Chapter XXII Atomic Structure

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Chapter XXII Atomic Structure

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Rutherford (also Geiger-Marsden) Experiment (1911): Measured angular dependence of   particles (He ions) scattered from gold foil. The results: • Mostly scattering at small angles. But… • Occasional scatterings at large angles (even 90o)  Something massive in there ! Conclusion: Most of atomic mass is concentrated in a small region of the atom Recall some history:

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Nội dung Text: Chapter XXII Atomic Structure

GENERAL PHYSICS III Optics & Quantum Physics
Chapter XXII Atomic Structure §1. Hydrogen atom §2. Angular momentum of electron §3. Electron spin §4. Many-electron atoms
At present it is well known for you that The Atom - electrons confined in Coulomb field of a nucleus Recall some history: Au Rutherford (also Geiger-Marsden)   Experiment (1911): v Measured angular dependence of  particles (He ions) scattered  from gold foil. The results: • Mostly scattering at small angles. But… • Occasional scatterings at large angles (even > 90o )  Something massive in there ! Conclusion: Most of atomic mass is concentrated in a small region of the atom a nucleus!
Early sugession for the quantum nature of atoms: Discrete emission and absorption spectra: (nm) • When excited in an electrical discharge, atoms emitted radiation only at discrete wavelengths • Different emission spectra for different atoms. Atomic hydrogen This could not be explained by classical physics. -e F There is also a BIG PROBLEM with the classical picture: +Ze As the electron moves in its circular orbit, it is ACCELERATING  radiates electromagnetic energy  would continuously lose energy and spiral into the nucleus (in about 10 -9 sec). We must apply QUANTUM MECHANICS to ATOMS
U(r) §1. Hydrogen atom: r 1.1 Potential energy of the electron 0 in the hydrogen atom: atom 2 e U ( r )  1 r    9 Nm 2 / C 2 9 10 4 0   (r) U 2 e This potential function implies • Make a test: f    2 actually the Coulomb force r r between the electron & the nucleus • The origin of coordinate system is at the center of mass of   the system (nucleus & electron). • U depends on r, but not on directions in the space. r z y Such a potential is called spherically symmetrical. • The problem is 3-D. Due to the spherical symmetry x   it is more convenient to use the spherical coordinates, than Cartersian.
1.2 The time independent SEQ:  d 2 (x) 2 • Recall 1-D time independent SEQ: 2  ( x) x)   x) U ( E ( 2m dx • The 3-D time independent SEQ is as follows:   2( x ) ( x ) ( x )  2  2 2     2   2  2  U ( x , y , z ) E   2m  x y z  • For the hyrdrogen atom, the potential function U(x,y,z) = U(r) , and the SEQ in the spherical coordinate system is 2   2   1 r  1     1   2   2   2     r sin     sin   2 2   ( r ) E  U 2m   r  r r    r sin     2  2 e Solving and finding the eigenvalues for E and the U ( r )  r eigenfunctions ψ requires some mathematical preparation. In our course of general physics, we will concentrate on the physical implications of mathematical solutions.
1.3 Energy levels of the electron in the hydrogen atom: As you have seen, in the solving the SEQ, the physical conditions for the wave function (finiteness, single valuedness, continuosness) lead to the consequence that the energy of the system can not have continuum values, but must accept a discrete set of values: “the energy levels”. The energy levels of the electron in the hydrogen atom is found to be  2 1 e 13 . 6 eV r/a0 En   n  , 2 , 3 ,... 1 0 0 5 10 15 20 20 2 2 2 ao n n 00 2 E3 a0  = “Bohr radius” = 0.053 nm m2 e -5 E2 * Notes: Before Schrödinger, N. Bohr E (eV) U(r) proposed the same scheme of energy -10 levels for the hydrogen atom in his theory (1913). But in the Bohr’s theory -15-15 E1 the concept of discrete energy levels was introduced as a postulate. The values of the energy levels are determined by fitting in the experimental discrete spectra of the hydrogen atom.
Recall that the discrete spectra of the hydrogen atom is related to the optical transitions of between the electron’s energy levels. Example: An electron, initially excited r/a0 to the n = 3 energy level of the hydrogen 0 0 5 10 15 20 20 atom, falls to the n = 2 level, emitting a 0 0 photon in the process. What are the  E3 energy and wavelength of the photon -5 E2 emitted? E (eV) 13.6 eV U(r) From En  2 -10 n we have 1 1   ni n f  13.6  2  2  E  eV -15 E1 i n  nf  -15  Atomic hydrogen  1 1 E photon  E3 2  13.6       eV 1.9eV  4 9 hc 1240eV nm   656nm  (nm)  E photon 1.9eV Note: The ionization energy of atomic hydrogen at its ground state is  En  1   1  .6eV E E 0 E 13
1.4 The eigenfunctions and physical interpretation: Due to the spherical symmetry, the solution to the SEQ in spherical coordinates is found   by the method of separation of variables, and has the form: r z y nlm ( r , , )  R nl ( r )Y lm ( , ) x with quantum numbers: n l and m  principal orbital magnetic (angular momentum) • Rnl (r) is the ‘radial part’ of wave function. • Ylm( ) are the angle-dependent functions called “spherical harmonics.” , • The principal quantum number n = 1,2, 3,… This number is that in the formula of the energy levels. • For every n, the values of l are 0, 1, 2,…, (n-1). • For every l, the values of m are -l, -(l-1), …, -2, -1, 0, 1, 2,…, (l-1), l.
The spherically symmetric states: We write here the wave functions for wavefunctions with no angular dependence. These are called “s-states”. For them, l = 0 and m = 0, the function Y00 = constant, and the part Rn0(r) obeys the ‘radial SEQ’ which takes the form: 1 2 2 2 e    n0 ( r )  E n Rn0 ( r ) s-state 2m r 2 r  r R wavefunction:  r  (l=0)  r , )  n0 ( r ) ( , R R10 R 20 R 30 0.5 0 0 0 0 0 10a 0 0 r 15a0 r 4a0 r  / ao  2r r  / 3a 2 R1,0 ( r ) e r R3, 0 (r )     r 0 3 2  e  r  r / 2a   a0 a0  3   R2,0 (r )  1 e  o    2a0 
P ( r ) dr  dV Rn 0 ( r ) 4 2 dr  2 Probability density of electrons: 2 r P(r) is plotted below for the two s-states with n=1,2: “1s state” “2s state” The horizontal axis is in (r/a 0) units. At r = a0 (the Bohr radius) the probabilty of finding the electron is highest for 1s-state.
 The angle dependent part of the eigenfunctions: The investigation of the angle dependent part of the eigenfunctions of the electron in the hydrogen atom gives the following results: This part can be represented in the following form: Y lm (  lm ( m ( , ) ) ), l 0 ,1, 2 , 3 ,... m  ,..., 0 ,...,  l l (m takes (2l+1) interger values from –l to +l). where m ( e im , ) lm ( sin Fl m (cos  → the associated Legendre function m ) . ) Examples: Y00 (,  1, → the spherically symmetrical (angle independent) ) “s-state” (l=0) Y1 ,  (,  sin e im , Y10 (  cos  Y1 ,1 (,  sin e , 1 ) . , ) , ) . im  → the “p-states” (l=1)
§2.Angular momentum of electron: The results of solving SEQ for the hydrogen atom with the conditions for the wave functions give also the eigenvalues for the angular moment vector L which is specified by L  L and L Z . 2.1 The LZ component of the angular momentum vector:  Angular momentum L Z around the z axis depends on how fast the phase changes as you rotate around the z-axis. The phase change is expessed by the factor exp(im in the angle ) dependent part of the eifenfunction  A state with an exact value of L Z is of the form:  Re() LZ   where (r ) Yl ,m ( eim m , ) r e.g. LZ can have values in a discrete set  LZ   ,   0,  2 etc m is called the ‘orbital’ magnetic quantum number, or shortly, magnetic quantum number.
2.2 Magnitude of orbital angular momentum: The index l of the eigenfunctions is directly related to the magnitude of the angular moment L  L L  l (l  )  where 1 l  , 1, 2, ..., (n  ). 0 1 Examples: l 0 , m 0  L 0 (the spherically symmetrical “s-state”) l 1, m  , 0 ,   L  2  L Z  , 0 ,  1 1 ,   (the “p-states”) l 2 , m  ,  , 0 ,  ,  2 1 1 2  L  6  L Z   , 0 , 2  , 2 ,  , . (the “d-states”)
Thus, when l ≠0 the angular momentum vector L can have only certain orientations in space. This phenomenon is called space quantization. 2.3 The Zeeman effect: • The Zeeman effect is the splitting of spectrum lines when the atoms placed in a magnetic field. • This effect confirms experimentally the quantization of the angular moment. It is known that a moving electron correspons to a current loop with the magnetic moment  e   L (e, m – the electron’s charge, mass) 2m e e g  is called the gyromagnetic ratio 2me
When a magnetic moment is placed in a magnetic field B the interaction potential energy is given by the following equation:  U  . B  B cos  When the external magnetic field is uniform, and the z-axis is oriented in the same direction with the field : The splitting of the state l=1 e  U  Z B m B m  B , ( m  ,...,  , 0 ,1,..., l ). 0 l 1 2m e e   0 9 . 2741  24 J / Tesla 10 Increase B  2me  is called the Bohr magneton. 0 the splitting is wider Due to the interaction between the electron’s magnetic moment and the external field, the energy of the state with the orbital moment number l splits into (2l+1) states with different values m.
2.4 Selection rules: m The splitting of the states with 2 the quantum number l in 3d 1 0 a magnetic field corresponds -1 -2 to the splitting of spectral lines. The experimental observations showed that not all combinations of initial and final levels are possible. m The allowed transitions are 1 only that obey the following rules: 2p 0 -1 l   and m  0 , 1 allowed 1 forbidden These are called the selection rules. • The Bohr theory could not give an explanation for these rules. From the form of the eigenfundtions it is found that the transition probabilities from the initial to final states, which is given by an integral of the corresponding eigenfuntions, cancel out unless l   & m  0 , 1 . 1
Allowed transition for H s p d f g n Energy (eV) l=0 1 2 3 4  0.00 4 -0.85 3 -1.51 2 -3.40  .6 eV 13 En  n2 1 -13.6 eV
Hydrogen Atom States: Summary  Key Points: 0 5 10 15 20 0  n: principal quantum number n=3 r/a0  l: orbital quantum number n=2  m: orbital magnetic quantum number -5  Energy depends only upon n  e2 1  e e4 1  m 2 U(r) En   E 2a0 n 2 2 n 2 2 -10  For a given value of n, there are n possible ang. momentum n=1 states: l = n-1, n-2,..., 2, 1, 0 -15 For a given angular momentum Therefore, a level with  state l, there are 2l + 1 possible orientations with z- quantum number n has axis m = -l, -(l -1),…0,…(l -1), l n2 degenerate states.
Hydrogen Atom States: Summary r/a0 0 5 10 15 20 (n,l,m) 0 n=3 (3,0,0), (3,1,-1), (3,1,0), (3,1,1), (3,2,-2), (3,2,-1), (3,2,0), (3,2,1), (3,2,2) n=2 (2,0,0), (2,1,-1), (2,1,0), (2,1,1) -5 n = 1, 2, 3,... E l = 0, 1, 2, .. n-1 U(r) -10 m = -l, -(l - 1), ..0,… (l -1), l n=1 (1,0,0) -15