Evans SPH 4U1

Unit 9: Quantum Theory

Note 8: Atomic Absorption and Emission Spectra

Reference: Chapter 12.4, 12.5

Atomic Emission:

When a gas is excited (say by applying an electric potential) then the gas will emit photons of light.  Each gas will have its own specific spectrum of emitted colours (photons) that it emits. For example, the gas in a neon sign.

Atomic Absorption:

When light passes through a gas (non-ionized/unexcited), photons from the light will be absorbed in the gas at specific wavelengths.  The dark lines in the absorption spectrum will occur at the same wavelengths as the emission lines for the gas.

See Figure 3 in text.

Rutherford's atom did not explain these specific emission and absorption spectra so James Franck and Gustav Hertz set out to do this..

The Franck-Hertz Experiment on Emission:

Objectives of this experiment were

1. to show that electrons in atoms have discrete energy levels

2. to find values for these excitation energy levels measured from the ground state.

Experiment setup - see diagram from class

Electrons are accelerated by a voltage V across parallel plates gaining kinetic energy

They then pass through a hole in one of the plates into a chamber of mercury vapour. After passing through the mercury vapour, the electrons enter a vacuum chamber where their energy can be measured using a mass spectrometer.

They found the following:

1. When the electrons have an initial energy less than 4.9 eV, they pass through the mercury vapour without losing any energy.

2. When the electrons have an initial energy of between 4.9eV and 6.7eV, those that lose energy to the mercury atoms, lose only 4.9 eV of their energy.

3. As the energy of the electrons is increased, it is found that the electrons lose only definite amounts of energy (4.9 eV, 6.7 eV, 8.8 eV, 10.4 eV and multiples of these values in the case of multiple collisions) up to a point above which they can lose all of their energy.

Refer to the energy level diagram for mercury atoms from class. (also figure 10 pg. 269)

The energy required to remove an orbiting electron completely from its atom is called the IONIZATION ENERGY.

Further Franck-Hertz Observations on Emission:

If an electron with Ek = 4.9eV collides with a mercury atom, it loses all its energy to the atom and doesn't reach the anode plate. Similarly if an electron has Ek =6.7eV or 8.8eV it loses all its energy to the atom.

BUT if it has ,say, 6.0 eV it will have left over Ek of 1.1eV (6.0eV-4.9eV) and it will reach the plate.

They found that the energy the electron lost in the collision with the atom is emitted in the form of light (photons) where,

Ephoton= Einitial-Efinal

Also recall:

Ephoton= hf = hc/wavelength

From the above two formulae one can determine the wavelength of the emitted light.

Hydrogen Energy Levels:

Frank and Hertz did their experiment for mercury but scientists were anxious to study hydrogen since it is the lightest and simplest atom.

All the energy levels of hydrogen can be calculated from:

where n = 1,2,3,....   Also see energy level diagram figure 1 on page 640.

Example 1:

See figure from class which shows the energy level diagram for cesium. The vapour of this element is bombarded with electrons having 3.00 eV of energy.

a) If the cesium atoms are in the ground state, what are the possible energies of the electrons after passing through these atoms?

(Answer: 3.00eV no collision, 0.70eV, 1.62eV, 0.24eV double collision)

b) What are the possible energies of the photons emitted? ( Answer: 2.3 eV, 0.92 eV, 1.38 eV)

c) Calculate the possible wavelengths of the photons emitted. ( Answer: 5.4 x 10-7m, 1.4 x 10-6m, 9.0 x 10-7m)

Example 2:

If an electron in a hydrogen atom moves from n=6 state to the n=4 state, what's the wavelength of the emitted photon?

Suggested Text Problems:

Page 630#1-3

Page 633 #4,5,7

Page 638 #8

Solution to #8 on page 638

a) The electron will bounce off the atom, emerging with 4.0eV

b) Nothing will happen because the photon energy must be exactly equal to the atom energy to change the energy level.

c) The photon will be completely absorbed by the atom.

d) The atom will be ionized.

Evans SPH 4U1

Unit 9: Quantum Theory

Note 8B: Atomic Absorption and Emission Spectra

Observations on Absorption:

Recall that when light passes through a gas, photons from the light are absorbed in the gas at specific wavelengths.

The photon will only be absorbed if its energy corresponds exactly to the difference between two energy levels in the atom, the lower being occupied by an electron.

Ephoton= Efinal-Einitial

So in summary, atoms can receive energy in 2 different ways:

1. By Collisions (refer to the example 1 that we worked out in class and that we drew energy diagrams for): EMISSION

•  An incident electron passing through a vapour undergoes collisions with the vapour's atoms. In the collision the electron can ONLY lose an amount of energy corresponding to one of the energy levels of the vapour's atom. (see example 1a)

• When atoms absorb energy in collisions they quickly emit photons.

• The photons will have energy equal to the difference between any 2 energy levels of the atom.........(see example 1b)

•      where Ep is the energy of the photon, Ei in the energy of the higher level, Ef is the energy of the lower level

2. By Absorbing a Photon: ABSORPTION

• When light passes through a gas and photons from the light are absorbed in the gas at specific wavelengths.

• The photon will only be absorbed if its energy corresponds exactly to the difference between two energy levels in the atom, the lower level being occupied by an electron.

Suggested Text Problems:

Page 634 #9-11