ATOMIC STRUCTURE

Lesson by Dr.Chris University of Phayao, August 2016

1

WHAT WE WILL LEARN … PART 1:

2

ATOM SYMBOLS

Ne 20

10

Atomic number Z

= no. of protons = no. of electrons

Mass number A

= no. of protons + neutrons

3

EXAMPLES

How many protons, electrons and neutrons are in:

4

SOLUTION

Cl can have 18 or 20 neutrons

35.45 is a mix of 2/3 35Cl and 1/3 37Cl

Element number

= no. of protons = no. of electrons

Mass number, not integer !

=> mix of ISOTOPES with different no. of neutrons !

Z

A

5

ISOTOPES

Nearly all elements have isotopes,

that means the same elements

(no. of protons = Z) has different no. of

neutrons, and therefore different mass

Example:

Copper exists to 69.2% of 63Cu and the rest

of 65Cu with masses 62.93 and 64.93

what is the atomic mass of the mixture ?

6

REVIEW BOHR ATOM MODEL

How do we know the structure of an atom ? 7

All waves have: frequency and wavelength

symbol: n (Greek letter “nu”) l (Greek “lambda”)

units: “cycles per sec” = Hertz “distance” (nm)

• All radiation: l • n = c

where c = velocity of light = 300’000 km/sec

ELECTROMAGNETIC RADIATION

Note: Long wavelength

small frequency

Short wavelength

high frequency increasing

wavelength

increasing

frequency

8

Example: Red light has l = 700 nm.

Calculate the frequency n .

= 3.00 x 10

8 m/s

7.00 x 10 -7

m = 4.29 x 10

14 Hz n =

c

l

• Wave nature of light is shown by classical

wave properties such as

• interference

• diffraction

ELECTROMAGNETIC RADIATION (2)

What is the energy in cm-1 ?

9

ENERGY UNITS

10

WHERE ARE THE ELECTRONS IN AN ATOM ? WE CAN KNOW FROM LIGHT SPECTRA !!

3 kinds of spectra:

11

BRIGHT COLORS FROM

HYDROGEN MATCH THE MISSING

COLORS IN SUNLIGHT Hydrogen spectrum

Solar spectrum

12

Li

Na

He

K

Cd

H

A

B

C

D

Identify the elements in the mixture A – D !

13

BOHR’S IDEA

“Allowed” orbits

14

WHERE DO THE LINES COME FROM ?

Bohr (1913)

emission spectra of hydrogen gas

Lines correspond to energies that are emitted by electrons:

emitted 15

ELECTRONS ARE “FIXED” ON ORBITS !

Electrons can move between distinct energy levels,

they cannot exist just anywhere in the atom = quantum

16

17

How many lines in the emission spectrum and at which energies (in cm-1) ?

18

Solution: 3 levels 3 lines

Transition A:

∆E = E3 – E2 =

-20’000 + 50’000 cm-1 =

30’000 cm-1 =

λ = 1/30’000cm-1 * 107 nm/1 cm = 333 nm

We can express energy as wavenumber, because h and c are constant:

= const * 1/λ = const * ν 19

THE ENERGIES OF ORBITALS

There are 2 forces acting on a circling

electron:

• Electrostatic attraction

• Centripetal force

They must be equal to keep the

electron stable:

The kinetic energy is then: 20

BOHR’S POSTULATE

A circling electric charge would emit energy, what is not

the case of real electrons.

Bohr postulated that electrons can live in certain energy

levels without loss of energy:

He assumed that the orbital angular momentum L = r *

p must be a multiple of h:

L = me * v * r = n * h/2 = n * h

21

From this we can calculate the orbital radius r:

with a0 as Bohr Radius

= 52.4 pm

-> the energy levels of the electron in a H atom are:

E1 = 13.6 eV

( 1eV = 8065.6 cm-1)

Calculate:

• an energy diagram for the H-atom electron

• the energy difference between the ground

and first excited state

• the wavelength of light emitted

22

1 eV => v = 8065.6 cm-1

l = 1240 nm

10.2 eV : v = 82.270 cm-1

l = 121.6 nm

23

RYDBERG EQUATION

From which energy level does an electron

come to n=2 when visible light of 410 nm

is emitted ?

What is the Ionization energy of hydrogen from this formula ?

-1

24

n = 2, RH = 1.0974 10-2 nm-1

1/λ = 1.0974 10-2 * ( 1/4 – 1/n2) = 1/410

1/n2 =1/4 - 1/(410 * 0.011)

n2 = 36

=> n =6

The electron goes from n=6 to n=2 and releases an energy that equals 410 nm

25

“HYDROGEN LIKE” ELEMENTS

The Bohr Theory can applied to all atoms

with one valence electron, like He(+)

Compare the energy of a transition

n=6 n=2 for H and He(+) in cm-1

http://www.fxsolver.com/browse/formulas/ Rydberg+formula+for+any+hydrogen-like+element

26

DETAILS

Summary

27

LESSON 2

QUANTUM THEORY: THE ELECTRON AS A WAVE

28

WHY A NEW THEORY ?

The Bohr Theory could explain the nature of

the Hydrogen atom quite precisely

BUT: elements with more than one electron

could not be handled by this theory

AND: the classical theory cannot explain

why there are only certain orbits that are

“allowed” for electrons. 29

ELECTRON BEAM INTERFERENCE ELECTRONS SHOW WAVE-BEHAVIOUR !

For example, if two slits

are separated by 0.5mm

(d), and are illuminated

with a 0.6μm

wavelength laser (λ),

then at a distance of 1m

(z), the spacing of the

fringes will be 1.2mm.

n λ / d

d

30

ELECTRONS AS WAVES

DeBroglie: (“ de Broy “) 1924

The energy of a photon from Einstein’s

theory:

E = c * p = h *c / l

31

RELATION PARTICLE - WAVE

Means every moving particle can be

considered as wave.

In the macroscopic world, the wavelength is

so small that it is not measurable.

Example:

a gun projectile with m=10 g moves with

800 m/sec => what is its wavelength ?

Compare to an electron with m = 9.1 10-31 kg

(h = 6.626 x 10-34 J*s / 1J = 1 kg m2/s2) 32

CALCULATION

λ = (6.626 x 10^-34 J*s)/(0.01 kg)(800 m/s)

1 J = 1 kg*m^2/s^2

λ = 0.83 x 10^-34 m = 8.3 10-24 nm

(compare: x-ray around 1 nm)

But for an electron it is:

λ = (6.626 x 10-34 J*s)/(9 10-31kg)(800 m/s)

= 9.2 10-7 m = 920 nm similar red light

33

LIGHT WAVES = PHOTONS

When light is emitted (like from the sun or

a heated metal), the energy of the photons

is discrete - comes in “packets” of

E = h ν = h c/λ = m c2

Energy is not continuous, but quantisized !

(one quantum of energy is h ν )

34

EXPLAINS WHY ELECTRONS CAN ONLY EXIST ON CERTAIN ORBITS

35

CONSEQUENCES: UNCERTAINTY !

When we treat electrons as waves, then we

cannot determine the position of the electron

exactly. The position (as particle) depends on

its speed – we describe this relation by the

Heisenberg Uncertainty Relation:

∆x m ∆v >= h/2π ( h )

(“Energy multiplied by time is constant” Murphy’s Law)

36

ENERGY OF “WAVE-ELECTRONS”

The model based on electrons circling around a nucleus is not satisfying, even it can explain the hydrogen atom.

A new model based on standing electron waves was developed:

Model the behaviour of a particle in a restricted space (“particle in a box”)

Quantization comes from the fact that a wave has to “fit” into the boundaries

37

Only wavelengths are allowed which “fit” into the box length L:

Use in DeBroglie: (h = Planck constant)

Therefore the kinetic energy is:

n is the main quantum number indicating the energy level 38

EXAMPLE

Calculate the first 2 energy levels for an

electron in a box with L = 300 pm (ca.

circumference of H-atom).

me = 9.1 * 10-31 kg h = 6.6 * 10-34 m2 kg/s

39

APPLICATION

The simple box model can help us to estimate the color of linear chromophores

We can assume that 2 electrons

can move along this molecule

chain (“resonance”)

If C-C and N-C is 1.40 A, which

light will be absorbed by this molecule ? 40

ANSWER

The molecule “length” is about 7 * 1.4 A

We have 8 electrons, the transition should occur between level 5 and 4:

=> 352 nm

http://www.umich.edu/~chem461/Ex3.pdf

41

CALCULATIONS

In the box model, an electron goes from n=2

level to the ground state and emits red light

of λ = 694 nm . What is the length of the box ?

(h = 6.63 E-34 Js, me = 9.11 E-31 kg, c= 3 E8 m/s 1 J = 1 kg·m2/s2.)

λ = 694 nm

42

ANSWER

http://www.physics.umd.edu/courses/Phys270/Jenkins/Hwksolns13_Publishers.pdf

(compare: r(H) = 0.053nm => L is about the double of the circumference )

43

SUMMARY Summary

44