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Light as a ParticleThe photoelectric effect. In 1888, Heinrich Hertz discovered that electrons could be

ejected from a sample by shining light on it.

Note the effects of changing:

Intensity - No matter how low the intensity there is still a current??

Frequency - Must have sufficient energy to eject an electron

Threshold Frequency, no

Black Body Radiation

Matter will emitted radiation when heated. How?

The amount of radiation, and its frequency, depends on the temperature

The higher the T, the lower the l, i.e. higher n

Classical Physics predicts infinite energy emission, which violates the law of conservation of energy.

Plank was able to show that the light intensity decays exponentially with wavelength allowing for energy to be conserved, and in so doing discovered the constant, h.

White Light

White light : Continuous spectrum Contains all frequencies in equal amounts

Prism: disperses light into its components

Light Emission From Atomic Gas

Atomic emissions - Not continuous - Why?

- Result from changes in the electron motion around the nucleus.

- Types of changes in motion are restricted corresponding to specific frequencies – i.e. quantized

Emission from the atoms

Otherwise electron will collapse into the nucleus, losing energy as radiation

The electron can change to a lower orbit

A photon is emitted when the electron changes from higher orbit to lower orbit

The electron remains in a stable trajectoryaround the nucleus

i.e its kinetic energy is in balances with the electron nuclear potential energy.

Energy Levels

12345

12

3

4

5

Frequency

Ene

rgy

-ve

0 Energy of free electron

- ve energy change=> more stable than free electron

Atomic SpectraHydrogen Spectrum : Anders Ångström (1817-1874)

In 1885, Johann Balmer (1825-1898) showed that the wavelengths of H could be described by:

1/l = (1.0974*107 m-1)*(1/4 -1/n2)

This equation was later generalized by, Johannes Rydberg (1854-1919) to described all the spectral lines of H as:

1/l = R*|1/n12 -1/n2

2|

R = 1.0974*107 m-1 = Rydberg Constant

Rydberg Equation n1 final

n2 initial

Calculate the wavelength of a photon emitted when a hydrogen atom changes to the n = 4 state from the n = 5 state. What type of electromagnetic radiation is this?

Exercise

1/l = R*|1/n12 -1/n2

2|

1/l = (1.0974*107 m-1 )*|1/42 -1/52|

1/l = (1.0974*107 m-1 )*|1/16 -1/25|

1/l = (1.0974*107 m-1 )*|0.0225|

1/l = 246920 m-1

l = 0.0000040500 m = 4.0500*10-6 m = 4.0500 mm

Visible light

Final n1 = 4; initial n2 = 5

The Bohr Model of the Hydrogen Atom

In 1913, Neils Bohr (1885-1962) proposed an explanation H hydrogen based on three postulates:

1. The orbital angular momentum of electrons in an atom is quantized. Only those electrons whose orbitals correspond to integer multiples of h/2π are “allowed”.

2. Electrons within an allowed orbital can move without radiating (so that there is no net loss of energy).

3. The emission or absorption of light occurs when electrons ‘jump’ from one orbital to another

The energy of an electron in the n th orbital of a hydrogen atom

E= - RhC/n2

n = principle quantum number

For any atomic system:

Energy Level of Electrons

En=- Ry Z/n2 Z = atomic number

Ry =RhC= 2.179*10-18 J = Rydberg unit

Energy is negative, i.e. means its stabilized

E/RhC = -1/n2 = -1, -1/4, -1/9, -1/16 ….

Bohr calculated the radius of each orbital:

r = ao (n2/Z) ao= Bohr radius

Absorption, Emission and Energy Levels

Lowest energy state : Ground state

Electrons cannot stand still therefore have an absolute minimum energyWhen a photon of the correct energy passes by it is absorbed and the electron goes to a higher energy level.

i.e. An Excited state

Absorption, Emission and Energy Levels

The electron can relax back to the ground state.

Upon relaxation it releases a photon.

The energy of the photon absorbed or released has energy matching the difference between the energy levels involved:

DE = Eex.s. – E g.s

Calculate the energy of a photon absorbed by a hydrogen atom when an electron jumps from the ground state to the n = 3 excited state?

Exercise

DE = Eex.s. – E g.s

DE = E(n=3) – E(n=1)

DE = -RhC/32 – (-RhC/12)

E=-RhC/n2

DE = -(RhC)(1/9 –1)=-(Ry)(-8/9)DE = (8/9) Ry

DE = (8/9) (2.179*10-18 J)

DE = 1.937*10-18 J

Ionization LimitNotice that as you increase n the energy approaches 0 but does not quite get there.

The corresponding orbital radius would approach infinity.

What does this mean?

The electron is no longer in orbit

i.e. The atom has ionized

The ionization energy is therefore the limit of DE as n → ∞

IE = -RhC/∞2 – (-RhC/ninitial2)

IE = RhC/n2 = -E(n) > 0