Feb 3, 2014 ... 3.2 Hydrogen Atom. 2. 2. 2. 1 m d d. = Φ. Φ. -. ϕ. In section 3.1, we solved the two
differential equations below where Eq. 3.2.1 is dependent on ...
3.2 Hydrogen Atom Nicholas Brown ECE 6451 Introduction to the Theory of Microelectronics Fall 2005
3.2 Hydrogen Atom In section 3.1, we solved the two differential equations below where Eq. 3.2.1 is dependent on phi and Eq. 3.2.2 which is dependent on theta
1 d 2Φ 2 − = m Φ dφ 2
(3.2.1)
1 d dΘ m 2Θ sin θ + λΘ = sin θ dθ dθ sin 2 θ
(3.2.2)
In section 3.2, we will solve the radial dependent differential equation and analyze its results in the context of the Hydrogen atom.
3.2 Hydrogen Atom The differential equation below is the radial dependent component of the 3-D Schrödinger Equation in spherical coordinates
h2 ∂ 2 ∂ λh 2 R + V (r ) R = ER − r R + 2 2 2mr ∂r ∂r 2mr We will relate our equation to the Hydrogen atom by considering the form of V(r).
1 V (r ) ~ r
The potential of the Hydrogen atom, V(r) is inversely proportional to r
(3.2.3)
3.2 Hydrogen Atom We will define an operator called the radial-momentum operator
pr =
h d 1 + i dr r
(3.2.4)
Using this operator we can redefine the differential term of the kinetic energy operator in spherical coordinates as 2
pr 2m
(3.2.5)
For the skeptics out there, the next slide will illustrate the equivalence of Eq. 3.2.5 and the differential term of the kinetic energy operator.
3.2 Hydrogen Atom Proof that the radial-momentum operator is equivalent to the differential term of the kinetic energy operator: differential term of 2 2 d the kinetic energy 2 1 d 2 d 2 d −h 2 r = −h 2 + operator r dr dr dr r dr
d 1 d 1 2 pr = −h 2 + + dr r dr r d2 1 d 1 1 d 1 = −h 2 + − 2+ + 2 r dr r r dr r dr 2 2 d 2 d = −h 2 + r dr dr 2
radial-momentum operator same as the differential term of the kinetic energy operator
3.2 Hydrogen Atom So rewriting the kinetic energy operator in terms of pr results in 2
pr L2 + T= 2m 2mr 2
(3.2.6)
Using Eq. 3.2.6, Schrödinger's Equation is given by
HΨ (r ) = EΨ (r ) pr 2 L2 Ψ ( r ) = EΨ ( r ) ( ) V r + + 2m 2mr 2 pr 2 h 2l (l + 1) Ψ (r ) = EΨ (r ) V ( r ) + + 2 2m 2mr
L2 Ψ = λΨ where
λ = l (l + 1)
(3.2.7)
3.2 Hydrogen Atom Let
Ψ (r ) =
u (r ) r
where u(r) is an polynomial of r
Consider 2
u (r ) d 1 d 1 u (r ) = + + r dr r dr r r 1 d 1 1 du 1 − 2 u (r ) + 2 u (r ) = + r dr r r dr r
pr Ψ ( r ) = pr
2
d 1 1 du d 1 du 1 du = + + 2 = dr r r dr dr r dr r dr 1 du 1 d 2u 1 du 1 d 2u = − 2 + + 2 = 2 r dr r dr 2 r dr r dr
(3.2.8)
3.2 Hydrogen Atom Substituting Ψ (r ) =
u (r ) and Eq. 3.2.8 into Eq. 3.2.7 r
u − h 2 d 2 u h 2l (l + 1) u u ( ) E + + V r = 2mr dr 2 2mr 2 r r r − h 2 d 2u h 2l (l + 1) u + V (r )u = Eu + 2 2 2m dr 2mr
(3.2.9)
Define V(r) (in cgs units) we have
− Zq 2 V (r ) = r
(3.2.10)
where Z is the atomic number, q is the charge of an electron and r is the distance between force center and particle under observation.
3.2 Hydrogen Atom We can take some steps to make Eq. 3.2.9 dimensionless to simplify our analysis. Defining
r = ao ρ
and
E = Eo ε
(3.2.11)
no units Substituting Eq. 3.2.11 and Eq. 3.2.10 into Eq. 3.2.9 results in
− h 2 1 d 2u h 2l (l + 1) Zq 2 + u− u = Eoεu 2 2 2 2 ao ρ 2m ao dρ 2mao ρ
(3.2.12)
3.2 Hydrogen Atom Defining ao and Eo from Eq. 3.2.11 so that Eq. 3.2.12 is dimensionless 2
2
h Zq = = Eo 2 ao mao
h2 ao = mq 2 Z
(3.2.13)
Z 2q 4m Eo = h2
(3.2.14)
Substituting Eq. 3.2.13 and Eq. 3.2.14 into Eq. 3.2.12 we get the dimensionless expression of the Schrödinger's Equation
1 d 2 l (l + 1) 1 + − u = εu − 2 2 2ρ ρ 2 dρ
(3.2.15)
3.2 Hydrogen Atom ao is known as the Bohr radius which is considered to be the effective radius of a Hydrogen atom. To calculate the Bohr radius we must consider the cgs units for energy, length and electric charge. The cgs unit for length is the centimeter.
Conversion between SI and cgs.
100cm = 1m 1erg = 1g ⋅ cm 2 /s 2
The cgs unit for energy is the erg.
107 erg = 1J
The cgs units for electric charge is “electrostatic unit” or esu.
1C = 3 ×109 esu 2
107 erg (1.05 × 10 J ⋅ s) 2 1 J h ao = = 2 9 mq 2 Z 2 3 10 esu 1000 g × (1) 1.6 × 10 −19 C (9.11× 10 −31 kg ) 1C 1kg −34
(
2
)
ao = 5.29 × 10 −9 cm = 5.29 × 10 −11 m
3.2 Hydrogen Atom Now let us assume a power series solution to Eq. 3.2.15 of the form
u ( ρ ) = Ao ρ n e
−
ρ n
+ A1 ρ n ρ 1e
−
ρ n
+ ... + Aq ρ n ρ q e
−
ρ n
+ ...
(3.2.16)
Applying the second derivative to the first term in Eq. 3.2.16, ρ ρ ρ − − − d2 d 1 n n −1 n n n n A e A n e A e ρ ρ ρ = + − o o o dρ dρ 2 n ρ d n −1 ρ n − n = Ao nρ − e dρ n ρ − ρn n −1 n−2 = Ao 2 − 2 ρ + n(n − 1) ρ e n n
(3.2.17)
3.2 Hydrogen Atom Substituting this result into Eq. 3.2.15 and the first term of Eq. 3.2.16, we get
l (l + 1) n − 2 1 ρn 1 n−2 n n n ( 1 ) − − − ρ + ρ = ερ 2 n2 2 2
(3.2.18)
Equating the coefficients of like powers
1 1 n n − ρ = ερ 2 n2 Thus
1 ε =− 2 2n
Z 2q 4m 1 Z 2q 4m E = Eo ε = − 2 = − 2 2 2 h 2h n 2n
(3.2.19)
(3.2.20)
3.2 Hydrogen Atom Substituting the appropriate values for the physical constants into Eq. 3.2.20 and setting Z=1 since we are considering the Hydrogen atom, we get
13.6eV E=− n2
(3.2.21)
Equating the coefficients of ρn-2 leads to
n(n − 1) l (l + 1) = 2 2
(3.2.22a)
Solving for l in terms of n we get
l 2 + l − n(n − 1) = 0 l = n −1
(3.2.22b)
3.2 Hydrogen Atom Now, we have defined three quantum numbers – n, l, m – that are needed to specify the particles motion in a hydrogen atom.
What about the wave function of a particle in a hydrogen atom? To accomplish this we will redefine the power series solution for Schrödinger's Equation
u=ρ
s
∞
∑A ρ q =0
q
q
as
u ( r ) = g ( r )e
−
e
−
r nao
ρ n
(3.2.23)
(3.2.24)
3.2 Hydrogen Atom Also r
r
− − du 1 nao ge nao = g ′e − dr nao 2
d u = g ′′e 2 dr
r − nao
−
2 g ′e nao
r − nao
+
1 ge 2 (nao )
r − nao
(3.2.25)
Substituting the above result into Eq. 3.2.9 where V(r) is defined as Eq. 3.2.10 and E is defined as Eq. 3.2.20, dividing out the exponential terms and simplifying the result, we get
h 2l (l + 1) 2 Zq 2 h2 − g ′ + g− g =0 g ′′ − 2 2m nao 2mr r
(3.2.26)
3.2 Hydrogen Atom g(r) is defined as
g (r ) = r s ∑ Aq r q q
(3.2.27)
So the first and second derivatives of g(r) are
g ′(r ) = sr s −1 Ao + ( s + 1)r s A1 + ... + ( s + q ) Aq r s + q −1 + ... g ′′(r ) = s ( s − 1)r s − 2 + ( s + 1) sA1r s −1 + ...
(3.2.28)
... + ( s + q )( s + q − 1) Aq r s + q − 2 + ... We want to make Eq. 3.2.26 dimensionless so again we substitute Eq. 3.2.11 into Eq. 3.2.26
d 2 g 2 dg l (l + 1) 2 − − g + g =0 2 2 ρ ρ dρ n dρ
(3.2.29)
3.2 Hydrogen Atom Substituting Eq. 3.2.11 into Eq. 3.2.27 and Eq. 3.2.28 and substituting that result into Eq. 3.2.29 gives us
s ( s − 1)ao Ao ρ s − 2 + ... + ( s + q )( s + q − 1)ao s
[
− l (l + 1) Ao ρ s − 2 + ... + Aq ao +
s+q
s+q
Aq ρ s + q − 2 + ...
]
ρ s + q − 2 + ...
[
]
2 s s+q sAo ao ρ s −1 + ... + ( s + q ) Aq ao ρ s + q −1 + ... n s s+q + 2 Ao ao ρ s −1 + ... + Aq ao ρ s + q −1 + ... = 0
[
]
(3.2.30)
3.2 Hydrogen Atom Eq. 3.2.30 must equal zero for all ρ. Therefore the coefficients of the ρs-2 must sum to zero.
s ( s − 1) − l (l + 1) = 0
(3.2.31)
The coefficients of the general term, ρs+q-1 are
2 s+q ( s + q + 1)( s + q) Aq +1ao − Aq ao ( s + q) n s + q +1 s+q − l (l + 1) Aq +1ao + 2 Aq ao = 0 s + q +1
(3.2.32)
3.2 Hydrogen Atom Solving for s in terms of l using Eq. 3.2.31 we find
s = l +1
(3.2.33)
Substituting Eq. 3.2.33 into Eq. 3.2.32 and solving for Aq+1 we have
2 + + − l q ( 1 ) 2 n Aq + 1 = Aq + + + + − + l q l q l l ( 2 )( 1 ) ( 1 )
Note: 1/ao factor is absorbed by the Aq term.
(3.2.34)
3.2 Hydrogen Atom Eq. 3.2.34 can be reexpressed as
2 ( l 1 q ) 2 + + − n Aq +1 = Aq ( q 1 )( q 2 l 2 ) + + +
(3.2.35)
Eq. 3.2.35 can be used to find the coefficients of the higher order terms of Eq. 3.2.23. The general wave function for the Hydrogen atom is defined by Eq. 3.2.36.
Ψ (r , θ , φ ) = R (r )Ylm (θ , φ )
(3.2.36)
Eq. 3.2.35 and Eq. 3.2.23 are used to define R(r) in Eq. 3.2.36 and Ylm(θ,Φ) is defined in Section 3.1.
References •
Brennan, Kevin F, “The Physics of Semiconductors with applications to Optoelectronic Devices.”
• •
http://www.physics.utoronto.ca/~jharlow/teaching/Units.htm Halliday, D., Resnick, R., Walker, “Fundamentals of Physics”
