1. Title: Lesson 3 Emission Spectra, Bohr and Planck
Learning Objectives:
Carry out a flame test to show how emission spectra informs us about electron configuration
Understand the Bohr and Planck models

2. Flame Testing and Line Spectrum
Metals can be identified by the colour of the flame produced when their compounds are heated in a Bunsen burner. The light emitted by different atoms can tell us about the configurations within the atoms.

3. Carrying out a flame test Practical details
Clean a platinum or nichrome (a nickel-chromium alloy) wire by dipping it into concentrated hydrochloric acid and then holding it in a hot Bunsen flame. Repeat this until the wire doesn't produce any colour in the flame.
When the wire is clean, moisten it again with some of the acid and then dip it into a small amount of the solid you are testing so that some sticks to the wire. Place the wire back in the flame again.
If the flame colour is weak, it is often worthwhile to dip the wire back in the acid again and put it back into the flame as if you were cleaning it. You often get a very short but intense flash of colour by doing that.
Ions
flame colour Li
red Na
strong persistent orange K
lilac (pink) Rb
red (reddish-violet) Cs
blue? violet? (see below) Ca
orange-red Sr Ba
pale green Cu
blue-green (often with white flashes) Pb
greyish-white

4. The origin of flame colours
Flame colours are produced from the movement of the electrons in the metal ions present in the compounds.
For example, a sodium ion in an unexcited state has electrons in the 1st and 2nd energy level.
When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher energy levels, depending on how much energy a particular electron happens to absorb from the flame.
Because the electrons are now at a higher and more energetically unstable level, they tend to fall back down to where they were before - but not necessarily all in one go.
Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular colour.
As a result of all these jumps, a spectrum of coloured lines will be produced. The colour you see will be a combination of all these individual colours.
The exact sizes of the possible jumps in energy terms vary from one metal ion to another. That means that each different ion will have a different pattern of spectral lines, and so a different flame colour.

5. The Bohr model Bohr proposed that:
Electrons move into an orbit or higher energy level when energy is absorbed  EXCITED STATE
This is unstable so it will fall back down to GROUND STATE (eventually). Energy is emitted.
The energy emitted is EM radiation.
Each electron transitioning will emit one packet of energy (quantum) or a photon.
Energy is proportional to the frequency of the radiation.

6. Problem with the Bohr Model
Could explain the emission spectrum of Hydrogen but not the spectral lines of atoms with more than one electron.
Light can be described by frequency (v) which is a wave property or by the photons which make up a beam of light.
To explain spectra from more than one electron you need model related by the Plank equation, where energy is related to the frequency:

8. Energy of the photon of light emitted is equal to the energy change in the atom
It is also related to frequency of the radiation by the Planck equation:
This equation can be used to work out the difference in energy between various levels in the hydrogen atom.
V = frequency of the light (Hz or s-1)
H = Planck’s constant (6.63 X10-34 Js)
The equation and value of h (Planck constant are given in the IB Data booklet).
Atoms emit lines of certain frequencies because electrons can only occupy certain orbits.
The electron cannot change it’s energy in a continuous way.
(So this means…)

9. Combining Planck and Wavelength Equations
Where:
λ is the wavelength of the light (m)
c is the speed of light (3.0 x 108 ms-1)
The two equations E = hv and c = v λ can be combined:
NOTE: Energy of a photon of EM radiation is:
Directly proportional to frequency
Inversely proportional to wavelength

10. Question
If the frequency of the convergence limit is the Lyman series for hydrogen is 3.28 x 1015 Hz, calculate the ionisation energy of hydrogen in kJ mol-1.
Solution
E = hv
Therefore E = 6.63 x x 3.28 x 1015
= 2.17 x 10-18J
(Energy required to move 1 electron from 1 atom of hydrogen)
Energy required is for 1 mole of hydrogen so multiply by Avogadro’s constant.
E = 2.17 X X 6.02 X 1023 = 1.31 X 106 J mol-1
Divide by 1000 gives the answer in kJ mol-1 so…
E = 1.31 X 103 kJ mol-1.

11. Complete the Test Yourself Questions
Page 77
Questions 13a+b
Check answers on page 559