ATOMIC STRUCTURE AND BONDING The Atom and Sub-Atomic Particles………………………………………………………………….1 Relative Atomic Mass, Ar..…………………………………………………………………………….2 Determination of Ar values 1. The Mass Spectrometer.…………………………………………………………………………4 2. Statistical Calculations.………………………………………………………………………….6 Radioactivity

1. Particles emitted during decay…………………………………………………………………10 2. Deflection of particles in electric and magnetic fields and Detection of particles…….……….11 3. Decay Series……………………………………………………………………………………12 4. Half-life of Nuclides……………………………………………………………………………13 5. Artificially induced Nuclear Reactions – Fission and Fusion………………………………….14 6. Applications of Nuclear Energy………………………………………………………………..15

The Electromagnetic Spectrum………………………………………………………………………20 The Hydrogen Emission Spectrum…………………………………………………………………..21 Ionisation Energy……………………………………………………………………………………..26 Electronic Configuration……………………………………………………………………………..27 Electron Orbitals …………………………………………..…………………………………………30 Shape and Symmetry of Orbitals…………………………………………………………………….34 Factors affecting Ionisation Energies………………………………………………………………..37 Bonding 1. The Ionic Bond…………………………………………………………………………………40 2. The Covalent Bond..……………………………………………………………………………44 i) Molecular Orbital formation..……………………………………………………………..45 ii) The Coordinate Covalent Bond..………………………………………………………….48 iii) Electronegativity differences Bond Polarity..……………………………………………49 3. Fajan’s Rules..……………………………………………………………………………….....53 4. Expansion of Electron Octets..…………………………………………………………………54 5. Summary.………………………………………………………………………………………55 Shapes of Molecules

1. Valency Shell Electron Pair Repulsion Theory..………………………………………………57 2. Molecular Orbital Treatment..…………………………………………………………………60 3. Hybridisation...…………………………………………………………………………………61

Macromolecular Structures.…………………………………………………………………………65 Metallic Bonding.……………………………………………………………………………………..68 Intermolecular Forces 1. van-der-Waals’ Dispersion Forces.…………………………………………………………….72 2. Permanent Dipole/Permanent Dipole Forces.………………………………………………….74 3. The Hydrogen Bond……………………………………………………………………………76 4. Effect of Hydrogen bonding on Properties.………………………………….…………………78 5. Final Considerations and Summary.……………………………………………………………81 The Liquid and Gaseous States………………………………………………………………………85 1. Gas Laws and the Ideal Gas Equation………………………………………………………….86 2. Non-Ideal Behaviour of Gases…………………………………………………………………87

1

ATOMIC STRUCTURE AND BONDING

SECTION 1 THE ATOM An atom is the smallest particle of an element that can ever exist and still show the chemical properties of that element. Evidence for the existence of atoms is obtained from X-ray diffraction experiments and electron microscopy. In 1803, John Dalton, a British scientist, put forward an Atomic Theory, the basic postulates of which are:

1. Matter is composed of tiny particles called atoms, which cannot be created, destroyed or split.

2. All the atoms of any one element are identical. They have the same mass and the same chemical properties. They differ from the atoms of all other elements.

3. A chemical reaction consists of re-arranging atoms from one combination into another. The individual atoms remain intact. When elements combine to form compounds, small whole numbers of atoms combine to form “compound atoms” or molecules as they are now called.

There have subsequently been some modifications to the theory. It is now known that atoms can undergo fusion to produce larger atoms (this process fuels the sun and the hydrogen bomb), and can be split (fission) into smaller particles, which may include smaller atoms, helium nuclei, or the three basic sub-atomic particles – protons, neutrons and electrons. Use your Data Booklet to complete Table 1. Table 1

Particle Proton Neutron Electron Rest Mass symbol and value/Kg Relative Mass Actual Charge/ C Relative Charge

The protons and neutrons make up the central nucleus of the atom, and are called nucleons. The mass of the atom is, therefore, centred in the nucleus, which occupies only a tiny portion of the atom. The electrons are arranged in “shells” or “energy levels” in the space outside the nucleus. Being very diffuse, they occupy most of the volume of the atom. Symbols in use: A = Nucleon Number (formerly called mass number) is the sum of the number of

protons and neutrons in the nucleus of the particular atom. Z = Proton Number (formerly called atomic number) describes the number of protons in the

nucleus of the atom. N = Neutron Number, which indicates the number of neutrons in the nucleus of the atom. For any given atom: A = Z + N.

2

The standard notation for describing details of the particles present in a given atom X, is XAZ . The

carbon atom that contains 6 neutrons and 6 protons would be described as C126 .

A particular nucleus is often referred to as a nuclide. Qn. 1: Using the notation above, write the symbol for the chlorine atom that has the following characteristics: 17 protons and 18 neutrons. Analysis of naturally occurring chlorine shows that while most chlorine atoms, Z = 17, have 18 neutrons in their nuclei, some have 20. Chemical behaviour is determined by the number of electrons in the atom, which in turn is determined by the number of protons in the atom. Since both species have 17 protons (and hence 17 electrons) they are chemically identical. Their different neutron numbers translates to different masses, and hence slight differences in physical properties, for the two species.

Atoms of the same element that differ in the number of neutrons in their nuclei are referred to as isotopes of the element. Qn. 2: The value of Z for calcium is 20. There are six calcium isotopes of nucleon numbers 40, 42, 43, 44, 46, 48. Calculate the numbers of protons and neutrons present in each of the given nuclei. Most elements that occur in nature consist of mixtures of isotopes in fixed proportions. The amount of a given isotope present in nature is referred to as its isotopic abundance. This may be expressed either as a % for each individual isotope, or as the fractional abundance of the isotope i.e. the abundance expressed as a fraction of 1.

Example 1: Naturally occurring bromine consists of 79Br and 81Br in approximately equal amounts, as a consequence, the abundance of each isotope is 50%, and the fractional abundance is 0.5. Example 2: There are three silicon isotopes, 28Si, 29Si and 30Si. Expressed as a % of natural silicon the abundances of these isotopes are: 28Si = 92.28%, 29Si = 4.67%, 30Si = 3.05%. As fractional abundances the values are: 28Si = 0.9228; 29Si = 0.0467; 30Si = 0.0305. It is clear, therefore, that most of the silicon atoms in nature have a nucleon number of 28. As can be seen, the percentages must always add up to 100 and the fractional abundances to 1.0000.

Atoms are very small particles, with diameters 2 x 10-10 m (0.2 nm) and masses in the range 10−27 to 10−25 Kg. Consequently, the commonly used mass scales would not be suitable for expressing atomic masses. Instead, masses of atoms are measured relative to each other, the scale in use being one atom of C12

6 = 12.0000 units. The unit sometimes used for atoms is the atomic mass unit (a.m.u.), but usually, no units are quoted

since atomic masses are ratios, i.e. 12.0000xatomCaofmass

atom1ofmass12

6

.

The relative isotopic mass of a single isotope is the relative mass of an atom of that isotope compared with an atom of 12

6C , which is assigned a relative mass of 12.0000 units, therefore, the

Relative Isotopic Mass is 0012.00xisotopeCtheofatom1ofmass

isotopeparticulartheofatom1ofmass12

6

.

Since naturally occurring elements generally consist of a mixture of isotopes in fairly constant proportion, the Relative Atomic Mass of an element represents the weighted average mass of an atom

3

(on the C126 = 12.0000 scale) and can be defined as “The average of the relative isotopic masses of

the different naturally occurring isotopes balanced in the proportions in which they occur.”

Chlorine has two naturally occurring isotopes, Cl3517 (75% of all chlorine atoms, so its fractional

abundance is 0.75) and Cl3717 (25% of all chlorine atoms, with a fractional abundance of 0.25). The

relative atomic mass of chlorine is:

100)2537()7535( = 35.50 or (35 x 0.75) + (37 x 0.25) = 35.50.

This value is a weighted mean value. The symbol for Relative Atomic Mass is Ar.

Qn. 3: (a) Calculate:

(i) the Ar of Silicon from the data given on page 2; (ii) an accurate Ar for chlorine, given the following abundances: Cl35

17 = 0.7553, Cl3717 = 0.2447.

(b) Express the relative abundances of the chlorine isotopes as percentages.

The Table that is derived by arranging the elements according to their “Z” values, keeping elements that exhibit similar chemical characteristics in the same vertical rows, is called the Periodic Table. The horizontal rows are called Periods, and the vertical rows are called Groups. Qn. 4:

(a) Use your Data Booklet to determine the A and Z values of the elements Tellurium (Te) and Iodine (I).

(b) Determine the numbers of protons, neutrons and electrons in what you think must be the commonest isotopes of Te and I.

(c) Why does Te come before I in the Periodic Table? (d) Suggest a reason why Te has a larger Ar value than I. (e) Find two other pairs of elements that are placed in the periodic table in the reverse order to

their Ar values. Relative masses can also be assigned to covalent compounds. For a given compound this is called its relative molecular mass, symbol Mr, and is equal to the weighted average mass of all the naturally occurring molecules of the compound on the 12

6C = 12.0000 scale. It can be determined by adding the Ar values of all the atoms present (e.g. Mr of methane, CH4, = {12 + (4 x 1)} = 16), or it may be determined by physical methods (e.g. by use of the mass spectrometer, pg. 4).

Ionic compounds consist of giant aggregates of ions of the different elements present. A sodium chloride crystal consists of billions of Na+ and Cl− ions in a 1:1 ratio. Its formula is given as NaCl, and it has a Relative Formula Mass (symbol also Mr), which is determined as for covalent molecules. For NaCl, therefore, Mr = 23 + 35.5 = 58.5.

Deflection of charged particles in electric and magnetic fields If streams of the three atomic particles were subjected to electric or magnetic fields, neutrons, being neutral, would not be deflected. The negatively charged electrons would be attracted to the positively charged plate (or bent in the appropriate direction in a magnetic field). The positively charged protons would be deflected in the opposite direction to that of the electron (Fig. 1 (a) and (b)).

4

Fig. 1 (a): Effect of an electrical field on Fig. 1 (b): Effect of a magnetic field on protons,

protons, neutrons and electrons neutrons and electrons The strength of the applied field would affect how much the particle stream is bent from its original path, that is, its degree of deflection. In a fixed electrical (or magnetic) field, the two factors that affect this degree of bending are

1. the mass of the particle (symbol ‘m’), and

2. the charge carried by the particle (symbol ‘e’).

The degree of deflection of particles in electric or magnetic fields is largest for particles of smallest m/e (i.e. mass/charge) ratio. For particles of larger mass or smaller charge, the deflection is reduced. Fig. 1 shows clearly that the degree of deflection of a proton beam is much smaller than that of a beam of electrons. Although the relative charge on both particles is 1, because of the much greater mass of the proton as compared to that of the electron, its m/e ratio would be much larger than that of the electron, and its degree of deflection would be correspondingly smaller.

Qn. 5: Arrange the following species in order of increasing deflection in a magnetic field: HHeCH 1

142

126

21 ,,, .

Determination of Ar Values - The Mass Spectrometer The rapid strides over the last eighty years in the development of technology for accurate mass measurement has resulted in a swing away from chemical methods and towards physical methods for the determination of Ar values.

Accurate, Ar values are now determined by use of the Mass Spectrometer (Fig. 2). In this instrument, which operates under very high vacuum (i.e. at very low pressures), the gaseous sample fed into the machine is changed into positive ions. This occurs when sample atoms collide with a high-energy electron beam travelling at right angles to the sample beam. Next, the positive ions are accelerated to a high and constant velocity by plates maintained at high negative potentials, and then the stream of positive ions enters a magnetic field.

5

If the ion beam contains ions of different m/e ratios (because the particles have different masses, and/or charges), each of these is deflected to a different extent, and for particles of each m/e ratio, a specific magnetic field strength would be necessary to focus the particles on the collector plate. The field strength necessary to focus a given particle on the collector plate is translated into an m/e ratio, and the recorder produces a printout showing the m/e ratios of the different isotopes present in the sample.

When the spectrometer is operated to ensure that the charge on all the ions is +1, the scale gives the values of the relative isotopic masses of all the isotopes present.

Furthermore, the mass spectrometer also records the relative ‘abundance’ of each of the species (i.e. the relative amounts of the different isotopes present in the sample). This is proportional to the area under the peak for each isotope. Alternatively, the relative peak heights can be used as an approximation for the areas under the respective peaks.

Once the masses and abundances of the different isotopes are known, the Ar (or weighted mean of the isotopic masses) can be calculated, as was discussed previously. Qn. 6: The diagram given shows the mass spectrum of naturally occurring lead.

(a) How many naturally occurring isotopes of lead are there?

(b) Which is the most abundant of these isotopes?

(c) What do you think will be the approximate value of the Ar of lead?

(d) How would the mass spectrum change if some of the lead atoms acquired a 2+ charge?

(e) Suggest how this might be brought about.

203 204 205 206 207 208 209 m/e ratio

abundance

abun

danc

e

electron beam electromagnet

beam of positive ions

negatively charged plate (~ 8000 V)

vapour of sample in

to vacuum pump

m/e ratio recorder amplifier

Fig. 2: The Mass Spectrometer

collector plate

6

Qn. 7: Natural chlorine consists of two isotopes 35Cl and 37Cl with fractional abundances of 0.75 and 0.25 respectively. The beam containing 35Cl+ and 37Cl+ ions is split by the magnetic field into two beams, (1) and (2) as shown in the diagram.

(a) Suggest which of the two ions is responsible for each beam.

(b) The mass spectrum would contain lines at m/e values 35 and 37. (i) Which of the lines would be more intense? (ii) What would be observed if some of the

ions gained a second charge (becoming 35Cl2+ and 37Cl2+)?

(c) Why is the mass spectrometer operated under very high vacuum?

(d) Why must the particles be accelerated to a constant velocity?

(e) In the operation of the mass spectrometer each particle beam is focussed in turn on the detector plate. If the beam caused by the 35Cl+ ions is impinging on the detector, suggest what changes would be necessary to bring the 37Cl+ beam into focus on the detector.

Molecules, too, can be analysed in the mass spectrometer. Initially the molecules are ionised by bombardment with the energetic electron beam. Some of the energy of the bombarding electrons is transferred to the ions produced, and this causes the bonds within the molecules to weaken and some of the bonds may break resulting in fragmentation of the molecules. One fragment carries the +ve charge and the other is neutral. Only the charged fragment can be focussed on the detector. It is this process that produces the 35Cl+ and 37Cl+ ions from the initial molecules of chlorine, Cl2(g), that are fed into the spectrometer.

35Cl 35Cl + e- 2e- + (35Cl 35Cl)+ 35Cl + 35Cl+ (a) (b) (c) (d) In the equation above, species (b) and (d) will be focussed on the detector, but not (a) or (c). The mass spectrum will, therefore, contain lines at m/e 70 as well as the lines due to single 35Cl+ and 37Cl+ ions. The presence of isotopes of chlorine results in different molecular species, and consequently in several lines due to Cl2 molecules, as shown below. Because of this the mass spectrum of molecules can get quite complicated.

species (35Cl 35Cl)+ (35Cl 37Cl)+ (37Cl 37Cl)+

m/e value 70 72 74 The abundances of the different molecular species may be determined from the mass spectrum, or may be calculated statistically. 1. Statistical calculation of abundances of diatomic molecules of elements with two isotopes: Using chlorine as an example, the following steps are carried out. (a) The fractional abundance of each isotope is determined: 35Cl = 0.75, 37Cl = 0.25. (b) For molecules containing only one type of isotope (e.g. 35Cl 35Cl and 37Cl 37Cl) the

fractional abundance of such molecules is equal to (fractional abundance of that isotope)2.

37Cl+ and 35Cl+ ion beam

electromagnet

(1) (2)

Deflection of ions in a magnetic field

7

The fractional abundance of 35Cl 35Cl molecules = (0.75)2 = 0.5625 [or ( 43 )2 = 16

9 ].

The fractional abundance of 37Cl 37Cl molecules = (0.25)2 = 0.0625 [or ( 41 )2 = 16

1 ].

(c) Since the total of all fractional abundances must equal 1, then the fractional abundance of 35Cl 37Cl molecules = 1 – (0.5625 + 0.0625) = 0.375

or 1 – ( 169 + 16

1 ) = 166

The ratio of abundances of the three molecules may be expressed in different ways.

species 35Cl 35Cl 35Cl 37Cl 37Cl 37Cl m/e value 70 72 74 % abundance 56.25 37.50 6.25 fractional either 0.5625 0.3750 0.0625 abundance or 16

9 166 16

1

Qn. 8: Using the axes provided, draw the line spectrum of chlorine in the region m/e 0 to m/e 80 (assume unit charge only on each species).

Qn. 9: The mass spectrum of dichloromethane, CH2Cl2, contains peaks at m/e 84, 86 and 88, having intensities in the ratio 9:6:1. (a) Explain these data. (b) Calculate a value for the Mr of

dichloromethane. (c) Suggest two other peaks that might be

present in the mass spectrum of CH2Cl2 at m/e values above 48, and explain their origins.

Qn. 10: The diagram given illustrates the ion paths in a mass spectrometer using bromine gas. (a) Bromine, Ar 79.91, consists entirely

of mass numbers 79 and 81. Calculate the relative abundance of each of the two isotopes.

(b) Each of the groups of lines, A, B and C, is caused by one of the ions, Br+(g),

2Br (g) and Br2+(g). (i) State which ions cause the lines in

each of the three groups, A, B and C.

(ii) Identify each of the lines in groups B and C, and indicate which line in each group would be the most intense.

abun

danc

e

m/e value Line spectrum of Chlorine

ion beam enters analyzing field

1 2 1 2 1 2 3

A B C

Ion Paths for Bromine

8

Qn. 11: The mass spectrum of C2H5Cl shows peaks corresponding to 1H, 2H, 12C, 13C, 35Cl and 37Cl. Calculate the mass numbers of the most abundant molecular ion and the heaviest molecular ion (the most abundant isotopes of each element being 1H, 12C and 35Cl). Write the formulae of four of the possible ions that would contribute to the peak at a mass number of 66. 2. Statistical Calculation of abundances of molecules of diatomic elements with three isotopes: Problem: Calculate the relative abundance of the molecules that would be present if equal quantities of atoms of 16O, 17O and 18O are allowed to combine with each other, and draw the line spectrum that would be obtained from analysis of this sample. Solution: Abundance of 16O = 17O = 18O, each being 3

1 of any sample, therefore, the chance that two identical atoms would combine would be the same for 16O as for 17O or 18O, and so the abundance of 16O2 = 17O2 = 18O2 = ( 3

1 )2 = 91 each. Consequently in the sample the total fraction of

molecules containing a single isotope only would be 3 x 91 = 9

3 .

The fraction that consists of combinations of different isotopes = (1 93 ) = 9

6 , and since there are equal quantities of the three different isotopes, there is equal chance of forming each of the following combinations of atoms: 16O 17O, 16O 18O and 17O 18O; therefore, each combination would be 3

1 of 96 = 9

2 of the whole.

species relative mass abundance 16O 16O 32 9

1 16O 17O 33 9

2 16O 18O 34 9

2 17O 17O 34 9

1 17O 18O 35 9

2 18O 18O 36 9

1 Qn. 12: On the axes provided draw the stick diagram for the oxygen mixture just discussed. Problem: Chlorine consists of two isotopes, 35Cl and 37Cl in the abundance ratio 3 : 1. Phosphorous is mono-isotopic, 31P. Apart from lines due to atomic ions, the mass spectrum of a chloride of phosphorous contains 9 lines arranged in 3 groups:

GROUP A B C m/e values of lines 66,68 101,103,105 136,138,140,142

(a) Identify the ions responsible for each group of lines.

(b) Predict the abundance ratios of the various m/e values within each group.

abun

danc

e

31 32 33 34 35 36

m/e ratio

9

Solution (a) GROUP A: 35ClP+ : m/e 66, 37ClP+ : m/e 68.

GROUP B: P35Cl2+: m/e 101; P35Cl37Cl+ : m/e 103; P37Cl2

+ : m/e 105. GROUP C: P35Cl3

+: m/e 136; P35Cl237Cl+: m/e 138; P37Cl2

35Cl+: m/e 140; P37Cl3

+: m/e 142.

(b) GROUP A: Since 35Cl : 37Cl = 3:1 the abundance ratio of 35ClP+ : 37ClP+ is also 3:1. GROUP B: An abundance ratio for chlorine of 3 : 1 means that ¾ of all Cl atoms is 35Cl

and ¼ is 37Cl. For the molecules of GROUP B that contain two Cl atoms the abundance ratio would be calculated as follows:

P35Cl2+ abundance = ( 4

34

3 x ) x 1 = 169

P35Cl37Cl+ abundance = ( 41

43 x ) x 2 = 16

6 .

[The “x 2” is necessary since in the ion that contains both isotopes there are two separate positions occupied by chlorine, and each position can be occupied by either a 35Cl or a 37Cl. This might be illustrated as 35ClP37Cl or 37ClP35Cl.] P37Cl2

+ abundance = 161

41

41 1x)x(

The ratio of 101:103:105 would be 9 : 6 : 1.

GROUP C: P35Cl3+ (m/e 136) abundance = 64

274

34

34

3 1x)xx( P35Cl2

37Cl+ (m/e 138) abundance = 6427

41

43

43 3x)xx(

P35Cl37Cl2+ (m/e 140) abundance = 64

94

14

14

3 3x)xx( P37Cl3

+ (m/e 142) abundance = 641

41

41

41 1x)xx(

[The “x 3” is necessary in the ions containing both isotopes since there are three separate positions occupied by chlorine in each of these ions, and each position can be occupied by either a 35Cl or a 37Cl.]

The ratio of molecules of m/e 136 : 138 : 140 : 142 would be 27 : 27 : 9 : 1.

Mass spectrometry is also used in the analysis of ionic compounds. Fig. 3 illustrates the mass spectrum of copper(II) nitrate. Again, fragmentation has occurred, forming a multiplicity of lines in the spectrum. The repeating pattern of groups of lines 2 units apart suggests that there are two isotopes of copper, of nucleon numbers 63 and 65. The highest m/e represents the molecular ion, and all the other lines are produced through fragmentation.

m/e ratio 63 65 79 81 125 127 187 189

abun

danc

e

Fig. 3

10

The different species causing the lines in Fig.8 are:

line 63 65 79 81 125 127 187 189 species 63Cu+ 65Cu+ 63CuO+ 65CuO+ 63CuNO3

+ 65CuNO3+ 63Cu(NO3)2

+ 65Cu(NO3)2+

Radioactivity In 1896, the French chemist Henri Becquerel by chance stored some photographic plates and some uranium salts in the same drawer. When the plates were developed it was seen that they had become fogged. This suggested to Becquerel that the uranium salts were spontaneously emitting radiations that could affect photographic plates. Rutherford showed that neither physical change (e.g. variation in temperature or pressure) nor variation in the nature of the uranium salt affected these radiations in any way. Becquerel called this phenomenon “radioactivity”. Subsequent investigations showed that there were several other elements that were radioactive.

It was found that the radiations emanated from the nuclei of radioactive elements. Three major types of radiations were discovered.

Table 2: Particles emitted during Radioactive Decay

α-particles These are fast-moving particles made up of 2 protons and 2 neutrons only; they are identical with helium nuclei, 24

2 He . Being charged particles, they are deflected by both electrical and magnetic fields.

β-particles These are high-speed electrons, e01 . Since these particles

are also charged, and have a very low mass, they are deflected by electric and magnetic fields, but to a much greater extent than α-particles.

γ-rays These are uncharged electromagnetic radiation similar to X-rays, but of very much smaller wavelength and, therefore, of higher energy. Electrical and magnetic fields do not affect these uncharged particles.

A nucleus will give off either α- or β- particles, but not both. γ-rays may also be given off by nuclei that undergo either α- or β- decay. α-particles are emitted with velocities in the range of 0.1 speed of light. For a particular disintegrating nuclide, all the α-particles have approximately the same speed. Their energies allow them to penetrate a few tens of mm of air, or a thin piece of aluminium foil. Although their range in air is very small, they may produce about 104 ion-pairs per mm of path at atmospheric pressure, as they collide with the molecules in the atmosphere. They are attracted to the negative pole of an electric field, and are also bent by a magnetic field. β-particles are emitted with very variable speeds – even for a particular disintegrating nucleus. Velocities are typically between 0.4 – 0.9 the speed of light. Because of this the penetrating power of these particles will not always be the same. For β-particles of low energy, the penetrating power may even be smaller than that of α-particles, but typically they have about 10 to 100 times the range of α-particles. While they are ionising radiations, they only produce about 100 ion-pairs per mm of path (i.e. about 100

1 that of α-particles). They are attracted to the positive pole of an electric field, and are

11

bent in a magnetic field. Although the charge on a β-particle is half of that on an α-particle, its degree of bending is much greater than that of α-particles because of its negligible mass.

γ-rays have variable energies. They travel at the speed of light. Air acts as an almost perfectly transparent medium to γ-rays. The ability of a material to absorb γ-rays depends on the density and the thickness of the absorbing material, and the most energetic radiation may penetrate up to several tens of mm of lead, or more than a metre of concrete. Being uncharged, γ-rays are unaffected by electric or magnetic fields.

For obvious reasons these diagrams are very similar to those shown in Fig. 1, for the deflection of protons, neutrons and electrons.

Detection of α- and β-particles, and γ-rays

1. Radiations may darken photographic emulsions. 2. The Geiger-Muller counter makes use of the ability of α- and β-particles and (indirectly)

γ-radiation to produce ion-pairs in gases through which they travel. 3. They may produce flashes of light by impinging on screens coated with radio-sensitive

materials (e.g.zinc sulphide phosphor). These are called “scintillation screens”. 4. They produce visible tracks of condensation on passing through super-cooled water vapour or

alcohol vapour (Wilson cloud chambers). Since these radiations originate in the nucleus, which contains protons and neutrons, one can understand how α-particles could be expelled from the nucleus, and γ-rays are simply electromagnetic radiation, emitted to release extra energy from the system. The origin of β-particles needs some explanation. β-particle emission accompanies the conversion of a neutron into a proton.

e p n 01-

11

10

Fig. 4 (a): Deflection of particles in an electric field

Fig. 4 (b) Deflection of particles in a magnetic field

γ

γ

β

β

α α

α, β, γ

α, β, γ

12

When this occurs the nuclear mass remains the same, but the number of neutrons decreases and the number of protons increases. It can be seen that in both α- and β-particle emission the number of protons in the nucleus changes, and so “transmutation” of the elements occurs, with new elements being produced from the original elements.

The question might be asked, why do some nuclei undergo radioactive decay and others do not? Except for 1H and 3He, a nucleus must contain at least as many neutrons as protons, that is, they have a neutron : proton (n/p) ratio of 1 or larger. For the elements with a proton number (Z) less than 20, the nuclei with n/p ratios of about 1 are stable.

For stable isotopes of elements of higher atomic numbers, the n/p ratio increases gradually, exceeding 1.5 in the region of lead and bismuth. In fact, for a nucleus to be stable, the ratio of neutrons to protons must lie within a restricted range for any given number of protons (or neutrons). Although there is no guarantee that nuclei having n/p ratios within that range will be stable, nuclei having n/p ratios appreciably outside the stability range will generally be radioactive – the type of decay occurring tending to adjust the n/p ratio towards more favourable values.

Furthermore because of the large repulsive forces that develop between similarly charged particles, all nuclei with 84 or more protons are unstable. α-emission removes two protons and so this is the method of decay that generally occurs for elements with large proton numbers.

On the other hand, β-particle emission results in a reduction in the number of neutrons and an increase in the number of protons, with consequent decrease in the n/p ratio. For example in the decay process:

Po e Bi 21084

01-

21083

the n/p ratio changes from 1.5301 83

83 210

- for the bismuth isotope to 1.5000 84

84 210

for the

polonium isotope.

When an element with an unstable nucleus undergoes radioactive decay, no further change occurs if the product nucleus is stable. However if the product nucleus is unstable, further decay occurs, until finally a stable nucleus is produced. In nature there are about 40 radioactive species of high proton number, and these fall into three distinct “decay series” called the Uranium, Thorium and Actinium Series. The example used above occurs as one step in the Uranium series. These series constitute what is called “natural radioactivity”.

Fig. 5: The Band of Stability

Num

ber o

f neu

trons

130

120

110

100

90

80

70

60

50

40

30

20

10

0 0 10 20 30 40 50 60 70 80

Number of protons

(1.41:1 ratio) Sn50

120

(1.25:1 ratio) Zr4090

(1.5:1 ratio) Hg80200

band of stability

1:1 neutron-to-proton ratio

13

Qn. 13: The first three steps in the Actinium decay series (the final product of which is 207Pb) are shown: PbAcPaThU cba 207

8222789

23191

23190

23592 .......................... .

Determine the identities of the three particles, a, b and c.

Qn. 14: In the Uranium series, the nucleus of U23892 is changed into a stable isotope in decay processes

that produce in total eight α-particles and six β-particles. Determine the nucleon and proton numbers of the stable isotope, and identify the element produced.

Another important point is that it has been found that for a spontaneous nuclear process to occur the mass of the parent nucleus must be greater than the sum of the masses of the daughter nucleus and of the particle emitted. This mass is changed into energy, the quantum of energy, E, being related to the mass deficit by the equation:

E = mc2

where m = the mass loss, and c = the velocity of electromagnetic radiation in vacuuo. This energy is carried by the α- or β-particle emitted, and where relevant, by the γ-ray photon that accompanies the emission of the other particle.

Half-Life of Radionuclides Unlike chemical reactions, physical conditions like temperature, pressure and catalysts, do not affect the rate of radioactive decay. However the rate of decay is affected by the number of nuclei present in the sample used, so that for any radionuclide the time it takes for a given sample to decay to one-half of its initial amount is the same, regardless of the amount of the sample initially used, and for the same species successive half-lives are constant. The half-life of a radioactive nucleus is defined as the time it takes for one-half of the nuclei in a sample to decay.

The decay process follows first order kinetics, and Fig. 6 is a graphical illustration of the decay process. Half-lives vary greatly, as the following sample shows:

Name and symbol Half-life

Polonium, Po21284 3 x 10-7 seconds

Oxygen, O158 124 seconds

Radon, Rn22286 3.82 days

Phosphorus, P3215 14.3 days

Tritium, H31 12.26 years

Radium, Ra22688 1590 years

Carbon, C146 5730 years

Potassium, K4019 1.3 x 109 years

Uranium, U23892 4.51 x 109 years

t½ {2 x t½} {3 x t½} {4 x t½} time

Fig. 6: Decay of a Radioactive Element

Table 3: Selected Half-lives

Frac

tion

of a

tom

s of r

adio

activ

e el

emen

t rem

aini

ng

1

½

¼ ⅛ 1/16

14

A 12 g sample of the 32P isotope of phosphorus, which decays by β-emission, would contain 6 g of these atoms after 14.3 days, and at 28.6 days after the original measurement was made there would be 3 g of 32P atoms in the sample. The activity (i.e. disintegrations per unit time) of the sample would decrease similarly.

It must be understood, however, that the total mass of the sample would not have changed. The mass of a beta particle is negligible compared to the mass of an atom. Each 32P atom that undergoes β-decay would produce another atom of the same mass! Qn. 15: Determine the nucleon and proton numbers, and hence the identity of the element produced by the β-decay of atoms of P32

15 , and write down its symbol.

Nuclear reactions can be artificially induced, by bombarding the nuclei of atoms with high-energy projectiles, which may be neutrons, protons or α-particles. In these nuclear reactions, the bombarding particle enters the target nucleus, often producing a very unstable species, which then expels another type of particle, and produces one or more different species. Four examples are shown below:

H O He N 4.Li He n B 3.

C He 3 .2 He2 H Li 1.

11

178

42

147

73

42

10

105

126

42

42

11

73

Less energy is required when the projectile is the neutral neutron, for to overcome the strong repulsion by the positively charged nuclei, protons and α- particle projectiles must have very high energies.

Note that in all examples the number of nucleons is unchanged. However whenever the mass of the products is less than the total mass of the reacting particles, huge quantities of energy are released.

The overall process occurring in our sun can be simplified to

He H 42

21 2

This type of process is described as “Nuclear Fusion”. This fusion process takes place in the hydrogen bomb. So much energy is needed to overcome the repulsive forces of the approaching nuclei that the hydrogen bomb has to be set off by the explosion of a small “fission” bomb (see below). Although much work has been and continues to be done, the technology to harness this source of energy in a controlled manner has not yet been developed.

In some cases, nuclei can be made to undergo fission, producing nuclei about half the size of the fissionable material. The nucleus of 235U, when struck by a slow moving neutron produces a very unstable particle, which may split in different ways – in each case releasing vast quantities of energy:

n2 Zr Te 10

9740

13752

U n 235921

0

n3 Kr Ba 10

9136 142

56

If each neutron produced in either reaction were to cause fission of another 235U nucleus, a chain reaction would ensue. The number of fission reactions and the energy released would quickly escalate. When uncontrolled, this is the source of energy in the nuclear fission bomb.

15

Uranium mined from the ground is less than 1 percent fissionable and must be enriched to 4 percent in order to be used in a nuclear reactor. The uranium would have to be 20 to 90 percent enriched before it could be used as a weapon. The controlled use of nuclear fission supplies much of the energy needed for the generation of electricity in power stations nowadays. The reactor core is usually graphite (or it may be “heavy water”, deuterium oxide, OH2

21 ), which acts as a “moderator” – slowing down the neutrons to the speed

required.

Nuclear fuel is processed into a hard ceramic pellet about the size of the tip of a little finger. The pellets are inserted into metal tubes, producing fuel rods. The rods are grouped into bundles to create fuel assemblies, which are loaded into the reactor core. Control rods of a Boron Cadmium Steel alloy inserted into the core absorb some of the neutrons, thereby maintaining the desired rate of fission, and preventing overheating of the reactor. In nuclear reactors, a coolant (e.g. liquid potassium or gaseous CO2) circulates throughout the core of the reactor, which contains the uranium fuel, and the heat removed is used to raise steam, which in turn drives the turbines that generate electricity. A nuclear fuel pellet can supply a lot of energy. One uranium nuclear fuel pellet is equivalent to the energy provided by 807.4 Kg of coal; or 564 litres of oil; or 48.14 cubic metres of natural gas. There are many socially beneficial activities for which use of radionuclides is suitable, as detailed below. However, the main disadvantage of the use of nuclear-powered devices is the problem of getting rid of the nuclear waste generated.

Applications of Nuclear Chemistry

1. Radioactive dating: Dating wood, campfire charcoal, parchment, jawbones and similar carbon-containing objects that are several thousand to fifty thousand years old can be done with radioactive carbon, carbon-14, which has a half-life of 5730 years. Carbon-14 is present in the atmosphere as a result of cosmic-ray bombardment of earth. Cosmic rays are radiations from space that consist of protons and alpha particles, as well as heavier ions. These radiations produce other kinds of particles, including neutrons, as they bombard the upper atmosphere. The collision of a neutron with a nitrogen-14 nucleus (the most abundant nitrogen nuclide) can produce a carbon-14 nucleus.

H C n N 11

146

10

147

Carbon dioxide containing carbon-14 mixes with the lower atmosphere. Its decay is as follows:

e N C 01

147

146

Because of the constant production of carbon-14 (due to a constant cosmic ray intensity for 40,000 to 50,000 years) and, in turn its constant rate of decay, a small, constant fractional abundance of carbon-14 is maintained in the atmosphere. Living plants, which continuously use atmospheric carbon dioxide, also maintain a constant abundance of carbon-14. Similarly, living animals, by feeding on plants, also have a constant fractional abundance of carbon-14. But once an organism dies, it no longer uses atmospheric CO2, and so its 14C decays but is not replaced. The ratio of carbon-14 to carbon-12 begins to decrease by radioactive decay of the carbon-14. The time of death of the organism can then be calculated by using the carbon-14 : carbon-12 ratio. This method is accurate up to 35,000 years ± 300 years.

16

The material being tested must be very pure, so that all of the carbon-14 comes from the test material. A sample of the material (about 30 g) is procured, carefully cleaned physically, and then chemically (using strong acid to remove decay products). The sample is then burned in purified oxygen to produce CO2, some of which will be radioactive, containing 14CO2. The products of combustion are then tested for radiation using a specially designed Geiger counter. Background radiation (from naturally occurring substances and radiation from cosmic rays or reactions caused by cosmic rays) is eliminated as best as possible, but some still filters through, and this is accommodated in a 2% error.

In using this method, it is assumed that the ratio of carbon isotopes in the lower atmosphere has remained at the present level for the last 50,000 years. Analyses of tree rings have shown that this assumption is not quite valid. Before 1000 B.C. the levels of carbon-14 were somewhat higher than they are today. Moreover, recent human activities (burning of fossil fuels and atmospheric nuclear testing) have changed the fraction of carbon-14 atmospheric CO2. This must also be taken into account when accurate values are being calculated.

For the age of rocks and meteorites, other similar methods of dating have been used. One method depends on the radioactivity of naturally occurring potassium-40, which produces argon when it decays. Potassium occurs in many rocks. Once such a rock forms by solidification of molten material, the argon from the decay of the potassium-40 is trapped. Since the half-life of potassium-40 is known, the amount of the remaining potassium-40 and the amount of argon present can be used to calculate the age of the rock. Use of this method of dating and consideration of other factors, leads to the belief that the age of the earth is 4.6 x 109 years.

Another method of dating rocks involves the use of uranium-238 and the final product of its disintegration, the nuclide lead-206. The concentrations of these two isotopes in a crushed sample of the rock, together with the calculated time needed for uranium-238 to finally be changed to lead-206, can be used to calculate the age of the rock specimen.

2. Chemical Analysis A radioactive tracer is a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system. Chemically, the radioactive tracer is no different to the non-radioactive isotope, but it can be detected in exceedingly small amounts by measuring the radiations emitted. Examples of the use of tracers in chemical analysis include:

a. The determination of the solubilities of relatively insoluble compounds (e.g. AgBr), and the vapour pressures of non-volatile substances.

b. Establishing that dissolving processes and other equilibria are dynamic and not static equilibria.

c. Determination of the mechanism of chemical reactions. i. In the late 50’s, a chemist Melvin Calvin, at the University of California, Berkeley,

was able to use radioactive carbon-14 to elucidate the steps in the mechanism of photosynthesis, the overall equation for which is

6CO2(g) + 6H2O(l) sunlight C6H12O6(aq) + 6O2(g) In 1961 he was awarded the Nobel Prize in chemistry for his work on photosynthesis. It was determined that the oxygen produced in photosynthesis is derived from the water, and not from the carbon dioxide.

ii. Tracers were also used in UWI in Jamaica in investigations to determine how the ackee fruit changed from being poisonous when green to edible when ripe.

17

iii. P3215 is formed in a nuclear reactor by neutron bombardment of a sulphur target:

p P n S 11

3215

10

3216

This isotope is a beta-particle emitter. It is made into phosphate fertiliser, and plant absorption of this fertiliser from the soil can be studied by tracing the phosphorus-32 absorbed. Either a Geiger-Muller counter or a photographic plate can be used to trace the progress of the radioactive phosphorus through the plant.

3. Industrial Uses

(i) Radiation is used to detect the thickness of metal sheets. Radiation is passed through the sheet, and the amount of radiation detected after passage through the sheet is used to determine the thickness of the sheet.

(ii) Monitoring the levels of liquid in tins, or in furnaces in the smelting of metal ores. The radioisotope and detector are placed on opposite sides of the system, and their movement up or down is synchronized. The amount of radiation detected at any level depends on whether liquid is present at that level or not.

(iii) Investigation of leaks in water or sewage systems. An appropriate compound containing a radioactive isotope whose emissions cannot pass through the intact pipe is added to the liquid in the pipes, and its presence in the soil is detected by use of a Geiger counter.

(iv) Gamma rays emitted from suitable radioactive isotopes are used in the sterilisation of surgical instruments, hospital blankets etc. This method of sterilisation is more effective than boiling.

(v) To check for engine wear. Piston rings are constructed using radioactive material. As the rings wear away the lubricating oil becomes radioactive. In this way the efficiency of various lubricating oils can be tested.

(vi) Assorted uses, including medical research; testing of new pharmaceuticals; non-destructive testing of pipes and welds; hardening of materials, like hardwood floors; breeding of new varieties of seed with higher crop yields; eradication of insect pests; production of ionisation-type smoke detectors and production of batteries using Plutonium-238 that power heart pacemakers. These batteries utilise the heat energy released in the decay process to generate electricity. They are compact, long lasting, dependable and very precise.

4. Medical Therapy and Diagnosis The use of radioactive isotopes has had a profound effect on the practice of medicine. Radioisotopes were first used in medicine in the treatment of cancer. This treatment is based on the fact that rapidly dividing cells, such as those in cancers, are more adversely affected by radiation from radioactive substances than are cells that divide more slowly. Radium-226 and its decay product radon-222 were used for cancer therapy a few years after the discovery of radioactivity. Today gamma radiation from cobalt-60, a -emitter with a ½ life 5 years is more commonly used.

The greatest advances in the use of radioactive isotopes have been in the diagnosis of disease. Radioactive isotopes are used for diagnosis in two ways. They are used to develop images of internal body organs so that their functioning can be examined; and they are used as tracers in the

18

analysis of minute amounts of substances, such as a growth hormone in blood, to deduce possible disease conditions.

Technetium-99m is the radioactive isotope used most often to develop images of internal body organs. This isotope is identical to technetium-99, except that it is specially produced to be of higher energy. Upon elimination of gamma rays (which is the only type of radiation it emits), it becomes technetium-99 in its nuclear ground state. It has a half-life of 6.02 hours and, therefore, must be prepared immediately before use. The image is produced by scanning parts of the body for gamma rays with a scintillation detector.

Being a gamma emitter only, all of the emitted radiation can leave the body and reach the detectors. Furthermore, the half-life of 6.02 hours is long enough that it does not decay before it can deliver diagnostic benefits. Yet its half-life is short enough that most of its decay occurs during diagnostic work, enabling the maximum use of its gamma radiation for investigative purposes.

The small half-life also means that very small samples will have a sufficiently high activity, while leaving behind a very low concentration of its decay product, technetium-99, which has a half-life so high (2.12 x 105 yr) that the activity introduced into the patient this way is very low. Moreover the body is able to eliminate technetium in any diagnostic form. All of these properties mean that the patient is minimally exposed to the dangers of radiation while enjoying the benefits.

One use of this isotope is in assessing the performance of a patient’s heart. A solution of a tin(II) ion and a solution of a special ion containing the radioisotope are successively injected into the patient’s vein. Under these conditions the ion binds to the red blood cells, and because of the large quantity of blood in the heart the heart becomes visible in gamma-ray imaging equipment. A slightly different technetium-99m species binds strongly to recently damaged heart muscle – which allows assessment of the extent of damage from a heart attack.

Thallium-201 similarly, is used to determine whether a person has heart disease (caused by narrowing of the arteries to the heart). Upon decay this isotope emits X-rays and gamma rays, which can be used to produce images. Diagnosis of heart disease depends on the fact that only tissue that receives sufficient blood flow binds thallium-201. When someone exercises strenuously, some part of the person’s heart tissue may not receive sufficient blood because of narrowed arteries. These areas do not bind thallium-201 and show up on an image as dark spots.

More than a hundred different radioactive isotopes have been used in medicine. These include iodine-131, (½ life of 8 days) used to measure thyroid activity, and in the treatment of liver and brain tumours; phosphorus-32, used to locate tumours, sodium-24, used to trace the flow of blood and to locate obstructions in the circulatory system and iron-59, used to measure the rate of formation of red blood cells.

Review Questions

Qn. 16 (a) Name, and give the compositions of the three kinds of radiations emitted from radionuclides.

(b) Draw a diagram to illustrate the effect of an electrical field on the particles you have named. Briefly explain the differ effects you have illustrated.

(c) How do these particles differ in penetrating power?

19

Qn. 17 (a) Determine the identities and compositions of the product nuclei formed when the following nuclei undergo beta particle emission: (i) neon-19, (ii) copper-58, (iii) barium-140, (iv) lead-211.

(b) State the meaning of the term “n : p ratio” and explain its significance. (c) What effect, if any, does beta particle emission have on the n : p ratios in the examples

used in part (a)?

Qn. 18 (a) Alpha particle emission is the type of decay that is common for elements with atomic numbers higher than 83. Suggest why this is so.

(b) Determine the identities and compositions of the product nuclei formed when the following nuclei undergo alpha particle emission: (i) Radon-220, (ii) Curium-245

Qn. 19 Write balanced nuclear equations for each of the following nuclear reactions: (a) alpha emission from plutonium-242; (b) beta emission from magnesium-28; (c) alpha emission from californium-251; (d) beta emission from potassium-42; (e) alpha emission from einsteinium-252; (f) beta emission from aluminium-30. Suggest why the magnesium, potassium and aluminium isotopes are beta emitters whereas the plutonium, californium and einsteinium isotopes are alpha emitters.

Qn. 20 Write the symbols, including proton and nucleon numbers, for the radionuclides that would give each of the following products by the decay mode specified: (a) fermium-257 by alpha emission; (b) bismuth-211 by beta emission; (c) antimony-121 by beta emission; (d) californium-253 by alpha emission.

Qn. 21 Iodine-131 is a radionuclide that decays by beta emission. (a) Determine the symbol, and proton and nucleon numbers of the product formed. (b) The half life of iodine-131 is 8.07 hr. An investigation is begun with 20 mg of a solid

sample containing 5.00 mg of iodine-131. Determine (i) the mass of iodine-131 remaining after the investigation has proceeded for 24 hours

and 12 minutes and (ii) the mass of solid sample remaining.

Qn. 22 Technecium-99m (half-life 6.02 hr) gives off gamma rays only as it changes to technecium-99 (half-life 2.12 x 105 yr). Both species of technetium can be eliminated from the body. It is used in medical investigation to develop images of internal body organs. From the information given, suggest why it is said to be ideal for this use.

Sources include “General Chemistry” by Ebbing, “Fundamentals of Chemistry” by Brady and Holum, “Chemistry The Central Science” by Brown and LeMay, “Chemistry” by Lewis and Berry, “Chemistry in Focus” by Andrew and Rispoli.

20

The Electromagnetic Spectrum – an Introduction

Research has shown that “electromagnetic radiation”, e.g. the light radiating from the sun, can be described in two ways:

1. As waves, and 2. As “quanta” – discrete packets of energy comprised of photons.

The fundamental characteristics of a wave include its wavelength and its frequency. The wavelength, symbol is the distance the wave travels to complete one up and one down cycle. The frequency, symbol , is the number of up and down cycles passing a fixed point in a given time interval.Electromagnetic radiation travels through different media at different speeds, in the same way that at the seaside one can walk faster along the beach than through the water. The speed through a vacuum is 3 x 108 metres per second – the symbol used to represent this value is ‘c’. Since the wavelengths vary, the frequency of rays of different wavelengths can be determined by the equation:

c

The frequency of the radiation can be connected to the energy of the photons (in Joules) by the equation:

E = h Where h is Planck’s Constant (6.626 x 10−34 Js).

Combining equations (1) and (2) gives E = hc

Large = Small = Low energy. Small = Large = High energy

The electromagnetic spectrum can be divided into different components according to their wavelengths.

When all the wavelengths of visible light are present it is seen as “white light”, but white light can be subdivided into different bands of wavelengths, which are perceived by our eyes as different colours: Violet Blue-

violet Blue Blue-

green Yellow-

green Yellow Orange Red

/nm 395-415 430 485 510 580 600 615 720

This separation is brought about when white light is passed through a prism or diffraction grating. Viewing such a system is facilitated by use of a simple optical system as in a spectroscope.

Cosmic Rays

-Rays

X-Rays

UV Rays

VISIBLE Infrared Radiation

Microwave Radiation

Radio Waves

/nm 10−3 10−1 10 350 – 770 105 107

Energy (J) 10−14 10−18 – 10−20 10−27

Table 4: The Electromagnetic Spectrum

Table 5: Colours and Wavelengths of Visible Light

21

Absorption and Emission Spectra Different parts of the electromagnetic spectrum can be utilised in various types of spectroscopic investigations. In producing an “absorption spectrum”, the chosen radiation is passed through the material and analysed upon exiting. As the radiation passes through the material, the atoms or molecules of the sample absorb specific wavelengths; the energy packets absorbed raising the absorbing particles to an excited state. The wavelengths that are absorbed are missing, or less intense, when the radiation reaches the detector. Alternatively, since the excited state produced by the absorption of the photons of energy being supplied is transitory, the energy is released when the material reverts to a lower, more stable energy state. Analysis of this energy produces an “emission spectrum”.

A particular species always produces the same spectrum when treated in the same manner. The reproducibility of these data allows the spectra to be used in analysis of samples.

In the early twentieth century, Max Planck, a renown physicist proposed his quantum theory, which suggested that while the electron was under the influence of the nucleus it would only absorb or emit energy in specific amounts called quanta. In 1913, Neils Bohr suggested that there were a restricted number of orbits at varying distances around the nucleus that the electron could occupy, and that each orbit was associated with a specific amount of energy. The further away from the nucleus a given orbit was the higher was its energy.

When an electron occupied a particular orbit it must possess the energy associated with that orbit. In other words, its energy is quantised. The orbits were assigned “quantum numbers”. The orbit nearest to the nucleus (and, therefore, of lowest energy) was assigned the quantum number 1, and consecutive orbits outwards from the nucleus increased in quantum number by 1 unit, from orbit to orbit.

If an electron were to occupy a given orbit, it could move in that orbit without requiring or giving out energy, but for an electron to move from an orbit near to the nucleus to one further away from the nucleus (and, therefore, of higher energy), it would have to absorb the difference in energy between that of the orbit it presently occupied (E1) an that of the orbit to which it transmits (E2).

E2 – E1 = E The excitation process that is brought about during generation of an absorption spectrum involves an increase in energy of electrons in the atoms of the gaseous sample, and the energy levels occupied by the affected electrons increase – i.e., atomic electronic transitions occur. Since the energy of the electron in an atom is quantised, the energy changes that occur are not arbitrary, but are very precise, and so the energy of the light (or other electromagnetic radiation) being absorbed or emitted would necessarily be specific, and would be prescribed by the relationship:

E = h The frequency of the radiation absorbed must be such that the exact energy required for the given electronic transition is provided, and so a line spectrum is produced by atoms of elements. Absorption of sufficient energy would allow the electron to be removed completely from the influence of the nucleus, and the atom would become ionised, at which point the electron could absorb and emit energy incrementally. Its energy is no longer quantised, and it generates a continuous spectrum.

The Hydrogen Spectrum When the single electron of the hydrogen atom is in the orbit of lowest energy (that of quantum number 1) it is said to be in its ground state. In any other orbit it would be in an excited state. For a sample of hydrogen containing trillions of atoms, not all of the sample will have their electron in the ground state. Atoms will be found with their electron in all the available higher energy states.

22

The emission spectrum of hydrogen produced by a spectrometer consists of a number of separate groups, or series, of lines, one in the ultraviolet, one in the visible and three in the infrared part of the spectrum. Two of these series are shown in Fig. 7. The series are named after their discoverers. The three series in the infrared part of the spectrum, as the wavelength increases, are the Paschen, Brackett and Pfund series.

In each series, the intervals between successive lines becomes smaller and smaller, towards the low (high ) end of the spectrum, until at the “convergence limit” they converge, forming a continuous band. The energy needed for each separate electron transition would require a quantum of light energy, a photon. For a given electron transition, the frequency of the light absorbed would be such that

E = h Each specific transition that the electron could undergo would require photons of different energy values, and, therefore, light of different frequencies and consequently different wavelengths. If there were no restrictions relative to the amounts of energy the electron in a hydrogen atom was able to

Energy Levels ∞ 65 4 3 2 1 Nucleus

Pfund

Brackett

Paschen

Balmer

Lyman

Fig. 8: Electronic Transitions generating lines in the Absorption Spectrum of Atomic Hydrogen

Continuous spectrum Continuous spectrum

Lyman Series Balmer Series

700 400 200 100 /nm

Convergence Limit

Visible Ultraviolet

Convergence Limit

770 350

Fig.7: The Lyman and Balmer series in the Emission Spectrum of Atomic Hydrogen

23

absorb (and consequently to release when it reverted to a lower energy state), the spectrum obtained would be a band spectrum, not a line spectrum.

For each allowable transition light of a specific wavelength is removed by the atoms in the sample, and analysis of the light leaving the sample would show black lines wherever large numbers of photons of those wavelengths had been removed. In each of the series of lines depicted in Fig. 8, the arrows with dashed lines indicate the transitions that would involve removal of the electron from the atom.

In the generation of an emission spectrum, the same energy must be released in a particular downward transition (e.g. from level 2 to level 1) that was absorbed when the electron made that particular upward transition (e.g. from level 1 to level 2), and so the emission spectrum would consist of lines of wavelengths that correspond with the dark spaces in the absorption spectrum.

The transitions from higher levels to level 2 produce a series of lines in the visible part of the spectrum. This is called the Balmer series.

As we have seen earlier, each series of lines has a “convergence limit”, where the line spectrum becomes continuous; the convergence limit for the Lyman series is indicated by the dashed arrow in

Principal quantum number, n

8 7

1

6 5 4

3

2

∞

Paschen Pfund Balmer Brackett Lyman

Fig. 9: Electronic transitions giving rise to Spectral Series in the Hydrogen Emission Spectrum

Continuous spectrum

486

434

397

364 /nm Convergence Limit

Fig. 10: The Balmer Series in the Emission Spectrum of Hydrogen

656

410

n = 3 4 5 6 7 colour of light red green violet ultraviolet

24

Fig. 9. Its value corresponds to the transition from n = ∞ orbit (in which the electron is not under the influence of the nucleus) to its ground state (n = 1). The energy value corresponding to this transition would be required to remove the electron from its ground state to the point at which it is no longer under the influence of the nucleus (i.e. to ionise the atom), and so the energy that corresponds to the frequency at the convergence limit of the Lyman series can be used to determine the ionisation energy of hydrogen.

The energy required to remove an electron from the ground state of each atom of a mole of hydrogen atoms to the point where the electrons are no longer attracted to the nucleus is the ionisation energy of hydrogen. It must be stressed here that while each series has a convergence limit, which represents removal of the electron from the atom, only the value for the convergence limit of the Lyman series could be used to determine the ionisation energy of hydrogen, for only level 1 is the ground state of the hydrogen electron. (See the definition of ionisation energy above.)

Knowledge of the frequencies of a number of the lines in the Lyman series, and the fact that the interval between successive lines decreases, makes it possible to determine the ionisation energy by a graphical method.

values are calculated, by subtracting the frequency of the first line from that of the second, and the second from the third etc. A graph of the lower of the two frequencies in each case () is plotted (y-axis) against (x-axis). The convergence limit is the point at which there is no interval between the lines, i.e. = 0, and so extrapolation of the curve obtained to 0 on the x-axis, allows the frequency of the convergence limit to be determined. (See Fig. 10) 0

Once the frequency has been determined, the corresponding energy value is obtained by using Planck’s equation, E = h However this energy relates to a single atom, and since the value of the ionisation energy refers to a mole of atoms, the value of E must be multiplied by the Avogadro’s constant (L), which is the number of atoms in a mole. Overall, therefore, when the frequency or wavelength refers to that at the beginning of the relevant continuum, the ionisation energy is obtained from:

E = hL (4) or Ehcx L

convergence limit

Fig. 11: Graphical determination of the of the Convergence Limit

25

The wavelength at the convergence limit of the sodium emission spectrum is found to be 242 nm (which is 2.42x10−7m). Calculation of the first ionisation energy of sodium would be done as follows: First the energy required per atom would be calculated using equation 5, Ehc:

E = )1042.2(

)10998.2()10626.6(7

1834

mxmsxxJsx

= 8.209 x 10−19 J

The energy needed per mole of atoms is now calculated by multiplying the energy per atom by the Avogadro constant L, 6.022 x 1023.

(8.209 x 10−19J per atom) x (6.022 x 1023 atom per mol) = 494320.3 J mol−1

The ionisation energy of sodium is 494.32 kJ mol−1 If the values of the wavelengths or frequencies of the lines are not known, they can be calculated, for it has been found that the lines in the hydrogen spectrum are given by the relationship

2

221

111nn

RH (6)

and since

c ,

c

1 and so

2

221

11nn

cRH (7)

In these equations, λ is the wavelength of the line, ν is the frequency of the line, c is the velocity of light in a vacuum, RH is the Rydberg constant (109700 cm−1), and n1 and n2 are the principal quantum numbers of the energy levels concerned in the electronic transitions. For the emission spectrum, n1 being the principal quantum number (PQN) of the level to which, and n2 the PQN of the level from which the electron is returning. The equations (6) and (7) can be manipulated in a number of ways in order to obtain a value for the energy of the line at the convergence limit, and consequently for the ionisation energy of hydrogen.

Qn. 23 The frequencies of the first six lines in the Lyman series (in 1015 Hz) are: 2.469; 2.926; 3.086; 3.160; 3.201; 3.225.

Using a graphical method, with the aid of values of the relevant physical constants in your Data Booklet, calculate a value for the ionisation energy of hydrogen. How does the value you have calculated compare with the value given in the Data Booklet?

Qn. 24 The emission spectrum of the element hydrogen contains several series of lines. Explain how these series arise.

Qn. 25 The wavelength, λ, of each of the lines in the emission spectrum of hydrogen may be calculated

using the expression

2

221

111nn

RH

where: RH is the Rydberg constant (109700 cm−1) and n1 and n2 are the principal quantum numbers of the energy levels concerned. Calculate the shortest wavelength for the series in which n1 = 2, and decide in which region of the electromagnetic spectrum this line appears.