15.0 Introduction to the interaction of electromagnetic radiation with matter

I use two abbreviations for electromagnetic radiation: emr  and  EM radiation

Sub-index for this page on spectroscopy

15.0.1 The electromagnetic radiation spectrum

15.0.2 The wave and particle nature of electromagnetic radiation (emr) and two equations

15.0.3 Example of calculations using the two formulae described above in 15.0.2 e.g. Planck's equation

15.0.4 What can happen when emr interacts with matter and how does this relate to spectroscopy?

15.0.5 Some extra questions for you to do based on Planck's equation (answers at end of page) - under development, but some exam practise questions available

15.0.6 Problem solving - deduction of molecular  structure from 4 given spectra - under development, but some exam practise questions available

See other important pages for

Spectroscopy and the hydrogen spectrum and explaining the emission and absorption spectra of elements

since there is little point in repeating this page!

15.0.1 The electromagnetic radiation spectrum

At pre-university level chemistry, most spectroscopy that you will study, apart from mass spectrometry, will involve the interaction of electromagnetic radiation and matter.

Here we are only dealing with the region from radio waves to uv radiation

Type of electromagnetic radiation ===>	Radio waves	Microwaves	Infrared radiation	Visible light	Ultraviolet light
Spectroscopy involved	NMR	microwave	infrared	emission and absorption	emission and absorption
Type of quantum level change	nuclear spin orientation	molecular rotation	covalent bond vibration	electronic	electronic
~wavelength range/m	>10-1	10-4 to 10-1	7 x 10-7 to 10-4	4-7 x 10-7	10-8  to 4 x 10-7
~typical wavelength/m	103	10-2	10-5	5 x 10-7	10-8
~wavelength range/nm	> 108	105 - 108	700 - 105	400 - 700	10 - 400
~frequency range/Hz	< 3 x 109	3 x 109  to 3 x 1012	3 x 1012 to 4.3 x 1014	4.0 x 1014 to 7.5 x 1014	7.5 x 1014 to 3 x 1016
Photon energy trend	=========== increasing energy of radiation photons  ========>
Frequency trend	=========== increasing frequency of radiation  (Hz) =========>
Wavelength trend	========== decreasing wavelength of radiation (m) =========>
'picture trend'	====================>

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15.0.2 The wave and particle nature of electromagnetic radiation (emr) and two equations

Since emr, like visible light, can behave as a wave, its motion can be described by the simple wave equation.

c = λ

where

c = velocity of light (or any emr) = 3.00 x 10⁸ ms^-1

λ = wavelength, m (be prepared to work in other units like nm = 10^-9 m)

= frequency, Hz (s^-1)

However, emr photons can behave as particles, in the sense they can act as little 'bullets' of energy called quanta (this term goes back to the origin of quantum physics).

The term photon can be applied to any part of the emr spectrum.

Each quanta or 'quantum of energy' is very small.

The energy of an emr photon is given by the Planck equation (named after the famous physicist Max Planck, one of the founding people of quantum physics.

E = h

E = energy of the photon (the quanta), J  (note that this is just for ONE PHOTON!)

h = Planck's constant = 6.63 x 10^-34 JHz^-1

= frequency, Hz (s^-1)

For any quantum level change you can express the Planck equation as

∆E = h

where ∆E represents the difference between two quantum levels, which equals the photon energy emitted or absorbed in that quantum level change e.g. molecular rotation (microwave), bond vibration (infrared) or an electronic energy level (visible or uv).

The above equation only deal with a single photon e.g. interacting with a single atom or molecule - that's why E or ∆E is so small.

Since chemists often work in molar quantities you can multiply the Planck equation by the Avogadro number/constant, N_A, to obtain the energy per mole of the matter-emr interaction.

E = ∆E = h x N_A  (Jmol^-1), where N_A = 6.02 x 10²³ mol^-1

You can then divide by 1000 if you quantity the energy change in kJmol^-1.

The measurement of these quantum level changes and their pattern for a molecule is the whole basis of the various types of emr spectroscopy.

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15.0.3 Example of calculations using the two formulae described above in 15.0.2

c = λ   and  E = h

There are three constants you need to be familiar with, but they are given in examinations.

(i) Speed of light for any emr:  3.00 x 10⁸ ms^-1

(ii) Planck's constant:  6.63 x 10^-34 JHz^-1

(iii) The Avogadro constant N_A = 6.02 x 10²³ mol^-1

 

Q1 If the wavelength of a visible blue light laser beam is 460 nm, calculate its frequency.

c = λ ,  so = c ÷ λ, c = velocity of light = 3.00 x 10⁸ ms^-1

= 3.00 x 10⁸ ÷ (460 x 10^-9) = 6.52 x 10¹⁴ Hz

 

Q2 The frequency of a uv light beam is 4.51 x 10¹⁵ Hz,

What is its wavelength in m and nm?

c = λ ,  so λ = c ÷ λ, c = velocity of light = 3.00 x 10⁸ ms^-1

λ = 4.51 x 10¹⁵ ÷ 3.00 x 10⁸ = 1.50 x 10^-7 m

λ = 4.51 x 10¹⁵ ÷ 3.00 x 10⁸ = 150 nm

 

Q3 An ultraviolet beam of light has a frequency of 2.10 x 10¹⁵ Hz.

(a) Calculate the energy of each photon

E = h, h = Planck's constant = 6.63 x 10^-34 JHz^-1

E_photon = 6.63 x 10^-34 x 2.10 x 10¹⁵ =  1.39 x 10^-18 J

(b) Calculate the energy of the photon in Jmol^-1.

You multiply the above value from (a) by the Avogadro number N_A.

N_A = 6.02 x 10²³ mol^-1.

E_photon = 1.39 x 10^-18 x  6.02 x 10²³ = 838000 Jmol^-1 (3 sig figs)

(c) The bond enthalpy for chlorine is +243 kJmol^-1.

Will chlorine molecules be split into radicals by this uv light? (3 sig figs)

Explain your answer.

The energy of the uv photons is 838000 ÷ 1000 = 838 kJmol^-1.

This is more than sufficient energy to excite the chlorine molecule (*) and heterolytically break the bond to give two chlorine radicals:

Cl₂  ===>  Cl₂^*  ===> 2Cl•

This is the first step of initiation in the chlorination of alkanes to make halogenoalkanes. For more details see Chlorination of alkanes

(d) What is the minimum frequency of light that can theoretically split a chlorine molecule? Use the bond enthalpy quoted in part (c).

(i) First calculate the bond enthalpy for one molecule

E_bond = (243 x 1000) ÷ 6.022 x 10²³ =  4.04 x 10^-19 J/molecule

(ii) The substitute your answer to (i) into a rearranged Planck's equation

E = h , = E/h,  h = Planck's constant = 6.63 x 10^-34 JHz^-1

= 4.04 x 10^-19 ÷ 6.63 x 10^-34 = 6.09 x 10¹⁴ Hz

This frequency is on the borderline between visible and uv light.

Which is why, when using chlorine, it should not be in contact with 'combustible' material in a sunlit laboratory unless under controlled darkened conditions!

 

Q4 Ozone (trioxygen, O₃) is considered a secondary pollutant because it is produced from primary pollutants like the oxides of nitrogen nitrogen(II) oxide, NO and nitrogen(IV) oxide, NO₂, formed from the combustion of fossil fuels.

(These were formally called nitrogen monoxide and nitrogen dioxide).

Ozone can be formed in the troposphere, the layer of air closest to the Earth's surface, by the action of oxygen atom radicals on the oxygen molecule (dioxygen O₂).

O(g)  +  O₂(g)  ===>  O₃(g)

BUT, where do the oxygen radicals come from?

One source is the photodissociation of nitrogen(IV) oxide and both reactions contribute to the serious atmospheric pollution known as photochemical smog.

NO₂(g)  ===>  O(g)  +  NO(g)  

(∆H = +307 kJmol^-1, very endothermic)

(a) Calculate the energy required to break one N-O bond in NO₂.

One N-O bond enthalpy = 307 x 10³ ÷ 6.02 x 10²³ = 5.10 x 10^-19 J

(b) What is the minimum frequency of EM radiation that will bring about photodissociation of NO₂?

You just use the above value and slot into Planck's equation.

E = h , = E/h = 5.10 x 10^-19 ÷  6.63 x 10^-34 = 7.69 x 10¹⁴ Hz

Note that any higher frequency will do, any excess energy is just converted into increased kinetic energy of the fragments.

(c) What is the wavelength in nm of this radiation?

c = λ ,  so λ = c ÷ λ = 3.00 x 10⁸ ÷ 7.69 x 10¹⁴ = 3.90 x 10^-7  m

3.90 x 10^-7 ÷ 10^-9 = 390 nm

(d) What region of the emr spectrum corresponds with this radiation?

The lower energy end of the ultraviolet region, so bright sunlight is quite sufficient to split the molecules of NO₂ in the troposphere

 

Q5 To move from one energy level to the next required 0.29 kJ mol^-1 for a hydrogen bromide molecule HBr.

(a) Calculate the energy change involved for one HBr molecule

∆E = 0.29 x 10³  ÷  6.02 x 10²³ = 4.82 x 10^-22 J

(b) Calculate the frequency of emr required to bring about the quantum level change.

∆E = h , = ∆E ÷ h = 4.82 x 10^-22 ÷ 6.63 x 10^-34 = 7.26 x 10¹¹ Hz

(c) Which part of the EM spectrum and quantum levels of the HBr molecule does this correspond to?

Microwave region of the emr and the rotational quantum levels of HBr.

 

Q6 A beam of EM radiation has an energy of 3.00 x 10^-19 J/photon.

(a) Calculate the frequency of the EM radiation.

E = h , = E ÷ h = 3.00 x 10^-19 ÷ 6.63 x 10^-34 = 4.52 x 10¹⁴ Hz

(b) Which part of the EM spectrum does this correspond to? Which quantum levels of a molecule might be excited to a higher level? and would a molecule be permanently changed?

This frequency corresponds to photons of visible light.

These photons can potentially excite electrons to a higher level e.g. in a coloured molecule.

There would be no permanent change to the molecule, the excited molecule would return to the lowest available electronic state.

 

Q7 Methane gas, CH₄, is a greenhouse gas.

(a) What quantum level changes occur when methane molecules absorb infrared radiation?

Infrared photons of sufficient energy will increase the C-H vibrations to higher quantum levels.

(b) Why does this cause a warming of the atmosphere?

The higher energy vibrating molecules will relax to their lowest vibrational state by losing energy that ultimately becomes thermal energy - increased kinetic energy of the molecules in the atmosphere.

 

Q8 A mug of coffee (250 cm³), in my favourite Periodic Table mug,  is placed in a microwave oven for 2 minutes to be heated from 20^oC to 60^oC. For the calculations assume all the coffee is water and has a density of 1.00 gcm^-3.

The microwave frequency of the oven 2.45 x 10⁹ Hz. The speed of light is 3.00 x 10⁸ ms^-1

The specific heat capacity of water is 4.18 Jg^-1K^-1.

(a) Calculate how much energy must be transferred to the coffee to heat it up from 20^oC to 60^oC.

∆H = mass x SHC_H2O x ∆T

Energy required = 250 x 4.18 x 40 = 4.18 x 10⁴ J

(b) Calculate the wavelength of the microwave radiation in cm.

c = λ ,  so λ = c ÷

λ = 3.00 x 10⁸ ÷ 2.45 x 10⁹ = 0.122 m

0.122 x 100 = 12.2 cm  (of similar magnitude to the interior of a microwave oven or the mug!)

(c) Calculate the energy per microwave photon.

E = h = 6.63 x 10^-34 x 2.45 x 10⁹ = 1.62 x 10^-24 J

(c) How many microwave photons did it take to make the coffee!?

Energy needed ÷ energy of photon = 4.18 x 10⁴ ÷ 1.62 x 10^-24

= 2.58 x 10²⁸ microwave photons

(d) How many moles of photons were needed to heat the coffee!?

2.58 x 10²⁸ ÷  6.02 x 10²³ = 4.28 x 10⁴ mol photons

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15.0.4 What can happen when emr interacts with matter and how does this relate to spectroscopy

When electromagnetic radiation impacts on, and is absorbed by a material, the effects are quite varied and are very dependent on the frequency (or wavelength) of the incoming radiation.

The first thing to realise is the interaction is all about quantum level changes which can be to do with, increasing quantum energy change:

translational kinetic energy < molecular rotation < bond/molecule vibration < electronic excitation < bond breaking

The diagram and table attempts to put this sequence the spectroscopy context where appropriate, using the simple molecule HBr hydrogen bromide.

Energy trend	Energy level required	A starter 'picture' view of HBr	Comments on the type of energy involved with the change in the molecule's energy state	Type of associated spectroscopy
↑ increasing energy involved or required	(a) Electronic		Ionisation, high energy uv or high energy electrons (as in a mass spectrometer), 1130 kJmol-1.	mass spectrometry
	(b) Electronic		Homolytic dissociation into free radicals by ultraviolet radiation, bond fission enthalpy 366 kJmol-1.	Not applicable here
	(c) Electronic		Electronic excitation of molecule by uv or visible light to a higher electronic quantum level, followed by relaxation and release of the absorbed energy.	Emission & absorption spectra
	(d) Vibrational kinetic		Vibrational quantum levels of bonds or the whole molecule. In this case it is the H-Br bond vibration.	Infrared spectroscopy
	(e) Rotational kinetic		Molecular rotation quantum levels. The whole molecule rotates within certain specific frequencies - rotational quantum levels.	Microwave spectroscopy
	(f) Translational kinetic		Kinetic energy of movement from place to place, the normal movement of free particles in a gas or liquid.	Not applicable here

Here (c) and (d) are the most important spectroscopic techniques to do with the interaction of electromagnetic radiation with materials for investigative or quantitative purposes.

The material is often a covalent molecule of an organic compound.

mass spectrometry is also very important but doesn't involve EM radiation.

Extra notes on the summary table content

(a) ===>   ===> +  e^-    A molecular ionisation process.

In a mass spectrometer a high energy electron beam knocks another electron off the molecule.

This involves the highest electronic energy level change - removal of an electron from the molecule and a permanent change.

e.g. HBr(g)  +  e^-  ===>  [HBr]⁺(g)  +  2e^-

(b) ===> ===> Homolytic bond fission - dissociation.

The production of two free radicals is facilitated by a uv light photon or a high kinetic energy particle at high temperature.

If the bond fission is promoted by high energy photons it is known as photodissociation or photolysis, which is part of a photochemical reaction.

If the molecule is sufficiently 'excited' to a high enough quantum level, the bond can break and the bonding pair of electrons shared between the two unionised fragments called free radicals.

This involves a high electronic energy level change in the bonding molecular orbitals and a permanent change to the molecule which is now split in two!

(c) ===>    Molecule absorbs energy and elevated to an electronically excited state.

This can happen at a high temperature or from a uv or visible light photon, however, the absorbed energy maybe insufficient energy for the molecule to undergo either bond fission or ionisation as in (a) or (b).

Never-the-less an outer electron, in a bonding molecular orbital, is promoted to a higher quantum level in the molecule which then 'relaxes' by releasing energy and there is no permanent change.

(a) and (b) involve a permanent change to the molecule, (c) to (g) do not.

This is the origin of uv and visible light absorption and emission spectra due to electronic quantum level changes in atoms or molecules.

Coloured materials (to our eyes) are made of substances that absorb visible light involving the outer electrons being promoted to a higher electronic quantum level.

Saturated organic molecules tend to be colourless because the molecular orbital electronic quantum levels are to high for excitation by visible light photons.

However, in unsaturated organic molecules with extensive delocalised electrons in pi orbitals, the electron levels are low enough for photon absorption to produce a colour effect.

e.g. is colourless but is a coloured dye

The molecular grouping causing the 'colour effect' is called the chromophore.

e.g. C₆H₅-N=N-C₆H₄- in the dye molecule shown above and gives a red, orange or yellow effect to the molecule.

β-Carotene is  strongly coloured red-orange organic pigment found in fungi, plants and fruits. It is synthesized biochemically from eight colourless isoprene units.

If the quantum energy is too great e.g. needing a uv photon, then, to us, the material is transparent, white or invisible!

Electronic, vibrational, rotational and translational energy levels of particles are all quantised.

(d) The more complex the molecule the more vibrational modes are possible.

This vibration creates an oscillating electric field which can interact with infrared waves to raise the vibrational quantum level of the bond - a resonance effect.

The stronger the bond, the higher the frequency (greater photon energy) of infrared radiation is required to raise the vibrational quantum level of the bond.

We experience infrared radiation as a heating effect on our skin, whose molecules increase in vibrational energy and then relax and convert the absorbed radiation into thermal energy (e).

Remember EM radiation is an oscillating electromagnetic field (see also microwaves).

(e) The fast rotation of the whole molecule creates an oscillating electric field which can interact with the oscillations of microwave radiation - a resonance effect.

The molecules gain rotational kinetic energy moving to higher rotation quantum levels.

The molecules relax to lower levels releasing energy, usually converting to translational kinetic energy -  thermal energy to cook your food in a microwave.

(f) All free moving particles in gases and liquids have translational kinetic energy.

Even KE this is quantised, but the quantum levels are so close together that the acceleration or deceleration of particles/objects seems to occur smoothly because you don't detect or notice the tiny incremental changes in the quantum levels.

(g) Nuclear magnetic resonance NMR involves the oscillation of the magnetic moment of a rotating nucleus.

The nuclear magnetic moment oscillates between two quantum levels in an applied magnetic field.

15.0.5 Some extra questions for you to do based on Planck's equation

(answers at end of page) - still under development, but some exam practise questions available

15.0.6 Problem solving - deduction of molecular  structure from 4 given spectra - still under development, but some exam practise questions available

Using molecular spectroscopy to investigate and deduce the structure of an organic molecule

INDEX OF SPECTRA QUESTIONS - molecular structure problem solving in organic chemistry

I have no objection to teachers or students doing a selective printout of these questions

You are given the infrared spectrum, mass spectrum, ¹H NMR spectrum and ¹³C NMR spectrum for an unknown molecule and one additional clue to help deduce the identity and molecular structure of the molecule

Question 01 Four spectra of molecule 01, empirical formula C₂H₄O *  ANSWER

Question 02 Four spectra of molecule 02, empirical formula C₃H₈O *  ANSWER

Email doc b with any query or comment?