2m r2 (~r )+V (~r ) (~r )=E (~r )

Transcription

1 Review of the Hydrogen Atom The Schrodinger equation (for 1D, 2D, or 3D) can be expressed as: ~ 2 2m r2 (~r, t )+V (~r ) (~r, The Laplacian is the divergence of the gradient: r 2 =r r The time-independent Schrodinger equation (TISE) can be found by assuming (~r, t )= (~r )exp( iet/~) This only works if the potential energy is independent of time ~ 2 2m r2 (~r )+V (~r ) (~r )=E (~r )

2 Review of Hydrogen Atom For hydrogen, the TISE becomes: ~ 2 r 2 ke 2 2m e r = E In spherical coordinates: r r @ + 1 r 2 sin 2

3 Review of Hydrogen Atom The TISE can be solved using separation of variables (r,, The spherical harmonics are given by Y m l Pl m )=R(r)Yl m (cos, ) = Pl m (cos )e im are associated Legendre functions The radial function can be expressed in terms of generalized Laguerre polynomials (and a decaying exponential): 2r R(r) =A exp( r/na 0 ) na 0 l L 2l+1 n l 1 2r na 0 Thus the overall eigenfunctions for the hydrogen atom are given by 2r nlm(r,, )=A nlm exp( r/na 0 ) na 0 a 0 = ~2 m e ke 2 l 2r L 2l+1 n l 1 Pl m (cos )exp(im ) na 0

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6 Radial Probability Distribution Recall that ψ 2 is the probability density (probability per unit volume) The radial probability distribution is given by P r (r) = Most likely radial distance is determined by where dp r (r) dr =0 Z Average radial distance is 0 Standard deviation: Z sin d d = r 2 R(r) 2 <r>= Z 1 0 rp r (r) dr r = r = p <r 2 > <r> 2

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8 Clicker Question For which n=3 state is the standard deviation in r the largest? 1. 3s 2. 3p 3. 3d 4. It is the same for all three!

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10 Angular Dependence Y m l = Pl m (cos )e im

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12 Eigenenergies Normalizable solutions to the Schrodinger equation are only found for specific energies. These are the Eigenenergies. The eigenenergies only depend on the quantum number n: E n = m e(ke 2 ) 2 2~ 2 1 n 2 = 13.6 ev1 n 2 ev For a given value of n, how many spatial quantum states have the same energy? Degeneracy = n 2

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14 from physchem.co.za/atomic/hydrogen%20spectrum.htm

15 Hydrogen-Like Atoms Hydrogen-like atoms are ionized atoms containing only one electrum. The nucleus has a nuclear charge Ze. For example, singly ionized helium is hydrogen-like with Z=2. How do the wave-functions differ from that of hydrogen? 1. For given quantum numbers, the hydrogen-like radial wave function causes the electron to be more likely closer to the nucleus. 2. For given quantum numbers, the hydrogen-like radial wave function causes the electron to be more likely farther from the nucleus. 3. The angular portion of the wave function is no longer described by spherical harmonics. 4. More than one of the above are true.

16 Hydrogen-Like Atoms The only difference to the TISE is that the potential energy term now is given by V (r) = Zke2 r This causes the electrons to be more tightly bound. a 0 a 0 /Z 100 = A exp Zr a 0 Electrons are closer to the nucleus a 0 = ~2 m e ke 2 The angular portion of the TISE doesn t change, so the angular portion of the wave functions are still spherical harmonics. Energy levels are given by E n = 13.6 ev Z2 n 2

17 Clicker Question For the hydrogen atom, we find that the energy levels are given by E n = (ke2 ) 2 m e 2~ 2 = 13.6 ev/n 2 For a hydrogen-like atom with atomic number Z, what are the energy levels? 1. E n = 13.6 ev Z n E n = 13.6 ev Z2 n 2 E n = 13.6 ev 1 Zn 2 4. None of the above

18 Orbital Angular Momentum Classically, the orbital angular momentum of a particle is given by L x = yp z zp y ~L = ~r ~p ) L y = zp x xp z L z = xp y yp x In quantum mechanics, ~p! ˆp = ~ i r Thus, for example, ˆL z = ~ @x In spherical coordinates, ˆL z =

19 Solutions to the L z operator: ˆL z = Want to find L z eigenstates, u, that satisfy u(, )= u(, ) Solution: u / e i ~ What are the eigenvalues λ?

20 L z for Electron in Hydrogen Notice that eigenstates of the Hamiltonian (energy) operator for hydrogen are also eigenstates of L z operator! ˆL z nlm = L z nlm Ĥ nlm = E n nlm nlm = R nl (r) lm ( )exp(im ) ) ˆL z nlm = m~ nlm Eigenvalues of L z operator are mħ. Possible outcomes of a measurement are the eigenvalues of the associated operator. If the particle is in an eigenstate, then the outcome of a measurement will be the corresponding eigenvalue. It is possible to both know E and L z values simultaneously to arbitrary precision.

21 L z for Electron in Hydrogen Space quantization: Possible values of L z are discrete! ) ˆL z nlm = m~ nlm m= -l,,0,, l L z = m~

22 Total Orbital Angular Momentum The total angular momentum squared is given by L 2 = L 2 x + L 2 y + L 2 z Using spherical coordinates and the appropriate operator for momentum, it turns out (after lots of math) that apple 1 ˆL 2 = ~ + sin 2 The possible (measured) values of the magnitude squared of the angular momentum (and the associated wave functions) come from solutions to the equation ~ 2 apple @ + 1 sin 2 u = L 2 u It turns out solutions to this equation are the spherical harmonics we saw for the Schrodinger equation!

23 Total Orbital Angular Momentum u lm = Y m l (, ) The resulting eigenvalues are L 2 = ~ 2 l(l + 1) Thus for the electron in a hydrogen atom, ~ L = ~ p l(l + 1)

24 Is it possible that L = L z? Clicker Question 1. Yes, for any value of l this is possible 2. No, never! 3. In general, no, but it is possible for one specific value of l.

25 Total Orbital Angular Momentum For the electron in a hydrogen atom, ~ L = ~ p l(l + 1) L z = m~ Notice that l=0) ~ L >L z (unless This is the most we can know about the angular momentum. The uncertainty principle prevents us from knowing the other two components If l =2, ~ L = p 6~ L z = 2~, ~, 0, ~, or 2~