The Hydrogen Atom. Nucleus charge +Ze mass m 1 coordinates x 1, y 1, z 1. Electron charge e mass m 2 coordinates x 2, y 2, z 2

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1 The Hydrogen Ato The only ato that can be solved exactly. The results becoe the basis for understanding all other atos and olecules. Orbital Angular Moentu Spherical Haronics Nucleus charge +Ze ass coordinates x, y, z Electron charge e ass coordinates x, y, z The potential arises fro the Coulob interaction between the charged particles. V Ze 4 r Ze 4 ( x x) ( y y) ( z z) Copyright Michael D. Fayer, 7

2 The Schrödinger equation for the hydrogen ato is T T T T T T ( ET V ) T x y z e x y z kinetic energy of nucleus kinetic energy of electron potential energy eigenvalues Can separate translational otion of the entire ato fro relative otion of nucleus and electron. Introduce new coordinates x, y, z - center of ass coordinates r,, - polar coordinates of second particle relative to the first Copyright Michael D. Fayer, 7

3 x x x y z y z y z center of ass coordinates rsin cos x x rsin sin y y relative position polar coordinates rcos z z. Copyright Michael D. Fayer, 7

4 Substituting these into the Schrödinger equation. Change differential operators. T T T x y z This ter only depends on center of ass coordinates. Other ters only on relative coordinates. T T T r sin r r r r sin r sin reduced ass [ E,, ] T V r T Try solution ( x, y, z, r,, ) F T ( x, y, z ) ( r,, ) Substitute and divide by T Copyright Michael D. Fayer, 7

5 Gives two independent equations F F F ( ) E Tr F x y z Depends only on center of ass coordinates. Translation of entire ato as free particle. Will not treat further. r r r r r sin r sin sin With E E E T Tr [ (,, )] E V r Relative positions of particles. Internal structure of H ato. In absence of external field V = V(r) Try ( r,, ) R( r) ( ) ( ) Substitute this into the equation and dividing by R yields Copyright Michael D. Fayer, 7

6 d d R d d 8 r d sin [ E V( r)] Rr dr dr r sin d r sin d d Multiply by r sin. Then second ter only depends. sin d d R d sin d d 8 r sin r sin [ E V( r)] R dr dr d d d Therefore, it ust be equal to a constant call constant. d d and d d Dividing the reaining equation by sin leaves Copyright Michael D. Fayer, 7

7 d dr d d r sin r E V( r) Rdr dr sin sin d d The second and third ters dependent only on. The other ters depend only on r. The ters are equal to a constant. Call it -. Multiplying by and transposing -, yields sin d sin d d d sin Replacing the second and third ters in the top equation by - and ultiplying by R/r gives d d R r R E V( r) R r dr dr r Copyright Michael D. Fayer, 7

8 The initial equation in 3 polar coordinates has been separated into three one diensional equations. d d sin d sin d d d sin d d R r R E V( r) R r dr dr r Solve equation. Find it is good for only certain values of. Solve equation. Find it is good for only certain values of. Solve R equation. Find it is good for only certain values of E. Copyright Michael D. Fayer, 7

9 Solutions of the equation d d Second derivative equals function ties negative constant. Solutions sin and cos. But can also use i e Must be single valued (Born conditions). and are sae point. i For arbitrary value of, e for but = for =. i e cos isin in e if n is a positive or negative integer or. Therefore, i e if i e if and wavefunction is single valued only if is a positive or negative integer or. Copyright Michael D. Fayer, 7

10 i e,,, 3 is called the agnetic quantu nuber. The functions having the sae can be added and subtracted to obtain real functions. cos sin,, 3 The cos function is used for positive s and the sin function is used for negative s. Copyright Michael D. Fayer, 7

11 Solution of the equation. sin d d sin d d sin Substitute z cos z varies between + and. Pz ( ) ( ) and sin z d d d P dz dz d dp dz sin dz sin d d dz. sin Making these substitutions yields Copyright Michael D. Fayer, 7

12 The differential equation in ters of P(z) d ( ) dp z dz dz z z P( z). This equation has a singularity. Blows up for z = ±. Singularity called Regular Point. Standard ethod for resolving singularity. Making the substitution ( ) ( ) ( ) Pz z Gz reoves the singularity and gives a new equation for G(z). ( z ) G ( ) zg ( ) G with dg G and G dz d G dz Copyright Michael D. Fayer, 7

13 Use the polynoial ethod (like in solution to haronic oscillator). Gz ( ) aazaz az 3 3 G and G are found by ter by ter differentiation of G(z). Like in the H. O. proble, the su of all the ters with different powers of z equals. Therefore, the coefficients of each power of z ust each be equal to. Let D ( ) Then { z } a Da { z } 6 a3 ( D( )) a { z } a4 ( D4( ) ) a 3 { z } a5 ( D6( ) 6) a3 odd and even series Pick a (a = ) get even ters. Pick a (a = ) get odd ters. a and a deterined by noralization. Copyright Michael D. Fayer, 7

14 The recursion forula is a ( )( ) a ( )( ) Solution to differential equation, but not good wavefunction if infinite nuber of ters in series (like H. O.). To break series off after ' ter ( ' )( ' ) ',,, Let ',,,3, Then ( ) s, p, d, f orbitals z ( ) This quantizes. The series is even or odd as is even or odd. Gz ( ) G(z) are defined by the recursion relation. z cos are the associated Legendre functions Copyright Michael D. Fayer, 7

15 Since ( ), we have d dr ( ) r EV( r) R r dr dr r Ze V( r) 4 r Make the substitutions E The potential only enters into the R(r) equation. Z is the charge on the nucleus. One for H ato. Two for He +, etc. Ze 4 Introduce the new independent variable r is the the distance variable in units of. Copyright Michael D. Fayer, 7

16 Making the substitutions and with S( ) Rr ( ) yields d ds ( ) S d d 4 To solve - look at solution for large, (like H. O.). r Consider the first ter in the equation above. d ds d S ds d d d d d d S ds d This ter goes to zero as r The ters in the full equation divided by and also go to zero as. r Copyright Michael D. Fayer, 7

17 Then, as r d S d S. 4 The solutions are S e / The full solution is / ( ) ( ) S e F S e / This blows up as r Not acceptable wavefunction. Substituting in the original equation, dividing by e / and rearranging gives ( ) F F F The underlined ters blow up at =. Regular point. Copyright Michael D. Fayer, 7

18 Singularity at = - regular point, to reove, substitute F( ) L( ) Gives L (( ) ) L ( ) L. Equation for L. Find L, get F. Know F, have S( ) Rr ( ). Solve using polynoial ethod. Copyright Michael D. Fayer, 7

19 L( ) a a a a Polynoial expansion for L. Get L' and L'' by ter by ter differentiation. Following substitution, the su of all the ters in all powers of equal. The coefficient of each power ust equal. { } ( ) a ( ) a { } ( ) a 4( ) a { } ( ) a 6( ) 6 a3 Note not separate odd and even series. Recursion forula a ( ) a [( )( ) ( )] Given a, all other ters coefficients deterined. a deterined by noralization condition. Copyright Michael D. Fayer, 7

20 a ( ) a [( )( ) ( )] Provides solution to differential equation, but not good wavefunction if infinite nuber of ters. Need to break off after the ν =n' ter by taking n or n with n n' integers n is an integer. n' n radial quantu nuber total quantu nuber n s orbital n', l n s, p orbitals n', l or n', l n 3 s, p, d orbitals n', l or n', l or n', l Copyright Michael D. Fayer, 7

21 Thus, Rr e L / ( ) ( ) with L() defined by the recursion relation, and n nn' integers Copyright Michael D. Fayer, 7

22 n n,, 3, Ze 4 E n Ze 3 4 E E n 4 Ze 8 hn Energy levels of the hydrogen ato. Z is the nuclear charge. for H; for He +, etc. Copyright Michael D. Fayer, 7

23 a h e a 5.9 Bohr radius - characteristic length in H ato proble. In ters of Bohr radius E n Ze 8 an Lowest energy, s, ground state energy, -3.6 ev. Rydberg constant RH - 9,677c E n Z n R hc H R e 8 hc 4 e 3 Rydberg constant if proton had infinite ass. Replace with e. R = 9,737 c -. Copyright Michael D. Fayer, 7

24 Have solved three one-diensional equations to get ( ) ( ) R ( r) The total wavefunction is n (,, r) ( ) ( ) R ( r) n n n,, 3 n, n,, Copyright Michael D. Fayer, 7

25 ( ) is given by the expressions in exponential for or in ters of sin and cos. i e cos,,3 sin Copyright Michael D. Fayer, 7

26 ( ) and R ( r) n can be obtained fro generating functions (like H. O.). See book. With noralization constants Associate Legendre Polynoials ( ) ( )! ( ) P (cos ). ( )! Associate Laguerre Polynoials 3 Z n! R r e L / n( ) ( ) 3 n na n[( n )!] r Z r an Copyright Michael D. Fayer, 7

27 Total Wavefunction (,, r) ( ) ( ) R ( r) n n s function / r R e 3 Z s(,, ) a s r/ a a 3 s function e for Z = No nodes. s(,, r) R ( r/ a ) e 3 4 a r/a Node at r = a. Copyright Michael D. Fayer, 7

28 H ato wavefunction - orbital s orbital ra / s Ae A a 3 a =.59 Å the Bohr radius The wavefunction is the probability aplitude. The probability is the absolute valued squared of the wavefunction. s / ra Ae This is the probability of finding the electron a distance r fro the nucleus on a line where the nucleus is at r = note scale difference r (Å) r (Å) Copyright Michael D. Fayer, 7

29 The s Hydrogen orbital s B( r/ a) e B r/a 4 a 3 Probability aplitude When r = a, this ter goes to zero. There is a node in the wave function. a =.59, the Bohr radius / ( / ) r a s B r a e Absolute value of the wavefunction squared probability distribution node..5. node r (Å) 3 4 r (Å) Copyright Michael D. Fayer, 7

30 Radial distribution function Probability of finding electron distance r fro the nucleus in a thin spherical shell. D ( r) 4 R ( r) r dr nl n For s orbital there is no angular dependence. Still ust integrate over angles with the differential operator sinθdθdφ = 4π.4 s 4[R n (r)]r dr s r/a Copyright Michael D. Fayer, 7

31 s orbitals - = s no nodes s node 3s nodes The nodes are radial nodes. Oxtoby, Freean, Block Copyright Michael D. Fayer, 7