Chap 13 - Chap 13 Atomic Structure Atkins: Elements of...

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Chap 13 Atomic Structure Atkins: Elements of Physical Chemistry 4ed
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Atomic structure – the description of the arrangement of electrons in atoms – is an essential part of chemistry because it is the basis of understanding molecular and solid structures and all the physical and chemical properties of elements and their compounds .
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Hydrogenic atom One electron atom or ion of general atomic number Z . This also applies to H, He + , Li +2 , C 5+ .. Exact solution of Schrodinger eq. can be found, H in particular. As the basis to describe structure of many-electron atoms
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Facts before QM Atom emits radiation (EM) when exposed to flame or excited by electric discharge. The emission spectra consist of distinct frequencies (lines)– discrete frequencies. Balmer (1885) found that in visible range the spectra (atomic hydrogen – electric discharges pass through hydrogen gas)
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/ E E hE Rydberg constant n 1 =1,2,3,. . n 2 =n 1 +1, n 1 +2,… energy is conserved for a transition Bohr frequency condition c v v / =
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13.2 The permitted energies of hydrogenic atoms Rutherford’s nuclear model ( r s : ) Potential function With a lot of work…… (as followings) For hydrogenic atom of atomic number with a nucleus of mass m N, the allowed energy levels are given as 2 2 n hcRZ E n - = 2 2 0 2 4 32 h e hcR ε π μ = N e N e m m m m + = n=1,2,… n = Principal quantum number
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2 1 hcRZ E - = 2 2 4 1 hcRZ E - = 2 2 2 1 n hcR n hcR E - = Energy level of hydrogen atom For hydrogen Z=1 Ground state E1 below the energy of Infinitely separated electron and nucleus Rydberg Eqn. The minimum energy needed to remove an electron completely frm an atom is called ionization energy I I =hcR H =2.18x10 -18 J 1312 kJ/mole
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QM description for hydrogenic atoms column potential B.C.: not infinite anywhere, same as we circle the nucleus over the poles or round the equator m=m e +m N Rutherford’s nuclear model ( L ± “S² )
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Center of mass Separation of variable again ψ( r , φ , θ29 = R ( r 29 Y , θ29 - 2 2 μ 1 r 2 1 sin 2 θ 2 Y ∂φ 2 + 1 sin θ ∂θ sin θ Y ∂θ = E o Y - 2 2 μ 1 r 2 d dr r 2 d R dr - Ze 2 4 πε 0 r R + E o R = E R I = μ r 2 1 sin 2 θ 2 Y ∂φ 2 + 1 sin θ ∂θ sin θ Y ∂θ = - 2 IE o E 2 Y
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Θ part Associated Legendra equation sin θ d d θ sin θ d Θ d θ + 2 IE o E 2 sin 2 θ - m 2 Θ = 0 ζ = cos θ d d θ = d ζ d θ d d ζ = - sin θ d d ζ sin 2 θ = ( 1 - cos 2 θ29 = ( 1 - ζ 2 29 2 IE o / 2 = l ( l + 1 29 ( 1 - ζ 2 29 d 2 Θ d ζ 2 - 2 ζ d Θ d ζ + l ( l + 1 29 - m 2 1 2 Θ = 0 l = 0,1, 2. . .and | m | ≤ l
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Exact solution associated Leguerre polynomial. n
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This note was uploaded on 11/11/2010 for the course CHE CH2005 taught by Professor 曹恒光 during the Spring '10 term at National Central University.

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Chap 13 - Chap 13 Atomic Structure Atkins: Elements of...

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