Lecture Physics A2: Atomic physics

Atomic Physics

Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology

History of Atomic Model

• Proposed an Atomic Theory which states that

all atoms are small, hard, indivisible and indestructible particles made of a single material formed into diﬀerent shapes and sizes.

• Aristotle did not support his atomic theory.

Democritus (460 BC - 370 BC)

History of Atomic Model

• Known as the “Father of Modern Chemistry”

• Was the ﬁrst person to generate a list of 23

elements in his textbook

• Devised the metric systems

• Was married to a 13-year-old Marie-Anne

Pierette Paulze, who assisted him much of his work

• Discovered/proposed that combustion occurs when oxygen combines with other elements

Antoine Lavoisier (1743 - 1794)

• Discovered/proposed the Law of

Conservation of Mass (or Matter) which states that, in a chemical reaction, matter is neither created or destroyed

History of Atomic Model

• In 1803, he proposed an Atomic Theory which

states that:

• All substances are made of atoms; atoms

are small particles that cannot be created, divided or destroyed.

• Atoms of the same element are exactly

alike, and atoms of diﬀerent elements are diﬀerent

• Atoms join with other atoms to make new

substances

John Dalton (1766 - 1844)

• He calculated the atomic weights of many

various elements

• He was a teacher at a very young age

• He was color blind

History of Atomic Model J. J. Thomson (1856 - 1940)

• He proved that an atom can be divided into

smaller parts

• While experimenting with cathode-ray tubes,

discovered corpuscles, which were later called electrons

• He stated that the atom is neutral

History of Atomic Model J. J. Thomson (1856 - 1940)

• He proved that an atom can be divided into

smaller parts

• While experimenting with cathode-ray tubes,

discovered corpuscles, which were later called electrons

• He stated that the atom is neutral

• In 1897, he proposed Plum

Pudding Model which states that atoms mostly consist of positively charged material with negatively charged particles (electrons) located throughout the positive material

• He won the Nobel Prize

History of Atomic Model Ernest Rutherford (1871 - 1937)

• In 1909, he performed Gold Foil Experiment

and suggested the characteristics of the atom:

• It consists of a small core, or nucleus, that

contains most of the mass of the atom

• This nucleus is made up of particles called protons, which have a positive charge

• The protons are surrounded by negative charged electrons, but most of atom is actually empty space

• He did extensive work on radioactivity (alpha, beta particles, gamma rays/waves) and was referred as the “Father of Nuclear Physics”

• He won the Nobel Prize

• He was a student of J.J. Thomson

History of Atomic Model Niels Bohr (1885 - 1962)

• In 1913, he proposed the Bohr Model, which suggests that electrons travel around the nucleus of an atom in orbits for deﬁnite paths. Additionally, the electron can jump from a path in one level to a path in another level (depending on their energy)

• He won the Nobel Prize

• He used to work with Ernest Rutherford

History of Atomic Model Erwin Schrodinger (1887 - 1961)

• In 1913, he further explained the nature of

electrons in an atom by stating that the exact location of an electron cannot be stated; therefore it is more accurate to view electrons in region called electron clouds.

• Electron clouds are places where the electrons

are likely to be found

• He did extensive work on the Wave formula ➔

Schrodinger equation.

• He won the Nobel Prize

History of Atomic Model James Chadwick (1891 - 1974)

• In 1932, he realized that the atomic mass of most elements was double the number of protons —> discovery of the neutron

• He used to work with Ernest Rutherford

• He won the Nobel Prize

Quantum Mechanical Atomic Theory

Schrödinger Equation in Three Dimensions

• Electrons in an atom can move in all three dimensions of space. If a

particle of mass m moves in the presence of a potential energy function U(x,y,z), the Schrödinger equation for the particle’s wave function ψ(x,y,z) is

+

~2 2m

@2 (x, y, z) @x2

@2 (x, y, z) @y2

@2 (x, y, z) @z2

✓

◆ +U (x, y, z) (x, y, z) = E (x, y, z)

2 + U (x, y, z)

(x, y, z) = E (x, y, z)

~ 2m r



• This is a direct extension of the one-dimensional Schrödinger equation

~2 2m

@2 (x) @x2 + U (x) (x) = E (x)

CHAPTER 41 Atomic Structure

1366

e-iEt

both sides by the factor

, leaving the time-independent Schrödinger equa-

tion in three dimensions for a stationary state:

x, y, z

+ U

0x 2

0y2

0z2

(41.5)

x, y, z

(three-dimensional time-independent

= Ec

Schrödinger equation)

The probability distribution function for a stationary state is just the square

e-iEt

of the absolute value of the spatial wave function:

x, y, z

ƒc

Uƒ

e+iEt

e-iEt

x, y, z

. Note that this doesn’t depend on

U = ƒc

time. (As we discussed in Section 40.1, that’s why we call these states stationary.)

Hence for a stationary state the wave function normalization condition, Eq. (41.3),

becomes

(normalization condition for a

(41.6)

x, y, z

2 dV = 1

ƒc

stationary state in three dimensions)

We won’t pretend that we have derived Eqs. (41.2) and (41.5). Like their one-

dimensional versions, these equations have to be tested by comparison of their

predictions with experimental results. Happily, Eqs. (41.2) and (41.5) both pass

this test with ﬂying colors, so we are conﬁdent that they are the correct equations.

An important topic that we will address in this chapter is the solutions for

Eq. (41.5) for the stationary states of the hydrogen atom. The potential-energy

function for an electron in a hydrogen atom is spherically symmetric; it depends

x 2

only on the distance

from the origin of coordinates. To

r =

+ y2

+ z2

take advantage of this symmetry, it’s best to use spherical coordinates rather than

the Cartesian coordinates

x, y, z

to solve the Schrödinger equation for the hydro-

gen atom. Before introducing these new coordinates and investigating the hydro-

gen atom, it’s useful to look at the three-dimensional version of the particle in a

box that we considered in Section 40.2. Solving this simpler problem will give us

insight into the more complicated stationary states found in atomic physics.

Test Your Understanding of Section 41.1 In a certain region of space the

potential-energy function for a quantum-mechanical particle is zero. In this region the

wave function

for a certain stationary state satisﬁes

x, y, z

0x 2

7 0

0y2

7 0

and

. The particle has a deﬁnite energy E that is positive. What can you con-

0z2

7 0

clude about

x, y, z

in this region? (i) It must be positive; (ii) it must be negative; (iii) it

❙

must be zero; (iv) not enough information given to decide.

41.2 Particle in a Three-Dimensional Box

x = L .

, and

41.1 A particle is conﬁned in a cubical x = 0 y = 0 box with walls at , y = L

z = 0

Particle in a Three-Dimensional Box , z = L

z 5 L

0 … y … L

0 … x … L

x, y, z

x 5 L x

y 5 L

ƒc

Consider a particle enclosed within a cubical box of side L. This could represent an electron that’s free to move anywhere within the interior of a solid metal cube but cannot escape the cube. We’ll choose the origin to be at one corner of the box, • In analogy with our inﬁnite square potential with the x-, y-, and z-axes along edges of the box. Then the particle is conﬁned to (U(x) = 0 inside, U(x) = ∞ outside), let us 0 … z … L the region (Fig. 41.1). What are the station- ary states of this system? consider a three-dimensional region space As for the particle in a one-dimensional box that we considered in Section (box) of equal sides of length L, with the same 40.2, we’ll say that the potential energy is zero inside the box but inﬁnite outside. c must be zero outside the box in order Hence the spatial wave function potential (U(x,y,z) = 0 inside, U(x,y,z) = ∞ x, y, z U x, y, z in the time-independent Schrödinger equation, that the term 1 2 2 x, y, z Eq. (41.5), not be inﬁnite. Hence the probability distribution function outside). ƒ is zero outside the box, and there is zero probability that the particle will be found

• We will consider the wave function as separable, that is can be written

as a product of the three independent dimensions x, y and z:

(x, y, z) = X(x)Y (y)Z(z)

• The Schrödinger equation inside the box becomes

@2X(x)

Y (y)Z(z)

= EX(x)Y (y)Z(z)

~2 2m

@x2 + Z(z)X(x)

@2Y (y) @y2 + X(x)Y (y)

@2Z(z) @z2



• OR dividing by X(x)Y(y)Z(z), we have:

@2X(x)

= E

~2 2m

1 X(x)

@x2 +

1 Y (y)

@2Y (y) @y2 +

1 Z(z)

@2Z(z) @z2



Particle in a Three-Dimensional Box • This says that the total energy is contributed by three terms on the left, each depending separately on x, y and z. Let us write E = Ex + Ey + Ez. Then this equation can be separated into three equations:

@2X(x)

~2 2m

@2Y (y) @y2 = EyY (y)

~2 2m

@x2 = ExX(x)

~2 2m

@2Z(z) @z2 = EzZ(z)

• These obviously have the same solutions separately as our original

particle in an inﬁnite square well, and corresponding energies

(nx = 1,2,3,…)

Xnx = Cxsin

Ex =

nx⇡x L

⌘

(ny = 1,2,3,…)

Xny = Cysin

Ey =

⇣ ny⇡y L

⇣

⌘

(nz = 1,2,3,…)

Xnz = Czsin

Ez =

nz⇡z L

x⇡2h2 n2 2mL2 n2 y⇡2h2 2mL2 n2 z⇡2h2 2mL2

⇣

⌘

Particle in a Three-Dimensional Box

• A particle’s wave function is the product of these three solutions:

(x, y, z) = X(x)Y (y)Z(z) = Csin

sin

nx⇡x L

ny⇡y L

nz⇡z L

⇣

⌘

⇣

⌘

⇣

⌘

• We can use the three quantum numbers nx, ny and nz to label the

stationary states (state of deﬁnite energy). Here is an example of a particle in three possible states (nx, ny, nz) = (2, 1, 1); (1, 2, 1) or (1, 1, 2)

• The three states shown here are degenerate: although they have diﬀerent values of nx, ny, nz, they have the same total energy E:

41.2 Particle in a Three-Dimensional Box

1369

x, y, z

for

equal to (a) (2, 1, 1), (b) (1, 2, 1), and (c) (1, 1, 2). The value

ƒ cnX,nY,nZ

n X, n Y, n Z is proportional to the density of dots. The wave function is zero on the walls of the box and on the midplane of the box, so

ƒ c ƒ 2

4⇡2h2 2mL2 +

⇡2h2 2mL2 +

⇡2h2 2mL2 =

3⇡2h2 mL2

at these locations.

41.2 Probability distribution function of ƒ c ƒ

E = Ex + Ey + Ez = 2 = 0

(b) !c1, 2, 1!2

(a) !c2, 1, 1!2

2, 1, 1

spots” where there is zero probability of ﬁnding the particle. As an example, con- . From Eq. (41.15), the probability distri- sider the case

n X, n Y, n Z

bution function for this case is

x, y, z

2 sin2 2px

sin2 py

sin2 pz

ƒc2,1,1

= ƒCƒ

As Fig. 41.2a shows, the probability distribution function is zero on the plane

2px

sin2

, where

. The particle is most likely to be

x = L

= sin2p = 0

found near where all three of the sine-squared functions are greatest, at

x, y, z

4, L

2, L

x, y, z

4, L

2, L

. Figures 41.2b and

1, 2, 1

41.2c show the similar cases

and

n X, n Y, n Z

1, 1, 2

. For higher values of the quantum numbers

, and

there are

n X

n Y

n Z

additional planes on which the probability distribution function equals zero, just

as the probability distribution function

for a one-dimensional box has

ƒc

more zeros for higher values of n (see Fig. 40.12).

Example 41.1

Probability in a three-dimensional box

x, y, z

integral of the probability distribution function

(a) Find the value of the constant C that normalizes the wave func-

ƒcnX, nY, nZ

over the volume within the box equals 1. (The integral is actually

tion of Eq. (41.15). (b) Find the probability that the particle will be

over all space, but the particle-in-a-box wave functions are zero

found somewhere in the region

(Fig. 41.3) for the

0 … x … L

outside the box.)

cases (i)

, (ii)

1, 2, 1

2, 1, 1

(n X, n Y, n Z

The probability of ﬁnding the particle within a certain volume

and (iii)

3, 1, 1

n X, n Y, n Z) =

within the box equals the integral of the probability distribution

function over that volume. Hence in part (b) we’ll integrate

SOLUTION

for the given values of

over the

x, y, z

n X, n Y, n Z

ƒcnX, nY, nZ

IDENTIFY and SET UP: Equation (41.6) tells us that to normalize

volume

0 … x … L

0 … y … L

0 … z … L

the wave function, we have to choose the value of C so that the

EXECUTE: (a) From Eq. (41.15),

x, y, z

2 sin2 n Xpx

sin2 n Ypy

sin2 n Zpz

41.3 What is the probability that the particle is in the dark-

ƒcnX,nY,nZ

= ƒCƒ

colored quarter of the box?

Hence the normalization condition is

z 5 L

2 dV

x, y, z

ƒcnX,nY,nZ

x = L

y = L

z = L

dx dy dz

sin2 n Xpx

sin2 n Ypy

sin2 n Zpz

= ƒCƒ

x = 0 L

y = 0 L

z = 0

x = L

y = L

sin2 n Xpx

sin2 n Ypy

= ƒCƒ

x = 0

y = 0

x 5 L

b a

z = L

y 5 L

sin2 n Zpz

= 1

z = 0

x 5 L/4

Continued

Energy Degeneracy

• For a particle in a three-dimensional box, the allowed energy levels are surprisingly complex. To ﬁnd them, just count up the diﬀerent possible states.

41.2 Particle in a Three-Dimensional Box

1371

• Here are the ﬁrst six states for an equal-side box:

6-fold degenerate

E1, 1, 1

(3, 2, 1), (3, 1, 2), (1, 3, 2), (2, 3, 1), (1, 2, 3), (2, 1, 3)

14 3

with that energy.

E1,1,1 =

nX, nY, nZ

not degenerate

3⇡2h2 2mL2

(2, 2, 2)

3-fold degenerate

(3, 1, 1), (1, 3, 1), (1, 1, 3)

4E1, 1, 1 11 E1, 1, 1 3

41.4 Energy-level diagram for a particle in a three-dimensional cubical box. We label each level with the quantum numbers of the states Several of the levels are degenerate (more than one state has the same energy). The lowest (ground) level,

nX, nY, nZ

, has energy

1, 1, 1

E1,1,1 =

3-fold degenerate

(2, 2, 1), (2, 1, 2), (1, 2, 2)

3E1, 1, 1

; we

2mL2

+ 12

= 3p2

• If the length of sides of

12 1 show the energies of the other levels as multiples of

E1,1,1

3-fold degenerate

box are diﬀerent:

(2, 1, 1), (1, 2, 1), (1, 1, 2)

2E1, 1, 1

not degenerate

(1, 1, 1)

E1, 1, 1

E =

n2 z L2

⇡2h2 2m

n2 x L2 x

n2 y L2 y

z !

E 5 0

(break the degeneracy)

Since degeneracy is a consequence of symmetry, we can remove the degener- acy by making the box asymmetric. We do this by giving the three sides of the . If we repeat the steps that we followed to box different lengths

, and

solve the time-independent Schrödinger equation, we ﬁnd that the energy levels

are given by

nX = 1, 2, 3, Á ; nY = 1, 2, 3, Á ;

EnX,nY,nZ =

2 +

(41.17)

nZ = 1, 2, 3, Á

(energy levels, particle in a three-dimensional box with sides

of length

, and

LZ)

, and

are all different, the states

2, 1, 1

1, 2, 1

nX, nY, nZ

and

1, 1, 2

have different energies and hence are no longer degenerate. Note

that Eq. (41.17) reduces to Eq. (41.16) if the lengths are all the same (

LX =

LY = LZ = L

Let’s summarize the key differences between the three-dimensional particle in

a box and the one-dimensional case that we examined in Section 40.2:

• We can write the wave function for a three-dimensional stationary state as

a product of three functions, one for each spatial coordinate. Only a single

function of the coordinate x is needed in one dimension.

•

In the three-dimensional case, three quantum numbers are needed to

describe each stationary state. Only one quantum number is needed in the

one-dimensional case.

• Most of the energy levels for the three-dimensional case are degenerate:

More than one stationary state has this energy. There is no degeneracy in

the one-dimensional case.

• For a stationary state of the three-dimensional case, there are surfaces on

which the probability distribution function

is zero. In the one-

ƒcƒ

dimensional case there are positions on the x-axis where

is zero.

ƒcƒ

We’ll see these same features in the following section as we examine a three-

dimensional situation that’s more realistic than a particle in a box: a hydrogen atom

in which a negatively charged electron orbits a positively charged nucleus.

Test Your Understanding of Section 41.2 Rank the following states

of a particle in a cubical box of side L in order from highest to lowest energy:

; (ii)

; (iii)

(i)

2, 3, 2

4, 1, 1

nX, nY, nZ

❙

; (iv)

2, 2, 3

1, 3, 3

nX, nY, nZ

CHAPTER 41 Atomic Structure

1372

41.3 The Hydrogen Atom

Let’s continue the discussion of the hydrogen atom that we began in Chapter 39.

In the Bohr model, electrons move in circular orbits like Newtonian particles, but

with quantized values of angular momentum. While this model gave the correct

energy levels of the hydrogen atom, as deduced from spectra, it had many con-

ceptual difﬁculties. It mixed classical physics with new and seemingly contradic-

tory concepts. It provided no insight into the process by which photons are

emitted and absorbed. It could not be generalized to atoms with more than one

electron. It predicted the wrong magnetic properties for the hydrogen atom. And

perhaps most important, its picture of the electron as a localized point particle

was inconsistent with the more general view we developed in Chapters 39 and

40. To go beyond the Bohr model, let’s apply the Schrödinger equation to ﬁnd the

wave functions for stationary states (states of deﬁnite energy) of the hydrogen

atom. As in Section 39.3, we include the motion of the nucleus by simply replac-

m ing the electron mass with the reduced mass m r.

The Schrödinger Equation for the Hydrogen Atom

We discussed the three-dimensional version of the Schrödinger equation in Sec-

tion 41.1. The potential-energy function is spherically symmetric: It depends

x 2 only on the distance from the origin of coordinates: r = + y2 + z2

(41.18)

e2 1 1 2 U r = - r 4pP0

1 2

, shown in Fig. 41.5; the spherically symmetric potential-energy func-

Hydrogen Atom

41.5 The Schrödinger equation for the hydrogen atom can be solved most readily using spherical coordinates.

Electron, charge 2e, at coordinates (r, u, f)

f u . The Schrödinger equation with this or 2

• For hydrogen atom, the three-dimensional potential energy depends only on the electron’s distance from the proton:

U (r) =

e2 r

1 4⇡"0

The hydrogen-atom problem is best formulated in spherical coordinates r, u, f tion depends only on r, not on 1 potential-energy function can be solved exactly; the solutions are combinations of familiar functions. Without going into a lot of detail, we can describe the most important features of the procedure and the results.

c as a product of three functions, each one a function of r, u, f

(41.19)

First, we ﬁnd the solutions using the same method of separation of variables that we employed for a particle in a cubical box in Section 41.2. We express the wave function only one of the three coordinates: 1 2 c u f r r, u, f = R ™ £

Nucleus, charge 1e, at the origin

• As before, we will seek separable wave functions, but this time, due to the spherical symmetry we will use spherical coordinates (r, θ, 𝜙) rather than (x,y,z). Then:

(r, ✓, ) = R(r)⇥(✓)()

f u r R depends only on depends only on and u, r, ™ £ 1 2 2 1 2 1 1 2 f. Just as for a particle in a three-dimensional box, when we 1 2 1 1 2 2

R r , a second

• Following the same procedure as before, we can

(41.20a)

• Energies are

2 U 2m rr 2

separate the problem into three separate equations

En =

l(l + 1)

(41.20b)

me4 2n2h2

1 (4⇡"0)2

–

R(r) = ER(r)

~2 2m

d dr

r2 dR(r) dr

~2 2m

✓

◆

(41.20c)

En =

sin✓

l(l + 1)

⇥(✓) = E⇥(✓)

13.6 eV n2

1 sin✓

d d✓

◆ d⇥(✓) d✓

r2 + U (r) m2 l sin2✓

✓

◆

✓

f f u u That is, the function depends only on substitute Eq. (41.19) into the Schrödinger equation, we get three separate ordi- nary differential equations. One equation involves only r and involves only and , and a third involves only and : ™ £ 1 2 dR r d U l + 1 1 2 1 2 r 2 r r R r - + = ER + U dr dr 2m rr 2 1 2 1 2 a b a 1 2 1 2b 2 1 u) d d™ m l u l sin u l + 1 - + = 0 ™ du du 1 sin u sin 2u 1 a b a 1 1 2 b 2 f d 2 £ f + m l £ = 0 df2 1 2 1 2 are constants m l

with the m l

Quantum numbers: n, l, ml

l () = E()

◆ d2() d2 + m2

In Eqs. (41.20) E is the energy of the stationary state and l and that we’ll discuss later. (Be careful! Don’t confuse the constant reduced mass ) m r.

We won’t attempt to solve this set of three equations, but we can describe how it’s done. As for the particle in a cubical box, the physically acceptable solutions

of these three equations are determined by boundary conditions. The radial func-

R r tion in Eq. (41.20a) must approach zero at large because we are describing r,

1 2

Hydrogen Atom: Quantum States

• Table below summarizes the quantum states of the hydrogen atom. For each value of the quantum number n, there are n possible values of the quantum number l. For each value of l, there are (2l + 1) values of the quantum number ml.

• Question: How many distinct states of the hydrogen atom (n, l, ml) are there for

n = 3 state? What are their energies?

Hydrogen Atom: Quantum States

• Question: How many distinct states of the hydrogen atom (n, l, ml) are there for

n = 3 state? What are their energies?

The n = 3 state has possible l values 0,1 or 2. Each l value has ml possible values of (0); (-1,0,1); or (-2,-1,0,1,2). The total number of states is then 1 + 3 + 5 = 9.

Each of these states have the same n, so they all have the same energy.

We will see later there is another quantum number s, for electron spin (±1/2), so there actually 18 possible states for n = 3.

Spectral Lines of the Hydrogen Atom

[eV]

En =

E1 n2 =

13.6 n2

En =

13.6 m2

13.6 n2

" = Em

✓

◆

Lyman series: transitions from m = 2,3,…∞ to n = 1: Ultraviolet series

Balmer series: transitions from m = 3,4,…∞ to n = 2: Visible series

Paschen: transitions from m = 4,5…∞ to n = 3: Infrared series

Brackett: transitions from m = 5,6…∞ to n = 4: Infrared series 2

Pfund: transitions from m = 6,7…∞ to n = 5: Infrared series 3

Homework

(1) Calculate wavelengths of Lyman, Balmer, Paschen, Brackett, Pfund series from n = 7 to lower levels?

(2) Of which can we see by human eyes?

Spectral Lines of the Hydrogen Atom

Hydrogen Atom: Degeneracy

• States with diﬀerent quantum numbers l and n are

often referred to with letters as follows:

n value

shell

l value

letter

• Hydrogen atom states with the same value of n but diﬀerent values of l and ml are degenerate (i.e. have the same energy).

• Figures at the right show radial probability

distribution for states with l = 0,1,2,3 and diﬀerent values of n = 1,2,3,4.

Hydrogen Atom: Probability Distribution

• States of the hydrogen atom with l = 0 (zero orbital angular momentum) have spherically symmetric wave functions that depend on r but not on θ or 𝜙. These are called s states. Figures below show the electron probability distributions for three of these states.

Hydrogen Atom: Probability Distribution

• States of the hydrogen atom with nonzero orbital angular momentum, such as p states (l = 1) and d states (l = 2) have wave functions that not spherically symmetric. Figures below show the electron probability distributions for three of these states as well as for two spherically symmetric s states.

Orbital Angular Momentum

• Motion of electron around the nucleus:

orbital motion

• Intrinsic motion of electron: spin motion

(rotation around its axis)

Orbital Angular Momentum ~L

1. The vector does not point in one

speciﬁc direction

2. The magnitude of orbital angular

momentum is quantized:

l(l + 1)

L = ~

p

~L 3. The projection of vector on a direction is quantized:

1,

2, ...,

l

Lz = ml~ ml = 0,

±

Quantization of Angular Momentum

• In addition to quantized energy (speciﬁed by principle quantum number n), the solutions subject to physical boundary conditions also have quantized orbital angular momentum L. The magnitude of the vector L is required to obey:

(l = 0, 1, 2,… n-1)

l(l + 1)

L = ~

where l is the orbital quantum number.

p

• Recall that the Bohr model of the Hydrogen atom also had quantized angels moment L = n(cid:15514), but the lowest

energy state n = 1 would have L = (cid:15514). In contrast, the

Schrodinger equation shows that the lowest state has L = 0. The wave function of this energy state is a perfectly symmetric sphere. For higher energy states, the vector L has only certain allowed directions, such that the z-component is quantized as

(ml = 0, ±1, ±2,…±l)

Lz = ml~

Orbital Angular Momentum and Quantum Numbers

Magnetic Moments and Zeeman Eﬀect

• Electron states with nonzero orbital angular momentum (l = 1,2,3,…)

have a magnetic dipole moment due to electron motion. Hence these states are aﬀected if the atom is placed in a magnetic ﬁeld. The result, called Zeeman eﬀect, is a shift in the energy of states with nonzero ml.

• The potential energy associated with a magnetic dipole moment ml in a

magnetic ﬁeld of strength B is U = mlµBB, and the magnetic dipole moment due to the orbital angular momentum of the electron is in units of the Bohr magneton

U = mlµBB

µB =

eh 2me

Zeeman Eﬀect and Selection Rules

• The ﬁgure shows the shift in energy for

the ﬁve l = 2 states (each with a diﬀerent value of ml) as the magnetic ﬁeld strength increases

• An atom in a magnetic

ﬁeld can make transitions between diﬀerent states by emitting or absorbing a photon ∆E transition is allowed if ∆l = 1 and ∆ml = 0, ±1. A transition is forbidden if it violates these selection rules.

Electron Spin and Stern-Gerlach Experiment

• The experiment of Stern and Gerlach demonstrated the existence of electron spin. The z-component of the spin angular momentum has only two possible values, corresponding to ms = +1/2 and ms = -1/2