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22
3.2 Hydrogen Atom Nicholas Brown ECE 6451 Introduction to the Theory of Microelectronics Fall 2005

Transcript of 3.2 Hydrogen Atomalan.ece.gatech.edu › ... › Brown_3p2_HydrogenAtom.pdf · 3.2 Hydrogen Atom (...

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3.2 Hydrogen Atom

Nicholas BrownECE 6451

Introduction to the Theory of MicroelectronicsFall 2005

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3.2 Hydrogen Atom

22

21 mdd

Φ−

φ

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

θλ

θθ

θθ 2

2

sinsin

sin1 Θ

=Θ+

Θ m

dd

dd

In section 3.2, we will solve the radial dependent differential equation and analyze its results in the context of the Hydrogen atom.

(3.2.1)

(3.2.2)

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3.2 Hydrogen Atom

ERRrVRmr

Rr

rrmr

=++

∂∂

∂∂

− )(22 2

22

2

2 hh λ

We will relate our equation to the Hydrogen atom by considering the form of V(r).

The differential equation below is the radial dependent component of the 3-D Schrödinger Equation in spherical coordinates

(3.2.3)

The potential of the Hydrogen atom, V(r) is inversely proportional to rr

rV 1~)(

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3.2 Hydrogen Atom

+=

rdrd

ipr

1h

We will define an operator called the radial-momentum operator

Using this operator we can redefine the differential term of the kinetic energy operator in spherical coordinates as

mpr2

2

(3.2.4)

(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.

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3.2 Hydrogen Atom

+−=

drd

rdrd

drdr

drd

r21

2

222

22 hh

Proof that the radial-momentum operator is equivalent to the differential term of the kinetic energy operator:

differential term of the kinetic energy operator

+−=

++−+−=

+

+−=

drd

rdrd

rdrd

rrdrd

rdrd

rdrd

rdrdpr

2

1111

11

2

22

222

22

22

h

h

h

radial-momentum operator same as the differential term of the kinetic energy operator

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3.2 Hydrogen Atom

2

22

22 mrL

mpT r +=

)()()(2

)1(2

)()()(22

)()(

2

22

2

22

rErrVmrll

mp

rErrVmrL

mp

rErH

r

r

Ψ=Ψ

+

++

Ψ=Ψ

++

Ψ=Ψ

h

Ψ=Ψ λ2L

So rewriting the kinetic energy operator in terms of pr results in

Using Eq. 3.2.6, Schrödinger's Equation is given by

where )1( += llλ

(3.2.7)

(3.2.6)

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3.2 Hydrogen Atom

rrur )()( =ΨLet where u(r) is an polynomial of r

Consider

2

2

22

2

2

2

22

22

1111

1111

)(1)(111

)(11)()(

drud

rdrdu

rdrud

rdrdu

r

drdu

rdrdurdr

ddrdurrdr

d

rur

rurdr

durrdr

drru

rdrd

rdrd

rruprp rr

=

++−=

+

=

+=

+−

+=

+

+==Ψ

(3.2.8)

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3.2 Hydrogen Atomrrur )()( =Ψ

EuurVumrll

drud

m

ruE

rurV

ru

mrll

drud

mr

=++

+−

=++

+−

)(2

)1(2

)(2

)1(2

2

2

2

22

2

2

2

2 2

hh

hh

Substituting and Eq. 3.2.8 into Eq. 3.2.7

(3.2.9)

Define V(r) (in cgs units) we have

rZqrV

2

)( −=

where Z is the atomic number, q is the charge of an electron and ris the distance between force center and particle under observation.

(3.2.10)

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3.2 Hydrogen Atom

ρoar = εoEE =

We can take some steps to make Eq. 3.2.9 dimensionless to simplify our analysis. Defining

no units

Substituting Eq. 3.2.11 and Eq. 3.2.10 into Eq. 3.2.9 results in

(3.2.11)

uEuaZqu

mall

dud

am oooo

ερρρ

=−+

+− 2

22

2

2

2

2

2

2)1(1

2hh

(3.2.12)

and

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3.2 Hydrogen Atom

ooo

Emaa

Zq== 2

22 h Zmqao 2

2h=

2

42

h

mqZEo =

Defining ao and Eo from Eq. 3.2.11 so that Eq. 3.2.12 is dimensionless

(3.2.13)

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

uulldd ε

ρρρ=

++−

12

)1(21

22

2

(3.2.15)

(3.2.14)

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3.2 Hydrogen Atom

m1cm100 =

J1erg10/scmg1erg1

7

22

=

⋅=

esu103C1 9×=

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.

The cgs unit for energy is the erg.

The cgs units for electric charge is “electrostatic unit” or esu.

Conversion between SI and cgs.

( )

m1029.5cm1029.5

)1(C1

esu103106.1kg1

g1000)kg1011.9(

J1erg10)sJ1005.1(

119

2921931

27234

2

2

−−

−−

×=×=

××

×

⋅×

==

o

o

a

CZmq

a h

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3.2 Hydrogen Atom

......)( 11 ++++=

−−−nqn

qnnnn

o eAeAeAuρρρ

ρρρρρρ

Now let us assume a power series solution to Eq. 3.2.15 of the form

(3.2.16)

Applying the second derivative to the first term in Eq. 3.2.16,

nnnn

o

nn

no

nno

nno

nno

ennn

A

en

nddA

en

AenAddeA

dd

ρ

ρ

ρρρ

ρρρ

ρρρ

ρρρ

ρρ

−−−

−−

−−−−

−+−=

−=

−+=

212

1

12

2

)1(2

1

(3.2.17)

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3.2 Hydrogen Atom

nnnn llnnn

ερρρρ=

++−−− −− 22

2 2)1()1(

21

21

nn

nερρ =− 2

121

221n

−=ε

Substituting this result into Eq. 3.2.15 and the first term of Eq. 3.2.16, we get

(3.2.18)

Equating the coefficients of like powers

(3.2.19)

Thus

22

42

22

42

221

nmqZ

nmqZEE o

hh−=

−== ε (3.2.20)

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3.2 Hydrogen Atom

2

eV6.13n

E −=

2)1(

2)1( +=

− llnn

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

(3.2.21)

Equating the coefficients of ρn-2 leads to

(3.2.22b)

Solving for l in terms of n we get

10)1(2

−==−−+

nlnnll

(3.2.22a)

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3.2 Hydrogen Atom

∑∞

=

−=

0q

nqq

s eAuρ

ρρ

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

(3.2.23)

as

onar

ergru−

= )()( (3.2.24)

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3.2 Hydrogen Atom

ooo

oo

nar

o

nar

o

nar

nar

o

nar

gena

egna

egdrud

gena

egdrdu

−−−

−−

+′−′′=

−′=

22

2

)(12

1

Also

(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

02

)1(22

2

2

22

=−+

+

′−′′− g

rZqg

mrllg

nag

m o

hh(3.2.26)

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3.2 Hydrogen Atom

∑=q

qq

s rArrg )(g(r) is defined as

(3.2.27)

So the first and second derivatives of g(r) are

...)1)((...

...)1()1()(

...)(...)1()(

2

11

2

11

1

+−+++

+++−=′′

++++++=′

−+

−−

−+−

qsq

ss

qsq

so

s

rAqsqs

rsAsrssrg

rAqsArsAsrrg(3.2.28)

We want to make Eq. 3.2.26 dimensionless so again we substitute Eq. 3.2.11 into Eq. 3.2.26

02)1(222

2

=++

−− ggllddg

ndgd

ρρρρ (3.2.29)

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3.2 Hydrogen AtomSubstituting 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

[ ][ ][ ] 0......2

...)(...2

......)1(

...)1)((...)1(

11

11

22

22

=++++

+++++

++++−

+−++++−

−++−

−++−

−++−

−++−

qsqsoq

ssoo

qsqsoq

ssoo

qsqsoq

so

qsq

qso

so

so

aAaA

aAqsasAn

aAAll

AaqsqsAass

ρρ

ρρ

ρρ

ρρ

(3.2.30)

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3.2 Hydrogen Atom

0)1()1( =+−− llss

Eq. 3.2.30 must equal zero for all ρ. Therefore the coefficients of theρs-2 must sum to zero.

(3.2.31)

The coefficients of the general term, ρs+q-1 are

02)1(

)(2))(1(

11

11

=++−

+−+++

++++

++++

qsoq

qsoq

qsoq

qsoq

aAaAll

qsaAn

aAqsqs(3.2.32)

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3.2 Hydrogen Atom

1+= ls

Solving for s in terms of l using Eq. 3.2.31 we find

Substituting Eq. 3.2.33 into Eq. 3.2.32 and solving for Aq+1 we have

(3.2.33)

+−++++

−++=+ )1()1)(2(

2)1(2

1llqlql

qlnAA qq

(3.2.34)

Note: 1/ao factor is absorbed by the Aq term.

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3.2 Hydrogen Atom

+++

−++=+ )22)(1(

2)1(2

1 lqq

qlnAA qq

Eq. 3.2.34 can be reexpressed as

(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.

),()(),,( φθφθ lmYrRr =Ψ (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.

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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”