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ATOMIC MODELS

Models are formulated to fit the available data. 1900 Atom was known to have certain size. Atom was known to be neutral. Atom was known to give off electrons. THOMPSON MODEL To satisfy the above conditions Thompson proposed Atom made of positive material with enough electrons embedded to make it neutral.

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RUTHERFORD Made measurements of alpha particles scattered from gold: Found 1. Most alpha particles passed through gold without being scattered. 2. Some alpha particles were scattered through large scattering angles. Data could not be explained by Thompson Model of Atom. 1. How could alpha particles pass through positive mass?

2. The force between charges was known to be

221

04

1

r

qqF

3

With the known charges and this force how could alpha particles be scattered through large angles?

RUTHERFORD MODEL

Small nucleus – all positive mass contained in small center part. Electrons in space surrounding nucleus.

SCALE MODEL

Nucleus Edge of Atom Radius = 1 mm 10 m

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LINE SPECTRA FIRST CONSIDER WHITE LIGHT THROUGH DIFFRACTION GRATING WHITE LIGHT GRATING RAINBOW COLORS

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BUT WITH HYDROGEN GAS AS SHOWN:

FIG 3.6 PAGE 93, T & R

6

7

FIG 3.7 PAGE 93, T & R

657

8

THE ONLY WAVELENGTHS PRESENT IN VISIBLE ARE 657 nm (red) 486 nm (blue-green) 434 nm (violet) 410 nm (violet) 397 nm Closely spaced lines out to 365 nm (series limit)

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BALMER Balmer found that he could predict the wavelength of these lines with one equation:

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1

2

11

nR

where 1710096776.1 mxR and .,.........5,4,3 etcn

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The Balmer lines are in the visible There are other lines: One set in the ultraviolet Three sets in the infrared All wavelengths can be determined by one equation:

22

21

111

nnR

RYDBERG EQUATION

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WHERE n1 = 1 and n2 = 2,3,4, ……… gives the uv series n1 = 2 and n2 = 3,4,5, ……… gives the visible series n1 = 3 and n2 = 4,5,6, ……… gives the 1st IR series n1 = 4 and n2 = 5,6,7, ……… gives the 2nd IR series n1 = 5 and n2 = 6,7,8, ……… gives the 3rd IR series

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The series have names for the physicists who did the early work in that part of the spectrum For n1 = 1 the series is named for Lyman For n1 = 2 the series is named for Balmer For n1 = 3 the series is named for Paschen For n1 = 4 the series is named for Brackett For n1 = 5 the series is named for Pfund

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THE BOHR MODEL OF THE ATOM Bohr proposed a modification of the Rutherford model that fit the data – all data before and including the Rutherford scattering data. The Bohr Atom Postulates: 1. Only orbits allowed are where angular momentum is integral multiple of

2

h

2. No atom radiates energy as long as electron is in one of the orbital states.

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DERIVATION OF RYDBERG EQUATION (thus validation of Bohr Model) Angular momemtum = mvr m = mass of electron v = velocity of electron in orbit r = radius of orbit From 1st postulate

2

hnmvr n = 1, 2, 3, …

Solving for r

mv

hnr

2 (1)

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Electron in orbit accelerates toward center

r

va

2

The force causing the acceleration is

2

2

04

1

r

eF

and

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maF So

r

vm

r

e 2

2

2

04

1

Solving for v

17

mr

ev

04

(2)

Put this in Eq. 1 above Solve for r

20

22

me

hnrn

Subscript n to remind one r for each n!

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For n = 1

011

20

2

1 100.5 amxme

hr

a0 is the Bohr orbit for the hydrogen atom in the lowest energy state. Then the other orbits

02anrn

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Bohr model is nucleus with electrons in only specific allowed orbits!

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ENERGY OF THE ATOM Total energy is KE + PE

2

2

1mvKE

r

ePE

0

2

4

and we found v

21

mr

ev

04

So

r

e

mr

emEyTotalEnerg

0

22

0 442

1

r

e

0

2

8

Or n

n r

eE

0

2

8

22

and

20

22

me

hnrn

Putting this in Total Energy equation

220

4 1

8 nh

meEn

where n= 1, 2, 3, ……

For: n= 1 E1 = -21.76 x 10-19 J = -13.6 eV n=2 E2 = - 5.43 x 10-19 J = - 3.39 eV n=3 E3 = - 2.41 x 10-19 J = - 1.51 eV n=4 E4 = - 1.36 x 10-19J = - 0.85 eV

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Plot these values on a vertical scale to get Figure 4.16 on page 145 of Thornton and Rex

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DERIVATION OF RYDBERG EQUATION Consider any two energy levels Ei with ni

and Ej with nj

If the atom is in ni state with total energy Ei and goes to the to the state nj with total energy Ej the atom will have less total energy. The difference in energy will be

ji EEE The energy is carried off by a photon with energy

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hf So

ji EEhf Or since

c

f

ji EE

c

f

1

27

and

230

4 1

8 ii nh

meE

230

4 1

8 jj nh

meE

So

2232

0

4 11

8

1

ij nnch

me

If

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1732

0

4

10097.18

mxRch

me

(and it does)

Then for nj = 1 We have the Lyman Equation nj = 2 We have the Balmer Equation nj = 3 We have the Paschen Equation nj = 4 We have the Brackett Equation nj = 5 We have the Pfund Equation STUDY FIGURE 4.16 ON PAGE 145 IN THORNTON AND REX!!