Atom hidrogen


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Atom hidrogen

  1. 1. The Hydrogen AtomSchrödinger EquationAngular MomentumSolution of the Schrödinger Equationfor Hydrogen Werner Heisenberg (1901-1976)The atom of modern physics can be symbolized only through a partial differentialequation in an abstract space of many dimensions. All its qualities are inferential; nomaterial properties can be directly attributed to it. An understanding of the atomic worldin that primary sensuous fashion…is impossible. - Werner Heisenberg
  2. 2. 1. Schrödinger EquationThe time-independent Schrodinger equation for atom hydrogen:Use the three-dimensional time-independent Schrödinger Equation.For Hydrogen-like atoms (He+ or Li++) Replace e2 with Ze2 (Z is the atomic number). Replace m with the reduced mass, m.. Our task is to find the wave functions ª(r; µ; Á) that satisfy this equation, and hence to find the allowed quantized energies E.
  3. 3. 2. Angular MomentumClassical definition of angular momentum:For circular orbits this simplifies to L = mvr, and in Bohr’s model, Lwas quantized in integer units of h.However, the full quantum treatment is more complicated, and requiresthe introduction of two other quantum numbers l and ml, as we shall nowsee.The components of L are given by
  4. 4. Angular Momentum …In quantum mechanics we know that the linear momentum bydifferential operators:Therefore, the quantum mechanical operators for the angularmomentum are given by:
  5. 5. Angular Momentum ...The magnitude of the angular momentum is given by:Now we therefore have the quantum mechanical operatorIt should be noted that this operators should be understood in termsof repeated operations:
  6. 6. Angular Momentum …It can be shown that the components of the angular momentumoperator do not commute, that isHowever, in fact we can show that:where the “commutator bracket” [ˆLx; ˆLy] is defined by Remember that: if two quantum mechanical operators do not commute, then it is not possible to know their values simultaneously.
  7. 7. Angular Momentum …For example, the operators for position and momentum in a one-dimensional system:Thus we have:
  8. 8. 3. Solution of the Schrödinger Equation for H satisfies the azimuthal equation for any value of mℓ. The solution must be single valued to be a valid solution for any f: mℓ must be an integer (positive or negative) for this to be true. Now set the left side equal to −mℓ2 and rearrange it [divide by sin2(q)]. Now, the left side depends only on r, and the right side depends only on θ. We can use the same trick again!
  9. 9. Solution of the Schrödinger Equation for HSet each side equal to the constant ℓ(ℓ + 1). Radial equation Angular equationWe’ve separated the Schrödinger equation into three ordinary second-order differential equations, each containing only one variable.
  10. 10. Solution of the Radial Equation for HThe radial equation is called the associated Laguerre equation and thesolutions R are called associated Laguerre functions. There areinfinitely many of them, for values of n = 1, 2, 3, …Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution.The radial equation becomes:The derivative of yields two terms.This yields:
  11. 11. Solution of the RadialTry a solution A is a normalization constant. Equation for H a0 is a constant with the dimension of length. Take derivatives of R and insert them into the radial equation. Tosatisfy this equation for any r, both expressions in parenthesesmust be zero.Set the second expressionequal to zero and solve for a0:Set the first expression equalto zero and solve for E:Both are equal to the Bohr results!
  12. 12. PrincipalQuantumNumber nThere are many solutions to the radial wave equation, one foreach positive integer, n.The result for the quantized energy is:A negative energy means that the electron and proton are boundtogether.
  13. 13. Quantum NumbersThe three quantum numbers: n: Principal quantum number ℓ: Orbital angular momentum quantum number mℓ: Magnetic (azimuthal) quantum numberThe restrictions for the quantum numbers: n = 1, 2, 3, 4, . . . ℓ = 0, 1, 2, 3, . . . , n − 1 mℓ = −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓEquivalently: n>0 The energy levels are: ℓ<n |mℓ| ≤ ℓ
  14. 14. Hydrogen Atom Radial Wave FunctionsFirst fewradialwavefunctionsRnℓSub-scriptson Rspecifythevalues ofn and ℓ.
  15. 15. Solution of the Angular and AzimuthalEquationsThe solutions to the azimuthal equation are:Solutions to the angular and azimuthal equations are linkedbecause both have mℓ.Physicists usually group these solutions together intofunctions called Spherical Harmonics: spherical harmonics
  16. 16. NormalizedSphericalHarmonics
  17. 17. Solution of the Angular and AzimuthalEquationsThe radial wave function R and the spherical harmonics Y determinethe probability density for the various quantum states. The total wavefunction depends on n, ℓ, and mℓ. The wave functionbecomes
  18. 18. Orbital Angular Momentum QuantumNumber ℓ Energy levels are degenerate with respect to ℓ (the energy is independent of ℓ). Physicists use letter names for the various ℓ values: ℓ= 0 1 2 3 4 5... Letter = s p d f g h... Atomic states are usualy referred to by their values of n and ℓ. A state with n = 2 and ℓ = 1 is called a 2p state.
  19. 19. Orbital Angular Momentum QuantumNumber ℓIt’s associated with the R(r) and f(θ) parts of the wave function.Classically, the orbital angular momentum with L = mvorbitalr.L is related to ℓ byIn an ℓ = 0 state, This disagrees with Bohr’s semi-classical “planetary” model of electrons orbiting Classical orbits—which do not a nucleus L = nħ. exist in quantum mechanics
  20. 20. The angle f is the angle from the Magnetic Quantumz axis. Number mℓThe solution for g(f) specifies thatmℓ is an integer and is related tothe z component of L:Example: ℓ = 2:Only certain orientations of arepossible. This is called spacequantization.And (except when ℓ = 0) we justdon’t know Lx and Ly!
  21. 21. Rough derivation of ‹L2› = ℓ(ℓ+1)ħ2We expect the average of the angular momentum componentssquared to be the same due to spherical symmetry:ButAveraging over all mℓ values (assuming each is equally likely):  because:  n   m 2  (  1)(2  1) / 3
  22. 22. 4. Magnetic Effects on Atomic Spectra—The Zeeman EffectIn 1896, the Dutch physicist Pieter Zeemanshowed that spectral lines emitted by atomsin a magnetic field split into multiple energylevels. It is called the Zeeman effect.Consider the atom to behave like a small magnet.Think of an electron as an orbiting circular current loop of I = dq / dtaround the nucleus. If the period is T = 2p r / v,then I = -e/T = -e/(2p r / v) = -e v /(2p r).The current loop has a magnetic moment m = IA = [-e v /(2p r)] p r2=  e [-e/2m] mrv: m  L 2mwhere L = mvr is the magnitude of the orbital angular momentum.
  23. 23.  e The Zeeman Effect m  L 2m The potential energy due to the magnetic field is:If the magnetic field is in the z-direction, we only care about the z-component of m: e e mz   Lz   (m )   mB m 2m 2mwhere mB = eħ / 2m is called the Bohr magneton.
  24. 24. The Zeeman EffectA magnetic field splits the mℓ levels. The potential energy is quantizedand now also depends on the magnetic quantum number mℓ.When a magnetic field is applied, the 2p level of atomic hydrogen issplit into three different energy states with energy difference of ΔE =mBB Δmℓ. mℓ Energy 1 E0 + μBB 0 E0 −1 E0 − μBB
  25. 25. TheZeemanEffectThe transitionfrom 2p to 1s,split by amagnetic field.
  26. 26. The Zeeman EffectAn atomic beam of particles in the ℓ = 1 state pass through amagnetic field along the z direction.   m B m (dB / dz )The mℓ = +1 state will be deflected down, the mℓ = −1 state up, andthe mℓ = 0 state will be undeflected.
  27. 27. 5. Energy Levelsand ElectronProbabilitiesFor hydrogen, the energylevel depends on the prin-cipal quantum number n.In the ground state, an atomcannot emit radiation. It canabsorb electromagneticradiation, or gain energythrough inelasticbombardment by particles.
  28. 28. We can use the wave functionsSelection Rules to calculate transition probabilities for the electron to change from one state to another.The probability is proportional tothe mag square of the  dipole moment:  d   er  *Allowed transitions:Electrons absorbing or emitting photons can change stateswhen Δℓ = 1 and Δmℓ = 0, 1.Forbidden transitions:Other transitions are possiblebut occur with much smallerprobabilities.
  29. 29. Probability Distribution FunctionsWe use the wave functions to calculate the probabilitydistributions of the electrons.The “position” of the electron is spread over space and is notwell defined.We may use the radial wave function R(r) to calculate radialprobability distributions of the electron.The probability of finding the electron in a differential volumeelement dτ is .
  30. 30. Probability Distribution FunctionsThe differential volume element in spherical polar coordinates isTherefore,At the moment, we’re only interested in the radial dependence.The radial probability density is P(r) = r2|R(r)|2 and it depends only on nand ℓ.
  31. 31. ProbabilityDistributionFunctionsR(r) and P(r)for the lowest-lying states ofthe hydrogenatom.
  32. 32. Probability Distribution FunctionsThe probability density for the hydrogen atom for three differentelectron states.
  33. 33. 6. Intrinsic SpinIn 1925, grad students, SamuelGoudsmit and George Uhlenbeck,in Holland proposed that theelectron must have anintrinsic angular momentumand therefore a magnetic moment.Paul Ehrenfest showed that, if so, the surface of the spinningelectron should be moving faster than the speed of light!In order to explain experimental data, Goudsmit and Uhlenbeckproposed that the electron must have an intrinsic spinquantum number s = ½.
  34. 34. Intrinsic SpinThe spinning electron reacts similarly tothe orbiting electron in a magnetic field.The magnetic spin quantum number mshas only two values, ms = ½. The electron’s spin will be either “up” or “down” and can never be spinning with its magnetic moment μs exactly along the z axis.
  35. 35.  e Intrinsic Spin Recall: m L   L 2mThe magnetic moment is .The coefficient of is −2μB and is aconsequence of relativistic quantum mechanics.Writing in terms of the gyromagnetic ratio, g: gℓ = 1 and gs = 2: andThe z component of .In an ℓ = 0 state: no splitting due to . there is space quantization due to the intrinsic spin.Apply ms and the potential energy becomes:
  36. 36. Multi-electron atomsWhen more than one electron is involved, the potential and thewave function are functions of more than one position:       V  V (r1 , r2 ,..., rN )    (r1 , r2 ,..., rN , t )Solving the Schrodinger Equation in this case can be very hard.But we can approximate the solution as the product of single-particle wave functions:        (r1 , r2 ,..., rN , t )  1 (r1 , t )  2 (r2 , t )   N (rN , t )And it turns out that, for electrons (and other spin ½ particles), allthe i’s must be different. This is the Pauli Exclusion Principle.
  37. 37. Molecules have many energy levels. A typical molecule’s energy levels: E = Eelectonic + Evibrational + Erotational 2ndexcitedelectronic state Lowest vibrational and rotational level of this electronic “manifold” Energy 1st excited Excited vibrational andelectronic state rotational level Transition There are many other complications, such as Ground spin-orbit coupling,electronic state nuclear spin, etc., which split levels.As a result, molecules generally have very complex spectra.
  38. 38. Generalized Uncertainty PrincipleDefine the Commutator of two operators, A and B:  A, B  AB  BAThen the uncertainty relation between the two correspondingobservables will be: A B  1 2  *  A, B  So if A and B commute, the two observables can be measuredsimultaneously. If not, they can’t.      Example:  p, x     px  xp     i   x   x  i    x   x   x        i   ix   x  i   i  x x   x So:  p, x  i and p x   / 2
  39. 39. Two Types of Uncertainty in QuantumMechanicsWe’ve seen that some quantities (e.g., energy levels) can becomputed precisely, and some not (Lx).Whatever the case, the accuracy of their measured values islimited by the Uncertainty Principle. For example, energies canonly be measured to an accuracy of ħ /t, where t is how longwe spent doing the measurement.And there is another type of uncertainty: we often simply don’tknow which state an atom is in.For example, suppose we have a batch of, say, 100 atoms,which we excite with just one photon. Only one atom is excited,but which one? We might say that each atom has a 1% chanceof being in an excited state and a 99% chance of being in theground state. This is called a superposition state.
  40. 40. Superpositions of statesStationary states are stationary. But an atom can be in asuperposition of two stationary states, and this state moves.    (r , t )  a1 1 (r ) exp(iE1t / )  a2 2 (r ) exp( iE2t / )where |ai|2 is the probability that the atom is in state i.Interestingly, this lack of knowledge means that theatom is vibrating:  2  2  2 (r , t )  a1 1 (r )  a2 2 (r )   *  2 Re a1 1 (r )a2 2 (r ) exp[i( E2  E1 )t / ]
  41. 41. Superpositions of statesVibrations occur at the frequency difference between the two levels.  2  2  2 (r , t )  a1 1 (r )  a2 2 (r )   *  2 Re a1 1 (r )a2 2 (r ) exp[i( E2  E1 )t / ] Excited level, E2  Energy E = hn Ground level, E1The atom is vibrating The atom is at least partially inat frequency, n. an excited state.
  42. 42. Calculations in Physics: Semi-classicalphysics The most precise computations are performed fully quantum- mechanically by calculating the potential precisely and solving Schrodinger’s Equation. But they can be very difficult. The least precise calculations are performed classically, neglecting quantization and using Newton’s Laws. An intermediate case is semi-classical computations, in which an atom’s energy levels are computed quantum- mechanically, but additional effects, such as light waves, are treated classically. Our Thomson Scattering calculation of a and n was an example of this.