Atomic Spectra•If you “engergize” an atom in a variety of ways, it will give off light with characteristic wavelengths•What wavelengths you get are a characteristic of the atom

Diffraction Grating

•There were patterns in the wavelengths, but no one understood why

•Triplets were not uncommon

•Later, more lines were discovered

1 1 1

A B C

2

2364.7 nm4

nn

1

2

1 191.17 nm4 n

•Hydrogen had the simplest pattern

, 1,2,3,m nn m

1

2 2

1 191.17 nmm n

Sample Problem:For what values of m and n will you get red light,

with wavelength between 625 and 740 nm?

1

2 2

1 1625 nm 91.17 nm 740 nmm n

2 2

1 10.123 0.146m n

•If m is 3 or bigger, you can’t get bigger than 1/9 =0.1111•If m is 1, then n is two or bigger, and the difference is no smaller than 0.75•m = 2•You need to subtract at least 0.1 n is not too large•n = 3

2

10.123 0.2500 0.146n

0.123 0.2500 0.1111 0.146

1

2 2

1 191.17 nmm n

1

2 2

1 16.86 8.12m n

Rutherford Scattering•Radioactive decay produces alpha particles

•Charge +2e•Mass = Helium atom mass•7000+ times heavier than electron

•Expected it would deflect – slightly – when it went through an atom

•1909: Geiger and Marsden perform this experiment under the direction of Ernest Rutherford

What they found:•Alpha particles would deflect, occasionally, by large angles

•Even backwards•Rutherford concluded that the positive charge was concentrated in a tiny region called the nucleus

+

b2

kqQFr

r2

cot2

m v bkqQ

q = 2e

Q = Ze

If you hit Gold (Au, Z = 79) with Ekin = 3.0 MeV -particles,1. How close do you have to get to be deflected by 90?2. What is the area of the target to be deflected by 90?

2 cot2

kqQbm v

2

2

2 cot 45kZem v

2

kin

kZeE

Continued . . .

b = impact parameter

Rutherford Scattering (cont.)

2

kin

kZebE

29 2 2 19

6 19

8.99 10 N m / C 79 1.602 10 C

3.0 10 eV 1.602 10 J/eV

+

143.79 10 m 37.9 fm

37.9

fm

deflects > 90

deflects < 90

Target area:

= b2 = 4512 fm2

If you hit Gold (Au, Z = 79) with Ekin = 3.0 MeV -particles,1. How close do you have to get to be deflected by 90?2. What is the area of the target to be deflected by 90?

The fact that this formula worked told Rutherford the nucleus was smaller than this, smaller than 38 fm in radius . . .

How small is the nucleus?How small a nucleus can one see by this method?•You can’t see the nucleus until you get close enough•Closest you can come happens when you are headed straight for the nucleus

+

•Conservation of energy gives you the closest approach•Rutherford realized he needed higher energies and smaller Z

•He used Aluminim (Z = 13) with Ekin = 7.7 MeV

kinmin

kqQER

2

min

2 ,ZkeR

2

minkin

2ZkeRE

29 2 2 19

6 19

2 13 8.99 10 N m / C 1.602 10 C

7.7 10 eV 1.602 10 J/eV

154.86 10 m 4.86 fm

He saw that the scattering changed, and concluded that the nucleus was a few fm across

The atom and the Solar System

The Solar system•Sun’s mass is 1048 times Jupiter’s mass•Radius of Neptune’s orbit is 6500 times radius of Sun•Sun stays put and all planets orbit it

Atoms are about a = 0.1 nm in radius Nucleus is about R = 10 fm in radius

The hydrogen atom•Nuclear mass is 1836 times electron’s mass•Radius of atom is about 10,000 times radius of nucleus

Could the atom be like a mini-solar system?

The Rutherford model of the atom1911: Rutherford develops his model•Nucleus at the center, nearly stationary•Electrons orbit nucleus in ellipses, like planets

+e-e-

e-

e-

Problems with the model:•Accelerating electrons should radiate, and drop into nucleus•What about electron-electron interactions? •No predictions of spectra of atoms

Equations you need

1

2 2

1 191.17 nmm n

2

2

minkin

cot2

2

m v bkqQ

ZkeRE

12

1 cos

2.426 10 m

e

e

hm c

hm c

maxeV hf

83.00 10 m/sf c E hf

2h 10

15

10 m 0.1 nm

5 10 m 5 fm

a

R

42

315Bk T

Uc

32.8978 10 m KT

Values of h, h-bar, e, k, kB, me

2 cosd m

The Bohr Model of the atom

E hf

+

1913: Extension of Rutherford model, with three new assumptions1. Electrons only go in circular orbits due to attraction of nucleus2. When the electron changes its configuration, energy is

emitted in the form of a single photon of energy3. The electrons must have angular momentum that is an integer multiple of

e-

L n em vr

Electrostatic Attraction:

2

2

keFr

Centripetal Force:2

em vFr

2 2

2em v ker r

e

nvm r

Subs

2 2

2e

e

m n ker m r r

2 2

2ne

nrm ke

2 2 2

3 2e

n kem r r

The Bohr Model of the atom (part 2)

+

e-

e

nvm r

2 2

2ne

nrm ke

Define:

Bohr Radius

2

0 2e

am ke

0.05297 nm

•Diameter of H-atom for n = 1 is 0.106 nm! 20nr a n

2

2 2e

e

m kenvm n

2ken

2kec

Define: Fine Structure Constant

v c n

1137

•Electrons are non-relativistic in hydrogen

But wait, there’s more!

The Bohr Model of the atom (part 3)2 2

2ne

nrm ke

Energy has two components:•Kinetic energy:•Potential energy*

2kevn

21kin 2 eE m v

2212 e

kemn

2 4

2 22em k e

n

potE Fdr2

2

ke drr

2ke

r

* Note: the sign of this equation assumes the force F is directed inwards.

22

2 2em ke

ken

2 4

2 2em k e

n

kin potE E E 2 4

2 22em k e

En

2kec

2 2

22em c

n

2

13.6 eVn

+

e-

The Bohr Model of the atom (part 4)

2 4

2 22em k e

En

What happens when the electron “jumps” from one level to another?2. When the electron changes its configuration, energy is emitted in

the form of a single photon of energy2 4

2 2 2

1 12em k eE

m n

hf 2 c

2 4

3 2 2

1 1 14

em k ec m n

Define:Rydberg Constant

2 4

34em k e

Rc

1 91.13 nmR

11

2 2

1 1Rm n

Compare:

1

2 2

1 191.17 nmm n

+

e-

level n

level m

Reduced Mass

+

e-

2 2

22em cE

n

1

137 2 20.5110 MeV /em c c

2

13.6 eVEn

•Calculated treating nucleus as stationary•Proton weighs 1836 times more than electron

•In fact, proton moves a very little bit as electron orbits it•Can be corrected for by using “reduced mass”•me 1 1 1

M m Mm

M m

•The real formula for energy levels of Hydrogen:•Spectroscopy can easily tell the difference•Slight corrections for isotopes

•0.027% for 1H vs. 2H•This is why wavelengths were slightly off

•Huge correction for “unusual” atoms

2 2

22cE

n

Sample Problem

1137

2 2

22cE

n

MmM m

Positronium consists of a bound state of an electron (m = 0.511 MeV/c2) and an anti-electron, or positron (same mass). What is

the binding energy of positronium in the ground state?2

122

ee

e

mm

m

2 2

24em c

n

5

2 2

5.11 10 eV4 137 1

6.81 eV

e-

e+

Hydrogen-like atoms•Atoms with one electron in them

•H, He+, Li+2, Be+3, . . . , U+91

•Bohr model could be easily modified to make them fit•e2 Ze2 Electrostatic

Attraction:2

2

keFr

2

2

kZeFr

+Ze

Approx. 2

2

13.6 eV ZE

n

20r n a Z

0 0.05297 nma

2 2 2

22Z cE

n

MmM m

Exact

v Zc n

Note: This ignores

relativity

Sample Problem

2

2

13.6 eV ZE

n

What is the energy and wavelength of light emitted when a Ca+19 (Z = 20 ) electron jumps from level 7 down to level 6?

2

2

2

2

13.6 eV 207

13.6 eV 206

111.0 eV

151.1 eV

40.1 eVE

cf

83.09 10 m hcE

15 84.136 10 eV s 3.00 10 m/s

40.1 eV

30.9 nm

Multiple Electron Atoms: X-raysCould it be used to explain other atoms, with multiple electrons?•Generally failed•Succeeded approximately for innermost electrons•Fast moving electron hits an inner electron and knocks it free•Outer electron falls to lowest level and fills in the void•Measure wavelength/energy of emitted photon

+e- e-

e-

e- e-

e-

e- e-

e-e-

e-

e-

Why it might work:•Innermost electron is near nucleus with large charge•Other electrons are outside of it

22

113.6 eV 1hf E Zn

?1/2

2

13.6 eV 11f Zh n

X-rays from inner electrons

•1913 Henry Moseley measures X-ray spectra•Finds this relationship almost works for inner electrons

?

1/2

2

13.6 eV 11 1f Zh n

1/2

2

13.6 eV 11f Zh n

•Why Z – 1?•Innermost electron slightly screened by other electron

•Also found a relationship for next level out

1/2

2

13.6 eV 1 1 7.44

f Zh n

+e- e-

e-

e- e-

e-

e- e-

e-e-

e-

e-

•There are more electrons screening the nucleus

Testing one of Bohr’s HypothesesE hf

2. When the electron changes its configuration, energy is emitted in the form of a single photon of energy

•If the atom has only discrete energy levels, then there should be a minimum amount of energy that can be added to it•Anything that collides with it with less energy must collide elastically•When it does collide inelastically, the energy will be transmitted to the atom•The atom then has to release a photon to lose a level

+e-

e-

e-

Predictions:•When you hit the threshold of energy, scattering (energy loss) will suddenly increase•The target atoms will start to glow at threshold energy

The Franck-Hertz Experiment

+– + –

e-

Experiment performed, 1914•At moderate voltage, current is high•Drops suddenly as you pass threshold•Drops periodically at higher voltage•Gas glows at predicted wavelength above first threshold

heaterthin gas wire screen

Vm

easure current

Assessing the Bohr ModelThe Good:•Accurately predicted hydrogen line spectrum•Predicted isotopic and other effects•Did pretty well at predicting X-ray lines for heavier elements•Clear evidence that atoms had “levels” of energy•Evidence that quantum effects would help describe the atom

The Bad:•No idea how to proceed for heavier elements•“Strength” of spectral lines was not predicted

The Ugly:•Arbitrary and unmotivated assumptions