1. 1 Light as a Particle The 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, o

2. 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.

3. White Light White light : Continuous spectrum Contains all frequencies in equal amounts Prism: disperses light into its components

4. 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

5. 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 trajectory around the nucleus i.e its kinetic energy is in balances with the electron nuclear potential energy.

6. Energy Levels 123 45 1 2 3 4 5 Frequency Energy -ve 0 Energy of free electron - ve energy change => more stable than free electron

7. Atomic Spectra Hydrogen Spectrum : Anders Ångström (1817-1874) In 1885, Johann Balmer (1825-1898) showed that the wavelengths of H could be described by: 1/ = (1.0974*10 7 m -1 )*(1/4 -1/n 2 ) This equation was later generalized by, Johannes Rydberg (1854-1919) to described all the spectral lines of H as: 1/ = R*|1/n 1 2 -1/n 2 2 | R = 1.0974*10 7 m -1 = Rydberg Constant Rydberg Equation n 1 final n 2 initial

8. 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/ = R*|1/n 1 2 -1/n 2 2 | 1/ = (1.0974*10 7 m -1 )*|1/4 2 -1/5 2 | 1/ = (1.0974*10 7 m -1 )*|1/16 -1/25| 1/ = (1.0974*10 7 m -1 )*|0.0225| 1/ = 246920 m -1 = 0.0000040500 m = 4.0500*10 -6 m = 4.0500  m Visible light Final n 1 = 4; initial n 2 = 5

9. 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

10. The energy of an electron in the n th orbital of a hydrogen atom E= - RhC/n 2 n = principle quantum number For any atomic system: Energy Level of Electrons E n =- R y Z/n 2 Z = atomic number R y =RhC = 2.179*10 -18 J = Rydberg unit Energy is negative, i.e. means its stabilized E/RhC = -1/n 2 = -1, -1/4, -1/9, -1/16 …. Bohr calculated the radius of each orbital : r = a o (n 2 /Z) a o = Bohr radius

11. Absorption, Emission and Energy Levels Lowest energy state : Ground state Electrons cannot stand still therefore have an absolute minimum energy When 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

12. 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:  E = E ex.s. – E g.s

13. 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  E = E ex.s. – E g.s  E = E(n=3) – E(n=1)  E = -RhC/3 2 – (-RhC/1 2 ) E=-RhC/n 2  E = -(RhC)(1/9 –1)=-(R y )(-8/9)  E = (8/9) Ry  E = (8/9) (2.179*10 -18 J)  E = 1.937*10 -18 J

14. Ionization Limit Notice 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  E as n → ∞ IE = -RhC/∞ 2 – (-RhC/n initial 2 ) IE = RhC/n 2 = -E(n) > 0