PY3P05 Lectures 3-4: One-electron atoms Lectures 3-4: One-electron atoms o Schrödinger equation for one-electron atom. o Solving the Schrödinger equation. o Wavefunctions and eigenvalues. o Atomic orbitals. o See Chapter 7 of Eisberg & Resnick. Ψ =Ψ( x , y, z , t)
Lectures 3-4: One-electron atoms. Schrödinger equation for one-electron atom. Solving the Schrödinger equation. Wavefunctions and eigenvalues. Atomic orbitals. See Chapter 7 of Eisberg & Resnick. The Schrödinger equation. One-electron atom is simplest bound system in nature. - PowerPoint PPT Presentation
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o One-electron atom is simplest bound system in nature.
o Consists of positive and negative particles moving in 3D Coulomb potential:
o Z =1 for atomic hydrogen, Z =2 for ionized helium, etc.
o Electron in orbit about proton treated using reduced mass:
o Total energy of system is therefore,
€
μ =mM
m + M
€
V = V (x,y,z) =−Ze2
4πε0 x 2 + y 2 + z2
€
KE + PE = E
1
2μ( px
2 + py2 + pz
2) + V (x,y,z) = E
PY3P05
The Schrödinger equationThe Schrödinger equation
o Using the Equivalence Principle, the classical dynamical quantities can be replaced with their associated differential operators:
o Substituting, we obtain the operator equation:
o Assuming electron can be described by a wavefunction of form,
can write
or
where, is the Laplacian operator.
€
px → −ih∂
∂x, py → −ih
∂
∂y, pz → −ih
∂
∂z
E → ih∂
∂t
€
−h2
2μ
∂ 2
∂x 2+
∂ 2
∂y 2+
∂ 2
∂z2
⎛
⎝ ⎜
⎞
⎠ ⎟+ V (x,y,z) = ih
∂
∂t
€
Ψ=Ψ(x,y,z, t)
€
−h2
2μ
∂ 2Ψ(x,y,z ,t)
∂x2+
∂ 2Ψ(x,y,z ,t)
∂y2+
∂ 2Ψ(x,y,z ,t)
∂z 2
⎛
⎝ ⎜
⎞
⎠ ⎟+V(x,y,z )Ψ(x,y,z ,t) = ih
∂Ψ(x,y,z ,t)
∂t
€
−h2
2μ∇ 2Ψ + VΨ = ih
∂Ψ
∂t
€
∇2 =∂ 2
∂x 2 +∂ 2
∂y 2 +∂ 2
∂z2
PY3P05
The Schrödinger equationThe Schrödinger equation
o Since V(x,y,z) does not depend on time, is a solution to the Schrödinger equation and the eigenfunction is a solution of the time-independent Schrödinger equation:
o As V = V(r), convenient to use spherical polar coordinates.
where
o Can now use separation of variables to split the partial
differential equation into a set of ordinary differential equations.
€
Ψ(x, y,z, t) =ψ (x, y,z)e−iEt / h
€
−h2
2μ∇ 2ψ (x,y,z) + Vψ (x, y,z) = Eψ (x,y,z)
€
−h2
2μ∇ 2ψ (r,θ,φ) + V (r)ψ (r,θ,φ) = Eψ (r,θ,φ)
€
∇2 =1
r 2
∂
∂r(r 2 ∂
∂r)+
1
r 2 sinθ
∂
∂θ(sinθ
∂
∂θ)+
1
r 2 sin2θ
∂ 2
∂φ 2
(1)
€
ψ(x, y,z)
PY3P05
Separation of the Schrödinger equationSeparation of the Schrödinger equation
o Assuming the eigenfunction is separable:
o Using the Laplacian, and substituting (2) and (1):
o Carrying out the differentiations,
o Note total derivatives now used, as R is a function of r alone, etc.
o Now multiply through by and taking transpose,
(2)
€
ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)
€
−h2
2μ
1
r2
∂
∂rr2 ∂RΘΦ
∂r
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
r2 sinθ
∂
∂θsinθ
∂RΘΦ
∂θ
⎛
⎝ ⎜
⎞
⎠ ⎟+
1
r2 sin2 θ
∂ 2RΘΦ
∂φ2
⎡
⎣ ⎢
⎤
⎦ ⎥+ V (r)RΘΦ = ERΘΦ
€
−h2
2μ
ΘΦ
r2
∂
∂rr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟+
RΦ
r2 sinθ
∂
∂θsinθ
dΘ
dθ
⎛
⎝ ⎜
⎞
⎠ ⎟+
RΘ
r2 sin2 θ
d2Φ
dφ2
⎡
⎣ ⎢
⎤
⎦ ⎥+ V (r)RΘΦ = ERΘΦ
€
−2μr2 sin2 θ /RΘΦh2
€
1
Φ
d2Φ
dφ2= −
sin2 θ
R
d
drr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟−
sinθ
Θ
d
dθsinθ
dΘ
dθ
⎛
⎝ ⎜
⎞
⎠ ⎟−
2μ
h2r2 sin2 θ[E −V (r)] (3)
PY3P05
Separation of the Schrödinger equationSeparation of the Schrödinger equation
o As the LHS of Eqn 3 does nor depend on r or and RHS does not depend on their common value cannot depend on any of these variables.
o Setting the LHS of Eqn 3 to a constant:
and RHS becomes
o Both sides must equal a constant, which we choose as l(l+1):
o We have now separated the time-independent Schrödinger equation into three ordinary differential equations, which each only depend on one of (4), (5) and R(6). .
€
1
Φ
d2Φ
dφ2= −ml
2 =>d2Φ
dφ2= −ml
2Φ (4)
€
−1
R
d
drr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟−
1
Θsinθ
d
dθsinθ
dΘ
dθ
⎛
⎝ ⎜
⎞
⎠ ⎟−
2μ
h2r2[E −V (r)] = −
ml2
sin2 θ
€
=>1
R
d
drr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟+
2μr2
h2[E −V (r)] =
ml2
sin2 θ−
1
Θsinθ
d
dθsinθ
dΘ
dθ
⎛
⎝ ⎜
⎞
⎠ ⎟
€
− 1
sinθ
d
dθsinθ
dΘ
dθ
⎛
⎝ ⎜
⎞
⎠ ⎟+
ml2Θ
sin2 θ= l(l +1)Θ
1
r2
d
drr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟+
2μr2
h2[E −V (r)]R = l(l +1)
R
r2
(5)
(6)
PY3P05
Summary of separation of Schrödinger equationSummary of separation of Schrödinger equation
o Express electron wavefunction as product of three functions:
o As V ≠ V(t), attempt to solve time-independent Schrodinger equation.
o Separate into three ordinary differential equations for and .
o Eqn. 4 for () only has acceptable solutions for certain value of ml.
o Using these values for ml in Eqn. 5, () only has acceptable values for certain values of l.
o With these values for l in Eqn. 6, R(r) only has acceptable solutions for certain values of En.
o Schrödinger equation produces three quantum numbers!
€
ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)
€
R(r),Θ(θ)
€
(φ)
PY3P05
Azimuthal solutions (Azimuthal solutions ((())))
o A particular solution of (4) is
o As the einegfunctions must be single valued, i.e., =>
and using Euler’s formula,
o This is only satisfied if ml = 0, ±1, ±2, ...
o Therefore, acceptable solutions to (4) only exist when ml can only have certain integer values, i.e. it is a quantum number.
o ml is called the magnetic quantum number in spectroscopy.
o Called magnetic quantum number because plays role when atom interacts with magnetic fields.
€
(φ) = e im lφ
€
e iml 0 = e iml 2π
€
1= cosml 2π + isin ml 2π
PY3P05
Polar solutions (Polar solutions ((())))
o Making change of variables (z = rcos, Eqn. 5 transformed into an associated Legendre equation:
o Solutions to Eqn. 7 are of form
where are associated Legendre polynomial functions.
remains finite when = 0, 1, 2, 3, ...
ml = -l, -l+1, .., 0, .., l-1, l
o Can write the associated Legendre functions using quantum number subscripts:
o Customary to multiply () and () to form so called spherical harmonic functions
which can be written as:
PY3P05
Radial solutions (R( r ))Radial solutions (R( r ))
o What is the ground state of hydrogen (Z=1)? Assuming that the ground state has n = 1, l = 0 Eqn. 6 can be written
o Taking the derivative
(7)
o Try solution , where A and a0 are constants. Sub into Eqn. 7:
o To satisfy this Eqn. for any r, both expressions in brackets must equal zero. Setting the second expression to zero =>
o Setting first term to zero =>
€
1
r2
d
drr2 dR
dr
⎛
⎝ ⎜
⎞
⎠ ⎟+
2μ
h2E +
e2
4πε0r
⎡
⎣ ⎢
⎤
⎦ ⎥R = 0
€
a0 =4πε0h
2
μe2
€
d2R
dr2+
2
r
dR
dr+
2μ
h2E +
e2
4πε0r
⎡
⎣ ⎢
⎤
⎦ ⎥R = 0
€
R = Ae−r / a0
€
E = −h2
2μa02
= −13.6 eVSame as Bohr’s results
PY3P05
Radial solutions (R( r ))Radial solutions (R( r ))
o Radial wave equation
has many solutions, one for each positive integer of n.
o Solutions are of the form (see Appendix N of Eisberg & Resnick):
where a0 is the Bohr radius. Bound-state solutions are only acceptable if
where n is the principal quantum number, defined by n = l +1, l +2, l +3, …
o En only depends on n: all l states for a given n are degenerate (i.e. have the same energy).
€
d2R
dr2+
2
r
dR
dr+
2μ
h2E +
e2
4πε0r
⎡
⎣ ⎢
⎤
⎦ ⎥R = 0
€
Rnl (r) = e−Zr / na0Zr
a0
⎛
⎝ ⎜
⎞
⎠ ⎟
l
Gnl
Zr
a0
⎛
⎝ ⎜
⎞
⎠ ⎟
€
En = −Z 2μe4
(4πε0)22h2n2
= −13.6Z 2
n2 eV
PY3P05
Radial solutions (R( r ))Radial solutions (R( r ))
o Gnl(Zr/a0) are called associated Laguerre polynomials, which depend on n and l.
o Several resultant radial wavefunctions (Rnl( r )) for the hydrogen atom are given below
PY3P05
Radial solutions (R( r ))Radial solutions (R( r ))
o The radial probability function Pnl(r ), is the probability that the electron is found between r and r + dr:
o Some representative radial probability functions are given at right:
o Some points to note:
o The r2 factor makes the radial probability density vanish at the origin, even for l = 0 states.
o For each state (given n and l), there are n - l - 1nodes in the distribution.
o The distribution for states with l = 0, have n maxima, which increase in amplitude with distance from origin.
€
Pnl (r)dr = Rnl (r)Ylm l(θ,φ)
2r2dV∫
= 4πr2 Rnl
2r2dr
PY3P05
Radial solutions (R( r ))Radial solutions (R( r ))
o Radial probability distributions for an electron in several of the low energy orbitals of hydrogen.
o The abscissa is the radius
in units of a0. s orbitals
p orbitals
d orbitals
PY3P05
Hydrogen eigenfunctionsHydrogen eigenfunctions
o Eigenfunctions for the state described by the quantum numbers (n, l, ml) are therefore of form:
and depend on quantum numbers:
n = 1, 2, 3, …
l = 0, 1, 2, …, n-1
ml = -l, -l+1, …, 0, …, l-1, l
o Energy of state on dependent on n:
o Usually more than one state has same
energy, i.e., are degenerate.
€
ψnlm l(r,θ,φ) = Rnl (r)Θ lm l
(θ)Φm l(φ)
€
En = −13.6Z 2
n2
PY3P05
Born interpretation of the wavefunctionBorn interpretation of the wavefunction
o Principle of QM: the wavefunction contains all the dynamical information about the system it describes.
o Born interpretation of the wavefunction: The probability (P(x,t)) of finding a particle at a position between x and x+dx is proportional to |Ψ(x,t)|2dx:
P(x,t) = Ψ*(x,t) Ψ(x,t) = |Ψ(x,t)|2
o P(x,t) is the probability density.
o Immediately implies that sign of wavefunction has no
direct physical significance.
Ψ(x,t) P(x,t)
PY3P05
Born interpretation of the wavefunctionBorn interpretation of the wavefunction
o In H-atom, ground state orbital has the same sign everywhere => sign of orbital must be all positive or all negative.
o Other orbitals vary in sign. Where orbital changes sign, Ψ = 0 (called a node) => probability of finding electron is zero.
o Consider first excited state of hydrogen: sign of
wavefunction is insignificant (P = Ψ2 = (-Ψ)2).
PY3P05
Born interpretation of the wavefunctionBorn interpretation of the wavefunction
o Next excited state of H-atom is asymmetric about origin. Wavefunction has opposite sign on opposite sides of nucleus.
o The square of the wavefunction is identical on
opposite sides, representing equal distribution
of electron density on both side of nucleus.
PY3P05
Atomic orbitalsAtomic orbitals
o Quantum mechanical equivalent of orbits in Bohr model.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
PY3P05
s orbitalss orbitals
o Named from “sharp” spectroscopic lines.
o l = 0, ml = 0
ψn,0,m = Rn,0 (r ) Y0,m (, )
o Angular solution:
o Value of Y0,0 is constant over sphere.
o For n = 0, l = 0, ml = 0 => 1s orbital
o The probability density is
-0.20
0.2
-0.20
0.2
-0.2
0
0.2
-0.2
0
0.2
€
Y0,0 =1
4π
€
ψ1,0,0 =1
π
Z
a0
⎛
⎝ ⎜
⎞
⎠ ⎟
3 / 2
e−Zr / a0
€
Pnlml(r,θ,φ) =ψ nlm l
(r,θ,φ)*ψ nlml(r,θ,φ)
€
=>P1,0,0(r,θ,φ) =1
π
Z
a0
⎛
⎝ ⎜
⎞
⎠ ⎟
3
e−2Zr / a0
PY3P05
p orbitalsp orbitals
o Named from “principal” spectroscopic lines.
o l = 1, ml = -1, 0, +1 (n must therefore be >1)
ψn,1,m = Rn1 (r ) Y1,m (, )
o Angular solution:
o A node passes through the nucleus and separates the two lobes of each orbital.
o Dark/light areas denote opposite sign of the wavefunction.
o Three p-orbitals denoted px, py , pz €
ψ2,1,0(r,θ,φ) = R2,1(r)3
4π
⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2
cosθ = 2 pz
€
Y1,0 =3
4πcosθ
PY3P05
d orbitalsd orbitals
o Named from “diffuse” spectroscopic lines.
o l = 2, ml = -2, -1, 0, +1, +2 (n must therefore be >2)
ψn,2,m = Rn1 (r ) Y2,m (, )
o Angular solution:
o There are five d-orbitals, denoted
o m = 0 is z2. Two orbitals of m = -1 and +1 are xz and yz. Two orbitals with m = -2 and +2 are designated xy and x2-y2.
-1-0.500.5
1-1 -0.5 0 0.5
1
-2
-1
0
1
2
-1-0.500.5
1
-2
-1
0
1
2
€
Y2,0 =5
4π
1
2(3cos2 θ −1)
€
dz 2 ,dxz,dyz,dxy,dx 2 −y 2
PY3P05
Quantum numbers and spectroscopic notationQuantum numbers and spectroscopic notation
o Principal quantum number: o n = 1 (K shell)o n = 2 (L shell)o n = 3 (M shell)o …
o If n = 1 and l = 0 = > the state is designated 1s. n = 3, l = 2 => 3d state.
o Three quantum numbers arise because time-independent Schrödinger equation contains three independent variables, one for each space coordinate.
o The eigenvalues of the one-electron atom depend only on n, by the eigenfunctions depend on n, l and ml, since they are the product of Rnl(r ), lml () and ml().
o For given n, there are generally several values of l and ml => degenerate eigenfunctions.
o Angular momentum quantum number:o l = 0 (s subshell)o l = 1 (p subshell)o l = 2 (d subshell)o l = 3 (f subshell)o …
PY3P05
Orbital transitions for hydrogenOrbital transitions for hydrogen
o Transition between different energy levels of the hydrogenic atom must follow the following selection rules:
l = ±1
m = 0, ±1
o A Grotrian diagram or a term diagram shows the allowed transitions.
o The thicker the line at right, the more probable and hence more intense the transitions.
o The intensity of emission/absorption lines could not be explained via Bohr model.
PY3P05
Schrödinger vs. Bohr modelsSchrödinger vs. Bohr models
o Schrodinger’s QM treatment had a number of advantages over semi-classical Bohr model:
1. Probability density orbitals do not violate the Heisenberg Uncertainty Principle.
1. Orbital angular momentum correctly accounted for.