Modern Physics 342

References :1. Modern Physics by Kenneth S. Krane. , 2nd Ed.

John Wiley & Sons, Inc.2. Concepts of Modern Physics by A. Beiser, 6th

Ed. (2002), McGraw Hill Com.3. Modern Physics for Scientists and Engineers by

J. Taylor, C. Zafiratos and M. Dubson, 2nd Ed, 2003.

Chapters: 5 (Revision), 6 (6.4), 7, 8, 10, 11 and 12

Ch. 5 (Revision)

Schrödinger Equation

Schrödinger Equation Requirements

1. Conservation of energy is necessary:

EUK

Kinetic energyPotential energy

Total energy

The kinetic energy K is conveniently given by

)2(m2

Pmv

2

1K

22

Where P is the momentum = m v

Schrödinger Equation Requirements (continued)

2. Consistency with de Broglie hypothesis

kPandh

P

Where, λ and k are, respectively, the wavelength and the wave number.

)3(m2

)k(K

2

Schrödinger Equation Requirements (continued)

3. Validity of the equation

The solution of this equation must be valid everywhere, single valued and linear.

By linear we meant that the equation must allow de Broglie waves to superimpose properly.

The following is a mathematical form of the wave associating the particle.

)4()xksin(A)x(

To make sure that the solution is continuous, its derivative must have a value everywhere.

)xkcos(kAdx

d

)5(k)xksin(kAdx

d 222

2

22 Km2

k

)6(Km2

dx

d22

2

Time-Independent Schrödinger Equation

)UE(m2

dx

d22

2

)7(EUdx

d

m2 2

22

Probability, Normalization and Average

• The probability density is given by

Which is the probability of finding a particle in the space dx.

The probability of finding this particle in a region between x1 and x2 is

dx)x(dx)x(P2

dx)x()x(P2

1

x

x

2

By normalization we mean the total probability allover the space is 1, so that,

1dx)x(

2

dx)x(

dxx)x(

dx)x(p

dxx)x(px

2

2

dxx)x(x2

The mean value ( the expectation value) of x,

If the wave function is normalized, therefore,

Applications

• The Free Particle ( a particle is moving with no forces acting on it)• U(x)= constant anywhere=0 (arbitrarily)

The solution of such deferential equation is given in the form

Ψ(x)=A sin(kx)+B cos(kx)

Edx

d

m2 2

22

22

2 mE2

dx

d

22

2

kdx

d2

2 mE2k

(8)

Particle in a one dimensional box

Lx0at)xkcos(B)xksin(A

Lxatand0xat0)x(

Finding A

Ψ(0)=A sin(kx) + B cos(kx)=0 and for this, B=0Therefore,

ψ(x)=A sin(kx) (9)

Ψ(L)=0, therefore

A sin(kL)=0 since A≠0, sin(kL)=0

kL= , 2 , 3 , .. .., p p p npL

nk

1)xL

n(sinA

L

0

22

LA

n

2

xL

n

4

)xL

n2sin(

2

L

0

L

2A

Normalization condition

(10)

The wave function is now given by,

The energy is given by using (8) and (10) together,

)xL

nsin(

L

2)x(

(11)

22 mE2

)L

n(

2

222

n mL2

nE

(12)

The ground state (lowest ) energy EO is given by

2

22

O mL2E

We used n=1

The allowed energies for this particle are

,.....E9,E4,EE OOOn

The wave function is shown here for n=1 and n=2

Ψ(x)

x

n=1

n=2

Example 5.2 P149An electron is trapped in a one-dimensional region of length 1X10-10 m. How much energy must be supplied to excite the electron from the ground state to the first excited state? In the ground state, what is the probability of finding the electron in the region from 0.09 X 10-10 m to 0.11 X 10-10 m? In the first excited state, what is the probability of finding the electron between x=0 and x=0.25 X 10-10 m?

Example 5.3 P151Show that the average value of x is L/2, for a particle in a box of length L, independent of the quantum state (not quantized).

)xL

nsin(

L

2)x(

Since the wave function is

And the average value is defined by

dxx)x(x2

L

0

2 dxx)]xL

nsin(

L

2[x

2

L

4

x

)L

n(8

)xL

n2cos(

L

n4

)xL

n2sin(x

L

2x

L

0

2

2

A particle in a two dimensional box

The Schrödinger equation in two dimensions is

)y,x(E)y,x()y,x(Uy

)y,x(

x

)y,x(

m2 2

2

2

22

U(x,y)=0 inside the box (0≤x≤L) & (0≤y≤L)

U(x,y)=∞ outside the box

The wave function ψ(x,y) is written as a product of two functions in x and y,

)y(g)x(f)y,x(

xkcosBxksinA)x(f xx

ykcosByksinA)y(g yy Since ψ(x,y) must be zero at the boundaries,

ψ(0,y) =0 ψ(L,y) =0 ψ(x,0) =0 ψ(x,L) =0

Therefore, A sin kx (0)+ B cos kx (0)=0 which requires B=0

xksinA)x(f xIn the same way

yksinC)y(g y

For x=L, f(x)=0 and y=L, g(y)=0

0LksinA x This requires kx L=nx p with n=1,2,3

xL

nsinA)x(f x and y

L

nsinC)y(g y

To find the constant A’, the wave function should be normalized

yL

nsinx

L

nsin'A)y(g)x(f)y,x( yx

1dydx)yL

nsinx

L

nsin'A( 2yx

L

0

L

0

This integration gives L

2'A

yL

nsinx

L

nsin

L

2)y,x( yx

The energy states of a particle in a two dimensional box

Substituting about the wave function ψ(x,y) in Schrödinger equation, we find

E)yL

nsin()x

L

nsin(

L

2)nn()y

L

nsin()x

L

nsin(

L

2

m2yx2

y2x

yx3

22

Which after simplification becomes

)nn(L

2

m2E 2

y2x3

22

Chapter 7The Hydrogen Atom Wave Functions

• The Schrödinger Equation in Spherical CoordinatesThe Schrödinger equation in three dimensions is

The potential energy for the force between the nucleus and the electron is

This form does not allow to separate wave function Ψ into functions in terms of x, y and z, so we have to express the whole equation of Schrödinger in terms of spherical coordinates, r, θ, and φ.

E)z,y,x(Uzyxm2 2

2

2

2

2

22

222

2

O

2

O zyx

e

4

1

r

e

4

1U

Cartesian and spherical coordinates

θ

φ

r

x

y

z

electron

r sin θ

r sin θ co

s φ

r cos

θ

r sin θ sin φ

x= r sin θ cos φy= r sin θ sin φz= r cos θAnd Schrödinger equation becomes

E),,r(Usinr

1sin

sinr

1

rr

2

rm2 2

2

2222

22

),,r(

)()()r(R),,r(

This wave function can be written in terms of 3 functions in their corresponding variables, r, θ and φ

Hydrogen wave functions in spherical coordinates

R(r) is called radial functionΘ(θ ) is called polar function and Φ(φ) is called azimuthal functionwhen solving the three differential equations in R(r), Θ(θ ) and Φ(φ), l and ml quantum numbers were obtained in addition to the previous principal quantum number n obtained before.

n

1

2

2

2

n the principal quantum number 1, 2, 3, …l angular momentum quantum number 0, 1, 2, ……±(n-1)ml magnetic quantum number 0, ±1, ±2, ……± l

The energy levels of the hydrogen atom

222O

2

4

n n

1

32

emE

The allowed values of the radius r around the nucleus are given by

22

2O

n nme

4r

2

2O

O me

4a

Bohr radius (r at n=1) is denoted by aO and is given by

The Radial Probability Density P(r)

The radial probability density of finding the electron at a given location is determined by

2

,n2 )r(Rr)r(P

The total probability of finding the electron anywhere around the nucleus is

0 0

2

,n2 dr)r(Rrdr)r(P

The limits of the integration depend on the conditions of the problem

Example 7.1 Prove that the most likely distance from the origin of an electron in the n=2, l=1 state is 4aO .

At n=2 and l =1, R2,1 (r) is given by

Oa2

r

O2/3

O

1,2 ea

r

)a2(3

1)r(R

The most likely distance means the most probable position. The maximum value of the probability is obtained if r=4aO .To prove that, the first derivative of P(r) with respect to r is zero at this value.

2

a2

r

O2/3

O

2 Oea

r

)a2(3

1r)r(P

0ea

r

)a2(3

1r

dr

d2

a2

r

O2/3

O

2 O

0a24

er

a6

er6O

a

r

4

5O

a

r

3 OO

Simplifying this result we get

Oa4r

Example 7.2 An electron in the n=1, l=0 state. What is the probability of finding the electron closer to the nucleus than the Bohr radius aO ?

The probability is given by

0 0

2

,n2 dr)r(Rrdr)r(P

Oa/r

2/3O

0,1 ea

2)r(R

323.0

ea

r2

a

r21dre

a

2r

OO

OO

ar

0r

a

0

a

r2

2O

2

O

2

a

r

2

3

o

2

32.3 % of the time the electron is closer than 1 Bohr radius to the nucleus.

Angular Momentum

)1(L

mL

We discussed the radial part R(r) of Schrodinger equation. In this section we will discuss the angular parts of the Schrodinger equation.

The classical angular momentum vector is given by prL

During the variables separation of wave functions in Schrodinger equation, angular momentum quantum number l was produced. The length of the angular momentum vector L is given by

The z-components of L are given by

where ml is the magnetic quantum number 0, ±l

The angular momentum vector components

For l=2, ml =0, ±1, ±2

The angle is given by

)1(

m

)1(

mcos

Intrinsic Spin

Angular momentum vector

Magnetic moment due to electric current i

Ai 2rA

T

qi

v

r2T

mvpmomentumlinear

prm2

q

Using q=-e the charge of the electron, and rp= L , we get

Lm2

eL

The negative singe indicates that µL and L work in opposite directions.

When the angular momentum vector L is inclined to the direction of the z-axis, the magnetic moment µL has a z-component given by

mm2

eL

m2

ezz,L

MgnetonBohrm2

eB

Bz,L m

T

J10X274.9 24

B

Remember, ml =0, ±l

An electric dipole in a uniform and non-uniform electric fieldA magnetic dipole in a non-uniform magnetic field

The electric dipole has its moment p rotates to align with the direction of the electric field

Two opposite dipoles in the same non-uniform electric field are affected by opposite net forces that lead to displacing each dipole up and down according to their respective alignments.

Similarly, the magnetic dipoles are affected in the same way.

When an electron with an angular momentum inclined to the magnetic filed, it may move up or down according to the direction of rotation around the nucleus.

A beam of hydrogen atoms is in the n=2, l= 1 state. The beam contains equal numbers of atoms in the ml = -1, 0, and +1 states. When the beam passes a region of non-uniform magnetic field, the atoms with ml =+1 experience a net upward force and are deflected upward, the atoms with ml =-1 are deflected downward, while the atoms with ml =0 are undeflected.

Stern-Gerlach Experiment

After passing through the field, the beam strikes a screen where it makes a visible image. 1. When the filed is off, we expect to see one image of the slit in the center of

the screen2. When the field is on, three images of the slit on the screen were expected –

one in the center, one above the center (ml =+1 ) and one below (ml =-1). The number of images is the number of ml values = 2l+1= 3 in our example.

• In the Stern - Gerlach experiment, a beam, of silver atoms is used instead of hydrogen.

• While the field is off, and instead of observing a single image of the slit, they observed two separate images.

• A new quantum number is introduced, which is specifies the electron, the spin quantum number s. It may have two values ±½. The magnitude of its vector is given by

)1s(s S Its z component to the magnetic field direction is

mSz

The experiment

Example 7.6

In a Stern – Gerlach type of experiment, the magnetic field varies with distance in the z direction according to The silver atoms travel a distance x=3.5 cm through the magnet. The most probable speed of the atoms emerging from the oven is v=750 m/s. Find the separation of the two beams as they leave the magnet. The mass of a silver atom is 1.8 X 10-25 kg, and its magnetic moment is about 1 Bohr magneton.

mm

T4.1

dz

dBz

3.5 cm

vO =750 m/s. The force applied to the beam must be obtained

The force is the change of potential energy U with distance z. The potential energy U is given by

zzBU Bμdz

dB

dz

dUF z

zz Bz

m

J10X29.1)

m

T10X4.1()

T

J10X27.9(F 20324

z

This is the vertical force due to the effect of the magnetic field on the electron magnetic dipoles.Using the law of motion at constant acceleration, D z=vO t + ½ a t2. The initial vertical speed was zero before the effect of the magnetic field. We can find the vertical deflection D z above the horizontal level of the beam.

24

25-

20

z

s

m10X7

kg 10 X1.8m

J10X29.1

m

Fa

s10X6.4

s

m750

m10X5.3

v

xt 5

2

D z= ½ a t2 =7.5X10-6 m, the beam separation is 2(7.5X10-5 m)=1.6X10-4 m

Energy Levels and Spectroscopic NotationThe notation for the quantum state of an electron is now described by the four quantum numbers n, l, ml and ms .

For example, the ground state of the hydrogen is labeled as (n,l,ml,ms )=(1,0,0,±½) which means the are two quantum states that can be occupied by the electron; (1,0,0,+½) or (1,0,0,-½).Degeneracy of the atomic levelsThe ground state energy level is now degenerated into two quantum states.The 1st excited state of the hydrogen atom has (n,l,ml,ms)= (2,1,1,+½), (2,1,1,-½), (2,1,0,+½), (2,1,0,-½), (2,1,-1,+½), (2,1,-1,-½), (2,0,0,+½), (2,0,0,-½). The degeneracy is 8, or 2n2 , where n is the principal quantum number.

Some degenerate states after spin quantum number

n=1,2,.. l=0,1,2, ..,n-1 ml =0,±l ms =±½ Degeneracy 2n2

1 0 0+½

2-½

2

0 0+½

8

-½

1

1+½-½

0+½-½

-1+½-½

Spectroscopic Notation

Magnetic quantum numbers, ml and ms have no necessity to be mentioned unless a magnetic field is applied to the atom. In normal cases, a certain notation is used to specify the energy levels of the electron in the atom. Also, it is important to specify the number of electrons occupying such state. This notation depends on the l values as follows:

Value of l 0 1 2 3 4 5

notation s p d f g h

For the electron in the n=1 level, it is in the state s and the level occupied by this electron is denoted as (1s1 )or generally (ns1 )

Electronic configuration of some elements

element Number of electrons

n 1 2 3

l 0 0 1 0 1 2

H 1 1s1

He 2 1s2

Li 3 1s2 2s1

Be 4 1s2 2s2

B 5 1s2 2s2 2p1

Ne 6 1s2 2s2 2p2

Na 11 1s2 2s2 2p6 3s1

K 18 1s2 2s2 2p6 3s2 3p6

Selection rules

Transitions between energy states of the atom are governed by the condition of Dl=±1.The transition from 4s state is possible to the 3p state because Dl=+1 , but not possible between 4s and 3s, 2s or 1s.In addition to this condition, there is a rule against ml differences (D ml ).

1D

1,0m D

n

43

2

1

l 0 1 2 3

s p d f

Zeeman EffectA hydrogen atom is prepared in a 2p (l =1) level, and is placed in an external uniform magnetic field B. The orbital angular momentum magnetic moment L interacts with the field and the potential energy of this interaction U is given by

Bμ LU BmBμU Bz,L

In addition to this, there is the energy of this level, EO , for example. So on turning the magnetic field on, the energy of this electron is given by

BmEE BO

Since ml in this example has the values +1, 0, and -1, there will be 3 energy values, 3 different wave lengths emitted.

BmEE BO OEE BmEE BO

Wavelength notation

chE D

D

22

chEd

chdE

BmE BD Ech

2

D

D Bmch B

2

D

+1 0 -1

0

0

D Bch B

2

Bch B

2

BB BBE

D D

Problem 22 p 233A hydrogen atom is in an excited 5g state, from which it makes a series of transitions, ending in the 1s state. Show on an energy levels diagram the sequence of transitions that can occur. Repeat the last steps if the atom begins in the 5d state.

1s

2s

3s

5g5f5p 5d5s

4f4d4s 4p

3p 3d

2p

1s

2s

3s

5g5f5p 5d5s

4f4d4s 4p

3p 3d

2p

Problem 23 p 233Consider the normal Zeeman effect applied to 3d to 2p transition. (a) sketch an energy-level diagram that shows the splitting of the 3d and 2p levels in an external magnetic field. Indicate all possible transitions from each ml state of the 3d level to each ml state of the 2p level. (b) which transitions satisfy the Dml =±1 or 0 selection rule? (c) Show that there are only three different transitions energies emitted.

3d

2p

ml

+2

+1

0

-1

-2

0+1

-1

Since there are three different values of Dml namely +1, 0 and -1, there will be three different energies emitted. Let the energy of the 2p state at ml=0 be denoted by Eop and that of the 3d state at ml=0 be denoted by Eod .

The following equation can be used to calculate any energy released from allowed transitions.

Bm)EE(E BOpOd DD

The difference Eod –Eop is constant, while Dml has 3 different values, therefore, DE has only three different values.

Many Electron Atoms

Pauli Exclusion PrincipleIt was believed that different atoms in the ground states have all their electrons dropped down in the 1s state. This means they all must have the same physical properties. This is not the case, in fact.A conclusion was drawn by Pauli that states that:

No two electrons in a single atom can have the same set of quantum numbers (n, l, ml , ms).

Inert GasesAlkali Metals

Alkaline Earth Metals Nonmetals Halogens

Transition Elements (Heavy Metals)

Lanthanide Series

Actinide Series