03 Atomic Physics.pptx

(1)

http://2011period6group4.wikispaces.com/2.+Thomson's+Model

JJ Thomson’s “Plum Pudding” Model

8 December 1856 – 30 August 1940

(2)

Thompson’s Model for the Atom

Electron Positive

pudding

Thompson’s plum pudding J.J. Thompson’s plum

pudding model consists of a sphere of positive charge with electrons embedded inside.

This model would explain that most of the mass was positive charge and that the atom was electrically neutral.

The size of the atom (10-10 m) prevented

(3)

http://sun.menloschool.org/~dspence/chemistry/atomic/ruth_expt.html

Ernest Rutherford’s Experiment

The Thompson model was abandoned in

1911 when Rutherford bombarded a thin

metal foil with a stream of positively charged alpha

particles.

The Thompson model was abandoned in

1911 when Rutherford bombarded a thin

metal foil with a stream of positively charged alpha

particles.

30 August 1871 – 19 October 1937

(4)

(5)

The Nucleus of an Atom

If electrons were distributed uniformly, particles would pass straight through an atom. Rutherford proposed an atom that is open space with positive charge concentrated in a very dense nucleus.

Gold foil Screen Alpha scattering

+

(6)

Electron Orbits

Consider the planetary model for electrons which move in a circle around the positive nucleus. The figure below is for the hydrogen atom.

Radius of Hydrogen atom F_C +

-Nucleus e

-r Coulomb’s law: Centripetal FC:

E 2 2 2 0

4

mv

e

r





r

(7)

Failure of Classical Model

v

+

-Nucleus

e- When an electron is accelerated

by the central force, it must radiate energy.

The loss of energy should cause the velocity v to de-crease, sending the electron crashing into the nucleus.

This does NOT happen and the Rutherford atom fails.

2

2 0

4

e

r

mv



(8)

(9)

(10)

Atomic Spectra

Earlier, we learned that objects continually emit and absorb electromagnetic radiation. In an emission spectrum, light is separated into characteristic wavelengths.

In an absorption spectrum, a gas absorbs certain wavelengths, which identify the element.

Emission Spectrum

Gas

l

₂

l

₁

(11)

Hydrogen Absorbtion Spectrum

400

nm 500 nm 600 nm 700_nm

(12)

http://annwn.tempwebpage.com/cookpot/bits/neon.jpg

Sodium

(13)

Bromine

Deuterium

Helium

Hydrogen

Krypton

Mercury

Neon

Water Vapor

Xenon

(14)

(15)

(16)

1 3 7 15 31 63 127 255

2 4 8 16 32 64 128 256

(17)

Johann Jakob Balmer

(18)

Emission Spectrum for H Atom

653

nm 486 nm 434 nm 410 nm

Characteristic wavelengths

n₌

3 n = 4 n = 5 n = 6

Balmer worked out a mathematical formula, called the Balmer series for predicting the absorbed wavelengths from hydrogen gas.

R

1.097 x 107 m-1

R

1.097 x 107 m-1

-2 2

1

;

3, 4, 5, . . .

2

R

n

l















(19)

Example 1: Use the Balmer equation to find the wavelength of the first line (n = 3) in the Balmer series. How can you find the energy?

R = 1.097 x 107 m-1

l = 656 nm

The frequency and the energy are found from:

c = fl and E = hf

-7 -1

1

0.361(1.097 x 10 m )

l



2 2

1

;

3

2

R

n

l



















2 2

1

(0.361);

2

3

0.361

(20)

http://education.jlab.org/qa/atom_model.html http://en.wikipedia.org/wiki/File:Niels_Bohr.jpg

7 October 1885 – 18 November 1962

(21)

Bohr's starting point was to realize that classical mechanics by itself could never explain the atom's stability. A stable atom has a certain size so that any equation describing it must contain some fundamental constant or combination of constants with a dimension of length. The classical fundamental constants--namely,

the charges and the masses of the electron and the nucleus--cannot be combined to make a length. Bohr noticed, however, that the quantum constant formulated

by the German physicist Max Planck has dimensions which, when combined with the mass and charge of the

electron, produce a measure of length. Numerically, the measure is close to the known size of atoms. This

(22)

The Bohr Atom

Atomic spectra indicate that atoms emit or absorb energy in discrete amounts. In 1913,

Neils Bohr explained that classical theory did not apply to the Rutherford atom.

+

Electron orbits

e

-An electron can only have certain orbits and the atom must have

definite energy levels which are analogous to standing waves.

An electron can only have certain orbits and the atom must have

(23)

(24)

(25)

Wave Analysis of Orbits

+

Electron orbits

e- Stable orbits exist for

integral multiples of de Broglie wavelengths.

2



r = n

l

n =

1,2,3, …

n = 4

𝑟

=

𝑛

h

2

𝜋 𝑚𝑣

2

π

𝑟

=

𝑛

h

(26)

Bohr’s Atom and Radiation

Emission

Absorption

When an electron drops to a lower level, radiation is emitted; when radiation is absorbed, the electron moves to a higher level.

Energy: hf = E_f - E_i

(27)

Radius of the Hydrogen Atom

Radius as function of energy level:

Classical radius

By eliminating r from these equations, we find the velocity v; elimination of v gives possible radii r_n: Bohr’s

radius 𝑟 =₂ _{𝜋 𝑚𝑣}𝑛h 𝑟 = 𝑒

2

4 𝜋 𝜀₀ 𝑚 𝑣2

2 0

2

n

e

v

nh





2 0 2

(28)

Example 2: Find the radius of the Hydrogen atom in its most stable state (n = 1).

m = 9.1 x 10-31 kg

e = 1.6 x 10-19 C

r = 5.31 x 10-11 m r = 53.1 pm

2 2 0 2 n

n

h

r

me







2 2

2 -12 _Nm 34 2

C

-31 -19 2

(1) (8.85 x 10

)(6.63 x 10 J s)

(9.1 x 10 kg)(1.6 x 10 C)

r







(29)

Total Energy of an Atom

The total energy at level n is the sum of the kinetic and potential energies at that level.

Substitution for v and r gives expression for total energy. But we recall that:

Total energy of Hydrogen atom for level n.

2 2 1 2 0

;

4

e

E K U

K

mv

U

r



 



2 0

2

n

e

v

nh





2 0 2

2 n

n

h

r

me







4

(30)

Energy for a Particular State

It will be useful to simplify the energy formula for a particular state by substitution of constants.

m = 9.1 x 10-31 kg

e = 1.6 x 10-19 C

_o = 8.85 x 10--12 C2/Nm2

h = 6.63 x 10-34 J s

Or

2 2

4 -31 -19 4

2 2 2 -12 _C 2 2 -34 2

0 _Nm

(9.1 x 10 kg)(1.6 x 10 C)

8 8(8.85 x 10 ) (6.63 x 10 Js) n

me E

n h n



   

-18

2

2.17 x 10 J

n

E

n

 

E

n

13.6 eV

₂

n

(31)

Balmer Revisited

Total energy of Hydrogen atom for level n.

Negative because outside energy to raise n level.

When an electron moves from an initial state

n_i to a final state n_f, energy involved is:

Balmer’s Equation:

𝐸

h𝑐=

1

𝜆

1

λ=

𝑚 𝑒4

8 𝜀₀ h3 𝑐

(

1

𝑛2_𝑓 −

1

𝑛₀2

)

R =

4

2 2 2 0 8 n me E n h    4 4

0 _{2 2 2} _{2 2 2}

0 0 0

1 1

;

8 8

f

hc me me

E E E

hc h n h n

l



 _      __  __   7 -1 2 2 0

1 1 1

; 1.097 x 10 m

(32)

𝐸

h𝑐=

1

𝜆

2.17 x 10-18

(33)

Energy Levels

We can now visualize the hydrogen atom with an electron at many possible energy levels.

Emission

Absorption

The energy of the atom increases

on absorption (n_f > n_i) and de-creases on emission (n_f < n_i).

Energy of nth level:

The change in energy of the atom can be given in terms of initial n_i and final n_f levels:

(34)

r

_A

r

_B

F_g = G

m₁·m₂

r_B2

F_g = G

m₁·m₂

r_A2

Work = average force · distance

= · (

G

r

_B

–

r

_A

)

m₁·m₂

r_Ar_B

= G·m₁·m₂

1 1

r_Ar_B

= G·m₁·m₂

1 1

∞r_B

zero

= - G·m₁·m₂

rB

Absolute Potential

Energy

U

G

=

G m1·m2

r_B

r_Ar_B

r_A

r_Ar_B

(35)

-2,000 Joules

-8,000 Joules

ΔU_G = Final - Initial

ΔU_G = -8,000 J – (-2,000 J)

ΔU_G = -6,000 Joules

Object LOST 6,000 Joules

ΔU_G = Final - Initial

ΔU_G = -2,000 J – (-8,000 J)

ΔU_G = +6,000 Joules

(36)

Energy of nth level:

-13.6 eV -3.40 eV -1.51 eV -.869 eV -.544 eV

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

2

13.6 eV

E

n

(37)

Spectral Series for an Atom

The Lyman series is for transitions to n = 1 level. The Balmer series is for transitions to n = 2 level.

The Pashen series is for transitions to n = 3 level.

The Brackett series is for transitions to n = 4 level.

n =2

n =6

n =1

n =3

n =4

n =5 2 2

0

1 1

13.6 eV

f

E

n n

 

  __  __

(38)

Example 3: What is the energy of an emitted photon if an electron drops from the n = 3 level to the n = 1 level for the hydrogen atom?

Change in energy of the

atom.

DE = -12.1 eV DE = -12.1 eV

The energy of the atom decreases by 12.1 eV as a photon of that energy is emitted.

You should show that 13.6 eV is required to move an electron from n = 1 to n = .

2 2 0 1 1 13.6 eV f E n n     __  __   2 2 1 1

13.6 eV 12.1 eV 1 3

E   _  _  

(39)

Modern Theory of the Atom

The model of an electron as a point particle moving in a circular orbit has undergone significant change.

• The quantum model now presents the

location of an electron as a probability distribution - a cloud around the nucleus.

• Additional quantum numbers have been added to describe such things as shape, orientation, and magnetic spin.

• Pauli’s exclusion principle showed that no

(40)

Modern Atomic Theory (Cont.)

The Bohr atom for Beryllium suggests a planetary model which

is not strictly correct.

The n = 2 level of the Hydrogen atom is