radii are r a μ > 2n 2 Classically excluded regions are indicated by gray regions. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

33 R n,l (r) for n = 5 and all possible values of l s 5p 5d 5f 5g s 5p 5d f 5g Radial function at origin, r = 0, is non-zero only for s states, where l = 0. Maximum in wave function at constant n is pushed further away from nucleus as l increases. This is consequence of centrifugal term in effective potential. V eff (r) = ħ2 2μ l(l + 1) r 2 Centrifugal Term Zq2 e 4πε 0 r P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

34 Hydrogen-Like Atom Wave Functions In summary, solutions to Schrödinger equation for single electron bound to positive charge nucleus are ψ n,l,ml (r, θ, φ) = R n,l (r)y l,ml (θ, φ). R n,l (r) is radial part of wave function and depends on quantum numbers n and l Y l,ml (θ, φ) is angular part of wave function and depends on quantum numbers l and m l. Radial part combined with spherical harmonics give full wave function, also called orbital since wave function describes electron s orbit around nucleus. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

35 Hydrogen wave functions for n = 1 to n = 3 Orbital Wave Function ( ) 3 2 1s ψ 1,0,0 = 2 Z e Zr a μ 4π aμ ( ) 3 2 ( ) 1 Z 2s ψ 2,0,0 = 2 Zr e Zr (2a μ ) 32π a μ a μ ( ) Z 2p 0 ψ 2,1,0 = re Zr (2a μ ) cos θ 32π aμ ( ) 5 2 2p ±1 ψ 2,1,±1 = 1 Z re Zr (2a μ ) sin θ e ±iφ 64π aμ Orbital Wave Function ( ) Z 3s ψ 3,0,0 = 162π a μ 6 6 ( 2Zr 3a μ ) ( ) 2 2Zr + 3a μ e Zr (3a μ ) ( ) 5 2 ( ( )) 3p 0 ψ 3,1,0 = 1 1 Z 2Zr 4 re Zr (3a μ ) cos θ 27 2π aμ 3aμ ( ) 5 2 ( ( )) 3p ±1 ψ 3,1,±1 = 1 1 Z 2Zr 4 re Zr (3a μ ) sin θ e ±iφ 27 4π a μ 3a μ ( ) 7 2 3d 0 ψ 3,2,0 = 2 1 Z r 2 e Zr (3a μ ) π aμ 2 (3 cos2 θ 1) ( ) 7 2 3d ±1 ψ 3,2,±1 = 1 1 Z r 2 e Zr (3a μ ) 3 cos θ sin θ e ±iφ 243 π a μ ( ) 7 2 3d ±2 ψ 3,2,±2 = 1 1 Z r 2 e Zr (3a μ ) 3 sin 2 θ e ±i2φ 486 π aμ P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

36 Hydrogen-Like Atom Wave Functions 2D cross sections through x-z plane of 3D probability density of selected hydrogen atom wave functions. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

37 Hydrogen-Like Atom Probability distributions Probability density for finding e in given volume element, dτ, is ψ(r, θ, φ) 2 dτ = R n,l (r) 2 Y l,ml (θ, φ) 2 r 2 dr sin θdθdφ Integrate over all values of θ and φ to calculate probability of finding e inside spherical shell of thickness dr at distance r from origin ψ(r, θ, φ) 2 dτ = R n,l (r) 2 π 0 0 2π Y l,ml (θ, φ) 2 r 2 dr sin θdθdφ = r 2 R 2 n,l (r)dr. Recall that spherical harmonic functions are already normalized. Probability density r 2 R 2 (r)dr is called radial distribution function. n,l P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

38 Cartesian (real) wave functions m l 0 complex wave function are hard to visualize. Sometimes easier to work with real wave functions by taking sum and difference of m l wave functions, ( ) ψ (±) = c ψ ml ± ψ m l Since ψ m l = ψ m l coefficient c is adjusted to make ψ(±) real and normalized. ψ px = 1 ( ) ψ p+1 + ψ p 1 2 ψ py = 1 2i ( ψ p+1 ψ p 1 ) Superposition principle tells us that these are also solutions to same wave equation. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

39 Cartesian (real) wave functions Can also form real wave functions with d orbitals nd z 2 = ψ n,2,0 nd xz = 1 2 ( ψn,2,1 + ψ n,2, 1 ) ndyz = 1 2i ( ψn,2,1 ψ n,2, 1 ) nd x 2 y 2 = 1 2 ( ψn,2,2 + ψ n,2, 2 ) ndxy = 1 2i ( ψn,2,2 ψ n,2, 2 ) As long as 2 wave functions are degenerate they will also be stationary states. Numerous illustrations of shape of the Cartesian H-orbitals can be found in elementary chemistry texts. P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

40 H-like Atom Transition Selection Rules For transitions between H-like atom energy states through absorption and emission of light we require non-zero electric dipole transition moment, μ n,l,ml,n,l,m l = V ψ n,l,m l (r, φ, θ) μ ψ n,l,m l(r, φ, θ)dτ For H-like atom electric dipole moment is μ = i q i r = Zq e r p q e r e Since r e = r + r p we can write μ = Zq e r p q e ( r + r p ) = (Z 1)q e r p q e r Substituting this into transition moment integral gives μ n,l,ml,n,l,m l = V ψ n,l,m l (r, φ, θ) (Z 1)q e r p ψ n,l,m l(r, φ, θ)dτ V ψ n,l,m l (r, φ, θ) q e r ψ n,l,m l(r, φ, θ)dτ P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41

41 H-like Atom Transition Selection Rules μ n,l,ml,n,l,m l = V ψ n,l,m l (r, φ, θ) (Z 1)q e r p ψ n,l,m l(r, φ, θ)dτ V ψ n,l,m l (r, φ, θ) q e r ψ n,l,m l(r, φ, θ)dτ 1st integral involving r p goes to zero for all transitions 2nd integral gives transition dipole moment Convenient to express r in spherical coordinates q e r = q e r ( sin θ cos φ e x + sin θ sin φ e y + cos θ e z ) Using H-like wave functions find that Δn can have any value and Δl = ±1, Δm l = 0 for μ z, Δm l = ±1 for μ x and μ y P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41