Atomic models THE NUCLEAR ATOM

Giant strides in our understanding of nature always go hand in hand with the emergence of new techniques - either experimental or theoretical. The refinements in techniques of making accurate measurements of mass, led to the laws of chemical combinations. These, in turn, led to the formulation of the first elaborate model of the atom as proposed by Dalton. Techniques for generating high voltage and high vacuum led to the cathode ray tube experiments. These experiments revealed the electrical nature of the atom and formed the basis for Thomson's model. The advent of the technique of controlled production of - rays and of the detection of - particles helped Rutherford to put forth the first nuclear model. Rutherford's Model In his famous experiment, Rutherford bombarded a thin gold leaf with - particles and observed their fate. He observed that most of the - particles (massive and positively charged) passed through the gold leaf without any deviation. However the few that were deviated, did so a lot! He concluded that atoms have large empty spaces in them. The nucleus which has most of the mass and all of the positive charge is extremely small in size. The negatively charged and almost massless electrons are placed at a very large distance from the nucleus. How do the negatively charged electrons stay far away from the positively charged nucleus? Rutherford defended his nuclear model by proposing an analogy with the solar system. According to him the electrons are moving around the nucleus just as the planets orbit at great distances from the sun without falling into it due to gravitational attraction. Let us study the motion of an electron around a proton as in the hydrogen atom.

mass of an electron, me (9.1 10-31 kg) or 0.00054858 amu charge of an electron, -e (-1.6 10-19 coulombs) charge of a proton, +e (+1.6 10-19 coulombs) velocity of the electron, v (tangent to the orbit) radius of the orbit, r (distance between the proton and the electron)

For the electron to revolve in a circular orbit around the nucleus (proton) with radius

r and velocity v it must experience a centripetal force which is given by This force is experienced due to the attraction between the proton and the

electron which is equal to .

( is the permittivity constant in vacuum, = 9.0 109 Nm2C-2) The negative sign denotes attraction, we can thus write

From this we get

(a) Kinetic energy of the electron:

(b) Relation between v and r : We can also evaluate the total energy of the electron. Total energy = Kinetic energy + Potential energy

= Explanation:

(a) We have used to evaluate the kinetic energy. (b) The potential energy is the standard expression for a charge -e placed at a distance r from another charge +e.

or, Total energy = Both total energy and potential energy carry a negative sign . Some important conclusions

1. This means larger the radius lower is the velocity and kinetic energy of the electron. Both will decrease to zero for infinite r. 2. Both total energy and potential energy are proportional to 1/r. But because of the negative sign, these values increase as r increases. The values increase to

zero as r goes to infinity! Also total energy = 1/2 PE 3. The magnitude of kinetic energy is half that of potential energy. Since kinetic energy must be positive we may write,

KE = |PE| 4. As r increases (a) kinetic energy decreases (b) potential energy increases at double the rate and (c) total energy increases. Assignment Consider the hydrogen atom to be a proton embedded in a cavity of radius a0(Bohr radius) whose charge is neutralised by the addition of an electron to the cavity in vacuum, infinitely slowly. Estimate the average total energy of an electron in its ground state in a hydrogen atom as the work done in the above neutralisation process. Also, if the magnitude of the average kinetic energy is half the magnitude of the average potential energy, find the average potential energy. (JEE '96) Answer In order to bring the electron to the cavity infinitely slowly from infinity, an external

force of has to be constantly applied opposing the attraction. The total work done in the process

= Note that due to the infinitely slow movement, no kinetic energy has been allowed to develop. If now this total energy is to be distributed between KE and PE, the potential energy has to decrease further in order to account for the development of KE so that total energy remains constant.

If KE =

Then, Total energy = KE + PE = PE + PE =

or

The Collapsing Atom - Limitation of the Rutherford Model The radius of the orbit determines the energy of the electron and vice versa. But from Maxwell's laws of electrodynamics, we know that the electron in an orbit being an accelerated (constant change of direction) charged particle must give out energy in the form of electromagnetic radiation. However, if such a thing happens, then it would mean that the radius of the orbit should constantly decrease and the electron should collapse into the proton following a spiral path. This does not happen! Before going into how developments provided a solution to this paradox, let us look into some other general aspects of the atom - specially its nucleus. THE CONSTITUTION OF THE ATOM: A GENERAL PICTURE The atom basically is made up of three types of fundamental particles: electrons, protons and neutrons. The physical data regarding these particles are summarised in a table below. Some other salient features are: Electrons 1. The electron (negatron) was discovered as cathode rays by Goldstein (1876) in discharge tube experiments. 2. Its charge/mass (e/m) ratio was determined by Sir J. J. Thomson (1897). 3. The charge of an electron was determined by Milliken through his oil drop experiment (1909). 4. It has an antiparticle called positron which may combine with it to give two photons of -radiation and vice versa.

e- + e+ 2 5. The charge of an electron is the smallest unit of charge.

6. The mass of an electron is sometimes used as a unit for calculating the relative mass of other fundamental particles. Protons 1. Protons were discovered in canal rays in hydrogen gas discharge tube experiments by Goldstein (1886). 2. The charge is the same as that of the electron in magnitude but opposite (positive) in sign. 3. The mass is ~ 1837 times greater than that of the electron. Neutrons 1. Neutrons were discovered by Chadwick (1932) while studying nuclear reactions. 2. They are very unstable and decay with a half-life of 920 s as: neutron proton + electron + antineutrino The unstability is the reason why they were discovered so late. 3. They have no charge but have a mass nearly the same as that of a proton.

Nucleons 1. Neutrons and protons are clustered in the nucleus and hence are called nucleons. 2. The attractive force between the nucleons (being baryon) is also called nuclear force which is much stronger than the coulombic repulsion force between protons. Thus many protons can co-exist within a small nucleus. 3. The strength of the nuclear forces decreases more rapidly than coulombic forces with increase in distance of separation. Hence large nuclei are unstable. 4. A large number amount of energy is released when nucleons come to form a nucleus. This energy is produced by the conversion of a part of the mass of the nucleons according to Einstein's formula E = mc2 and is called the binding energy. The decrease in mass is called mass defect. The binding energy per nucleon is an indicator of stability of the nucleus.

Nuclear Charge, Atomic Number and Chemical Identity 1. The charge of the nucleus is units of the electronic charge and is calculated from the number of protons in the nucleus. This is also referred to as atomic number Z and it determines the number of electrons in an atom. 2. The electrons are bound to the nucleus by coulombic forces which are much weaker than the nuclear forces. Chemical reactions involve rearrangement and transfer of electrons and the energy requirement are orders of magnitude less than that involved in nuclear reactions. Nuclear reactions are independent of the chemical state (bonding) of atoms. 3. The number of electrons (and hence the atomic number) determines the

chemical properties and chemical identity of an atom.

Nuclear Mass and Mass Number 1. The mass of the nucleus is calculated from the number of nucleons (neutrons and protons) and is therefore referred to as mass number A. 2. The mass of the nucleus is NOT equal to the sum of the masses of the nucleons, due to mass defect.

Nuclidic Symbols - The Nuclear Identity of Atomic Species The IUPAC term used to denote an atomic species of a given atomic number Z and mass number A is nuclide. The corresponding formula for a nuclide is

Thus, for describing a carbon atom which has 6 protons (Z = 6) and 6 neutrons (A =

6 + 6 = 12) the nuclidic symbol is . Writing and using nuclidic symbols: calculating the number of electrons, neutrons and protons in an atom or an atomic ion A nuclidic symbol gives more than the required information for finding out the number of electrons, protons and neutrons in an atomic species. The atomic number (or the number of the element) and the mass number (or the number of neutrons) along with the charge (if it is an atomic ion) is enough for obtaining all information. Example

Given the nuclidic symbol derive the following details: (a) atomic number (b) mass number (c) number of protons (d) number of electrons (e) number of neutrons (f) net charge. Answer (a) Z = 28 (given in symbol) (b) A = 90 (given in symbol) (c) number of protons = Z = 28 (d) number of electrons = Z - netcharge = 28 - 2 = 26 (e) number of neutrons = A - Z = 90 - 28 = 62 (f) Net charge is 2+ (given in symbol) The nuclidic symbols also help us to compare and classify atomic species in terms of isotopes, isobars, isotones and isodiaphers. Example Identify sets of nuclides which are (a) isotopes, (b) isobars, (c) isotones,

(d) isodiapheres from the following:

Answer 1. Isotopes: Nuclides having the same atomic number but different mass numbers i.e. those having the same number of protons but different number of neutrons are called isotopes. Sets of isotopes above are -

Isotopes have the same atomic number and hence same chemical identity. However, because of their different masses they sometimes may induce a difference in physical properties such as boiling point or even to some extent bond energies (due to difference in mass). Their nuclear properties may however be very different.

For example, is radioactive and gives out a - particle to give its isobar

. But is a stable nuclide. 2. Isobars: Nuclides having the same number of nucleons are called isobars. We

have already seen above that and are isobars.

When a - particle (an electron) is emitted from a nuclide, a neutron is changed to another nucleon i.e. to a proton. The daughter nuclide thus becomes an isobar of the mother. Isobars have different physical and chemical properties. 3. Isotones: Nuclides having the same number of neutrons (N) are called isotones. The number of neutrons is given by N = (A - Z). The sets of isotones in the example

above are: (i) and (N = 14 - 6 and 15 - 7 = 8)

and (ii) and (N = 13 - 6 and 14 - 7 = 7) Isotones also have different chemical and physical properties. 4. Isodiaphers: Nuclides having the same difference in their number of neutrons (N = A - Z) and number of protons (P = Z) are called isodiaphers. Thus isodiaphers have same value for N - P = (A - Z) - Z = A - 2Z The sets of isodiaphers in the example are:

(i) and (ii) (N - P = 13 - 12 = 15 - 14 = 1)

and (ii) and (N - P = 12 - 12 = 14 - 14 = 0) Isodiaphers are usually used to identify mother and daughter nuclides in an alpha,

decay which involves the simultaneous loss of two protons and two neutrons, thus maintaining the (N - P) value. Another important term involving nuclear identity is isomer. If two nuclides have the same members of protons and neutrons, respectively, but have different nuclear properties, they are called nuclear isomers.

Example Arrange the following species in order of increasing number of neutrons

Atomic models Answer

Example (a) Write the nuclidic symbol for a species with 29 protons, 34 neutrons and 27 electrons. (b) The mass number of an ion is 69. It has 38 neutrons and 28 protons. Give its nuclidic symbol. Answer

Example Fill in the following table. Look up the periodic table, if necessary, to find the chemical symbol for a particular atomic number. Table

Answer

The Atom and the Nucleus - Comparing Sizes (a) The size of the nucleus is of the order of 10-15 m (1 femto metre). The radius R of the nucleus of an atom having mass number A is given by R = A1/3 1.4 10-15 m = A1/3 Fermi (b) In contrast the average radius of an atom is of the order of 10-10 m (1 Angstrom) i.e. 105 times greater than that of the nucleus. This means there is a lot of empty space in an atom. Relative Atomic Mass, Mole Concept, Avogadro's Number and amu 1. The mass of an atom is the mass of the nucleus plus the mass of the electrons. Since a nucleon is about 2000 times heavier than an electron, we usually consider the mass of the nucleus as the mass of the atom and do not consider atoms and their ions to have different masses for stoichiometric calculations. 2. The mass of an atom is usually expressed relative to the mass of an arbitrarily defined standard atom, in atomic mass units (amu). 3. Because of mass defect, the average mass of a nucleon is different from the actual mass of the nucleon and also differs for different atoms. The amu is defined

as 1/12th of the mass of a nuclide. An atom n times heavier than a

atom has a mass 12n amu e.g. Mg is two times heavier than a and has a mass of 24 amu. 4. Moles and molar mass In most calculations, chemists are more interested in the number of atoms, ions or molecules present rather than in the mass of the sample or volume of the solution. Knowing the number of atoms is fundamental to quantitative chemistry, i.e. calculation of properties. In order to efficiently determine and report enormously large numbers of atoms, ions or molecules in a sample, chemists use a unit called mole. 1 mole is the number of atoms in exactly 12 g of carbon-12.

Since it is impractical to directly count atoms in a mole of a sample, we can use an indirect route based on the mass of one atom. The mass of a carbon-12 atom has been found, by mass spectrometry, to be 1.99265 10-23 g. Hence, the number of atoms in exactly 12 g of carbon-12 is

Therefore, 1 mo le of any objects, atoms, ions or molecules means 6.0221 1023 of those objects. The number of objects per mole, i.e. 6.0221 1023 mol-1 is called the Avogadro constant, NA .

N = nNA Molar mass: The molar mass M is the mass per mole of particles

The value for No using as standard is 6.023 1023 mol-1. If is taken as the standard for amu, then the volume of No would be slightly less, since mass of

Example Suppose the atomic weight scale was so changed that the atomic weight of Ne becomes 100 amu, what would the value of Avogadro's number change to? Given that in the current scale, the weight of Ne is 20.18 amu and Avogadro's number is 6.02 1023. Answer

Let new value of amu = amu' and new value of N0 = N0 ' Then we have 100 amu' = weight of one Ne atom = 20.18 amu

or From the equation in point 5 above, we can also write 1 amu N0 = 1 g = 1 amu' N0 '

or N0 ' = N0

Substituting the value of N0 and of we have

N0 ' = 6.02 1023 = 2.983 1024 Assignment If the atomic weight scale is changed as mentioned in the above example, what

would be the new atomic mass of in new amu? 6. The above relation embodies the mole concept and is used to convert the number of atoms into the mass of atoms and vice versa. According to this,

7. The mass of a atom is 12.0000 amu (by definition). The mass of 1 mole

of atoms is

12.0000 g of atom are therefore referred to as 1 gram mol of atoms or

simply 1 gram atoms of Example

If the mass of is 18.998 amu and Avogadro's number is 6.02 1023 mol-1, how

many atoms are there in 10 g of ? Answer

Weight of 1 g mole (or 6.02 1023 atoms) of is by definition 18.998 g.

Hence 10 g of contains:

atoms of In general:

and one mole contains atoms. Average Atomic Mass and Isotopic Masses

1. Most naturally occuring elements consist of a mixture of isotopes, having identical chemical properties, the composition of which is fairly constant. The masses of the isotopes are different from each other and different from the average mass of an atom. 2. When we refer to the atomic mass of an element we usually refer to the average atomic mass. Example Iron found in all iron ores, meteorites and iron compounds consists of

If we consider the atomic masses of the different isotopes to be equal to their respective mass numbers, calculate the average atomic mass (Av) of iron.

Av = 55.912 The actual average atomic mass of iron is 55.847. The discrepancy is because we have wrongly taken the mass of an isotope as equal to the mass number. The above example however illustrates, why for accurate stoichiometric calculation, we should use the average atomic mass. We can also understand why atomic masses are usually not whole numbers though mass numbers are. Assignment

The mass of and are 34.97 and 36.97, respectively. If the average atomic

mass of naturally occurring chlorine is 35.45 and if and are the only two isotopes of chlorine, calculate their percentage distribution in naturally occurring chlorine.

Answer 76% and 24% 3. The accurate mass of an isotope may be determined by using a mass spectrometer. When atoms of different isotopic atoms are ionised, the ions have the same charge but have different masses. They will thus have different e/m ratios which can be accurately determined in a mass spectrometer. Since the charge of an electron is accurately known, we may calculate the exact mass of the ion.

NATURE OF ELECTROMAGNETIC RADIATIONS, ATOMIC SPECTROSCOPY AND BOHR MODEL OF THE HYDROGEN ATOM Contrary to the classical theory of electrodynamics as applied to the 'Rutherford atom' - atoms are stable and their electrons do not spiral into the nucleus by giving out energy in the form of electromagnetic radiations. The clues towards solving this paradox came in the form of two developments in physics at the turn of the last century.

1. An understanding, mainly due to Max Planck (Black body radiation, 1900) and Albert Einstein (Photoelectric effect, 1905), of the dual (wave as well as particle) nature of light. 2. 'Atomic spectroscopy' and the quantitative analysis of spectral lines (J. J. Balmer, 1885) We will see how Bohr used these clues to 'solve' the paradox posed by the Rutherford model. But first a few concepts and some operational skills. The Wave Nature of Light and the Electromagnetic Spectrum Light is a wave. It consists of oscillations in electric and magnetic field that can travel through space (even in vacuum!). Visible light as well as X-rays and radio waves are forms of electromagnetic radiation and are characterised by their wavelength ( ) or frequency ( ). If for a wave, the wavelength is metres per cycle and the frequency is cycles per second, then in one second the 'wave front' moves

through which is the wave velocity of a propagating wave. The wavelength and the frequency of a propagating wave such as light is therefore related through its velocity c by the relation . The velocity of light in vacuum is an absolute physical constant independent of its frequency, wavelength or even the relative motion of an observer (!) and has a value

. Another way to characterise light waves (apart from its and ) is by its wave number ( ) which is equal to 1/ . It represents the number of wavelengths (the frequency of the wave) per unit distance and is expressed in SI units in metre-1. Students often make mistakes in the conversion of wave number from one unit to

another. e.g. (not ). After all, the number of waves in 1 cm will be 100 times less than the number of waves in 1 m. Let us now correlate wavelength, frequency and wave number o f electromagnetic radiations. Example The visible spectrum of hydrogen shows only four spectral lines corresponding to the colours red, blue-green, blue and violet. (a) One of them has a wavelength 6.56 10-7 m (i) What is the wavelength in nanometers? What is its frequency and wave number?

Velocity of light c =

(b) Another line corresponds to the wave number . Calculate its

wavelength in nanometer (nm) and angstroms ( ). Also calculate its frequency. (a) (i) (ii)

(iii) Wave number We first convert the wave number into SI units.

since

we have

(i) 1 nm = 10-9 m 4.86 10-7m = 486 10-9 m = 486 nm

(ii) 1 = 10-10 m

4.86 10-7m = 4860 10-10 m = 4860

(iii)

Alternatively

Assignment (a) A third line in the visible spectrum of hydrogen has a wavelength of

4340 .Calculate its (i) frequency and (ii) wave number. (b) The wave number of the fourth line in the same spectrum is given by the Balmer formula

Calculate the (i) wavelength and (ii) frequency.

(Answer: (a) (i) (ii)

(b) (i) (ii) ) Assignment

In the hypothetical electromagnetic wave pictured, the distance between two representative points is noted. (a) Calculate the frequency of this radiation. (b) What will be the energy per mol for this radiation? (c) What is the wave number for this radiation? (d) A 100 watt lamp sends out 20 percent of its energy output in the form of waves of the above frequency. How many photons of the above frequency are given out per second from the lamp?

(Answer: (a) (b) (c)

(d) The range of frequencies or wavelength of electromagnetic radiation is called the electromagnetic spectrum. The chart below will give you an idea regarding the wavelength of different types of electromagnetic radiations and also their approximate comparison with the sizes of material objects.

Objects for comparison Size/Wavelength Corresponding type of electromagnetic radiation

Diameter of an atomic nucleus cosmic rays

Radius of 1s electrons of hydrogen atom - rays

Radius of a carbon atom -rays

Thickness of a small enzyme molecule X-rays

Thickness of a spherical polio virus X-rays

Thickness of a Tobacco mosaic virus UV

UV

visible

Length of E coli bacterium near infrared

Length of a section of blue algae far infrared

Length of paramecium far infrared

Length of a mosquito far infrared

Length of a bee microwave, radar

Length of a small rabbit UHF TV

Height of a human being VHF TV

Length of a blue whale

Length of a foot ball field short wave

medium wave Distance from Parliament House to the Gateway of India

flight navigation

Atomic models Quantum Effects and the Particle Nature of Light (Photons) Originally Newton had given the idea of a beam of light as a stream of particles. In the beginning of the 19th century, Thomas Young used the properties of diffraction and interference of light to show the wave nature of light. A century later, Max Planck hesitatingly introduced the concept of quantisation of energy. Einstein extended this concept to include the structure of light itself. He postulated that light consisted of quanta (now called photons) or particles of electromagnetic energy, with the energy E proportional to the observed frequency of light.

Where h is a physical constant called Planck's constant and has a value

. He used this photon concept to explain the photoelectric effect: the ejection of electrons from the surface of a metal or other material when light shines on it. Calculating the energy of photons Example A 2000 watt sodium vapour lamp gives out half its energy in the form of yellow light with a frequency of 5.09 1014 Hz. (a) Calculate the (i) wave length and (ii) wave number of the yellow light. (b) What will be the energy of (i) each photon in joules and (ii) one mole of photons in kJ mol-1? (c) How many photons corresponding to the above mentioned radiation must be coming out of the lamp every second? Answer

(a) (i)

(ii)

(b) (i)

=

(ii)

= = 2.03 105 J mol-1 = 203 kJ mol-1

(c) Energy radiated due to the given frequency is J s-1 = 1000 J s-1

Each photon carries . Hence, the number of photons coming out is

Atomic Spectroscopy - Continuous versus Line Spectra The heated tungsten filament of an ordinary light bulb emits light which when dispersed using a prism gives a continuous spectrum i.e. a spectrum containing light of all wavelengths. Dispersion of light emitted from a hydrogen discharge tube gives only four lines in the visible region. This is an example of the atomic line spectrum. Balmer correlated the wave length of the four coloured lines using the Balmer formula.

where, the values of n were integral namely 3, 4, 5 and 6. Other spectroscopists have later analysed the radiations of the hydrogen atom spectrum in infrared and ultraviolet regions and have confirmed the existence of spectral lines in these regions. Rydberg gave a general correlation for the whole spectrum of the hydrogen atom,

where both n1 and n2 are integers and n2 > n1 and RH is the Rydberg's constant having a value 1.097 107 m-1 (the same as calculated by Balmer).

Bohr's Model of the Hydrogen Atom: Postulate I Why do we obtain line spectra in atomic spectroscopy and what is the significance of Rydberg's constant? Niels Bohr provided an explanation to these questions and also to the paradox of the

'collapsing atom' by proposing three postulates. 1. The postulates a) The electron in an atom has only certain definite stationary states of motion allowed to it. b) Each of these stationary states has a definite fixed energy. c) An atom in any of these states has its electrons moving in a circular fixed orbit around the nucleus quantitatively. The angular momentum of an electron in an atom is quantised. Bohr expressed this formula as

me = mass of an electron = velocity of the electron r = radius of the orbit n = integer - 1,2,3,..., etc. h = Planck's constant 2. Consequences of the quantisation of angular momentum We have seen that for an electron in a circular orbit around the nucleus, the total energy depends on r.

Total Energy We will see that Bohr's postulate automatically restricts the possible values of r and hence the possible values of total energy. 3. A little algebra From Bohr's postulate we see

where n = 1, 2, 3,..., etc. This means that the product of two variables and r can only have values which

are integral multiples of . In order to obtain the allowed values of r independent of we eliminate using the

earlier relation between kinetic energy and r.

By simple algebraic manipulations, we get We can now obtain the expressions for velocity ( ) and total energy (E),

respectively.

4. Some important conclusions If we substitute the values for the physical constants in the last three equations, we get the following results for an electron in the nth orbit. rn = a0n2 where a0 = 5.29 10-11 m and is the Bohr's radius. It is equal to the radius of the first Bohr orbit of the hydrogen atom corresponding to n = 1 and is used as the atomic unit for distances.

where and is the velocity of the electron in the first orbit (n =1) of the hydrogen atom.

where E0 = -2.18 10-18 J and is the energy of the electron in the first orbit of the hydrogen atom. This value of energy is also used to define the atomic unit of energy. -2E0 = 4.36 10-18 J is called Hartree (atomic unit of energy) Since n (hitherto referred to as quantum number) can take only integral values such as 1, 2, 3, etc. we see how Bohr's postulate generates the concept of specific allowed orbits with n = 1 corresponding to the first orbit, n = 2 corresponding to the second orbit, etc. The allowed values for r, v and E are also thus specified. 5. Rydberg's constant The most important conclusion comes in the form of quantisation of energy levels for the hydrogen atom. Corresponding to the electron being in orbits 1, 2, 3, etc the

atom will have the values of energy ..., etc. respectively. If we consider a photon corresponding to the absolute value of Eo i.e. 2.18 10-18

J, then using the relation we have a corresponding wave number

(since E0 has a negative value) Substituting the values of E0, h and c in the above equation we have

But this is the value of Rydberg's constant (RH) as shown in the equation.

For a given n = n1 we thus have Atomic models

or For values of n = 1, 2, 3, etc. we may thus write

Rydberg's constant in different units

Joules Kilojoules per mole 131.8 kJ mol-1 Electron volts 13.6 eV

Frequency Wave number

Bohr's Model of the Hydrogen Atom: Postulate II When an atom is in one of the stationary states, it does not radiate radiate. But when it changes from a higher energy state to a lower energy state, the atom emits a quantum of radiation whose energy is equal to the difference in energy of the two states. Mathematically we may represent this as

or, (for the emitted photon when the atom goes from Ei to Ef) =

=

or The similarity of this relation with the Rydberg's formula is evident. If we specifically look at the Balmer formula giving the wave numbers of the four radiation we can conclude that they were due to electronic transition from ni (ni = 3, 4, 5 and 6 respectively) to nf (nf = 2)

Evolution of Bohr's Theory Before analysing the significance of the Bohr model, the following remarks in the context of our current state of understanding are in order. a) All but one aspect of Bohr's postulates are essentially correct. b) For reasons explained later, the postulate regarding electrons in atoms moving in circular orbits is incorrect. c) Bohr's model explained most of the aspects of the spectrum of the hydrogen atom including the existence of different spectral series. Spectral Series The spectrum of the hydrogen atom has more lines in the infrared and ultraviolet regions where as the Rydberg's formula generalises all of them. They have been conveniently categorised as belonging to a different series similar to the Balmer series. A particular series has a fixed value for nf and the different lines in the series correspond to the transition of an electron from ni where ni takes the values (nf + 1), (nf + 2), (nf + 3), .... UV region Lyman series nf = 1 Visible region Balmer series nf = 2 IR region Paschen series nf = 3 Brackett series nf = 4 Pfund series nf = 5 Bohr's explanation for the emission spectrum of hydrogen can be extended also to understand the absorption spectrum. So far, we have considered cases where ni > nf so that an electronic transition causes the release of energy or an emission of photon. Naturally if ni < nf , the reverse will take place and we would expect to observe the absorption o f photons in the process. Example Fill in the blanks in the following table and convince yourself whether such a thing exists for all the five series. Hence speculate on why we talk of only five series and not more. Atomic models

Name of series nf Highest energy transition

Lowest energy transition

Lyman 1 -E0 -0.75E0

Balmer 2 -0.25E0 (1)

Paschen 3 (2) (3)

Bracket 4 (4) (5)

Pfund 5 (6) (7)

Answer

Lyman

(1)

Paschen

(2)

(3)

Brackett

(4)

(5)

Pfund (6)

(7) From the above it is evident that the Lyman, Balmer and the Paschen series have lines belonging to distinct regions of the electromagnetic spectrum. However, the Brackett lines overlap with the Paschen region and Pfund lines overlap with the Brackett region. Such intermixing of lines and close spacing of the lines for subsequent series make the study of series after Pfund complicated and unproductive. Number of Lines in the Spectrum Since there are an infinite number of energy levels there are an infinite number of transitions possible in each series. However, as the value of n increases the energy level spacing become smaller and smaller and it become very difficult to distinguish the lines in the far infrared region. If however, we have a sample with atoms excited to the nth energy level where n is finite we can calculate the number of possible spectral lines. Example

How many lines will you observe in the spectrum of a sample of hydrogen atoms excited to the level of n = 3 and for n = 4? Answer

Here we will have lines corresponding to the transition i.e. 3 lines are possible for n = 3.

For n = 4 the transitions are i.e. six lines.

The general formula is lines. APPLICATIONS OF BOHR'S THEORY Bohr's model can be used to correlate energy levels and transitions with spectral lines of not only the hydrogen atom spectra but also of all hydrogen like species. A hydrogen like species is any ion having only one electron. He+, Li2+, Be3+ etc. are examples of such one electron species. We can also do calculations regarding the velocities and radii of Bohr orbits for the electron in such species. Extension of Bohr's Theory to Hydrogen like Species We begin by looking at the general formulae for energies, velocities and orbits for electrons in hydrogen like species. The only change we need to make in equations is to write the charge of the nucleus as Ze instead of e where Z is the atomic number.

Accordingly becomes . Accordingly a more general form of the equations are

Calculating energies, velocities and radii of orbits of electrons in a hydrogen atom or hydrogen like species Example Calculate the energy of one mole of electrons in kJ mol-1 for the first excited state in Hydrogen. Given that Eo is - 13.6 eV and N0 is 6.02 1023 mol-1. Answer The first excited state of the hydrogen atom implies that the electron is in the second orbit that is in n = 2,

We have Substituting the values for Eo, Z and n

Energy of one mole of electron is given by Energy

= -327.49 kJ mol-1

Example Compare the radii of He+ and Li2+. Answer

We have rz, n = a0 For He+, Z = 2, n = 1

rHe+ and Li2+ , Z = 3, n = 1

Example (a) Calculate the energy required to remove an electron completely from n = 2 orbit

in Helium ion . (b) Compare this with the energies required to remove an electron completely from n = 1 and n = 2 orbit of the hydrogen atom, respectively. (c) What is the longest wavelength of light that can cause the changes in each case?

[Given:

Answer

(a) This corresponds to a transition from n = 2 to n = in the ion where Z = 2

We have

The energy required =

(b) Similarly, for hydrogen Z = 1

Removing the electron from n = 1 of hydrogen

= Removing the electron from n = 2 of hydrogen

= Thus, it requires the same energy to remove the electron from n = 2 of He+ as that required to remove the electron from n = 1 of H! However we require one fourth of this energy to remove the electron from n = 2 of hydrogen. Atomic models

(c) We have E = where corresponding to or the minimum energy of photon required to cause a transition. Rearranging the above equation

.

For removing the electron from n = 2 of and n = 1 of H we

have

Hence For removing the electron from n = 2 of H we require one-fourth of this energy

i.e. will be four times (note the inverse relationship). Hence,

Example The lone electron from He+ is made to go from n = 3 to n = 4. Calculate the percent increase or decrease in (a) the velocity of the electron (b) the number of revolutions per second (c) total energy.

Answer

(a) We have The velocities of the electron are given by for n = 3,

and for n = 4,

Percentage change = = -25% Velocity decreases by 25% (b) If the radius of an orbit is r metres and velocity is v metres per second then, the time taken per revolution is given by

revolution per second or rps =

We have rz,n = where Z = 2 (for He+)

and

Thus rps decrease by 57.81%

(c) We have Ez,n = where Z = 2 (for He+)

Percentage change

Since E0 is negative quantity, due to the change of -43.75%

energy increases by 43.75%. Correlating electronic energy levels with the atomic spectra of hydrogen and hydrogen like atoms Example a) Estimate the difference in energy between the first and the second Bohr orbit for a hydrogen atom in eV. b) What would be the wavelength of the resultant emission? c) At what minimum atomic number, a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with = 3.0 10-8 m. Which hydrogen like species does this atomic number correspond to? (Given RH = 1.097 107 m-1) Answer

We know that for hydrogen atom

or

E2 - E1 = -RH hc

(Note: (a) The values for constants such as h, c, value of electron volts in Joules, etc are normally supplied as general data at the top of the question paper. (b) You have to show the working only with the data provided. For example, you

cannot assume the value of E0 as -13.6 eV. You have to calculate it from the value of RH provided.) (b) The wavelength, of the emission can be calculated by using the relation

(c) We have

RH = 1.097 107 m-1 We have to find Z.

or Z = 2 and the species is He+. Example The first ionisation energy of H is 2.179 10 -18 J. Calculate the wavelength of emission for the lowest energy transition in the Paschen series. Answer Lowest transition for Paschen

n = 4 n = 3

E = 2.179 10 -18

Example On the basis of the above data, calculate the energy difference between the n = 3 to n = 2 levels of He+.

Answer

= 4 2.179 10-18 = 5.45 10-17 J THE RISE AND THE FALL OF THE BOHR MODEL - QUANTUM MECHANICS COMES OF AGE

Limitations of the Bohr Model Bohr's model initiated a spate of research activities along two lines. One was attempting to apply, and if necessary extend, Bohr's model of the hydrogen atom to all atoms. This led to the revelation of the many limitations of the Bohr model. The other was developing on Bohr's basic idea of quantisation of energy states in an atom. This was also experimentally proved by Frank and Hertz. They showed that atoms in a gas bombarded by accelerated electrons absorbed energy, if and only if the accelerated electrons had energy higher than a threshold energy. As the atoms absorb energy from the electron they begin to emit photons of definite energies. These latter efforts led to the development of a new theory - a new understanding of physical reality known as quantum mechanics. One of the tenets of this 'new' mechanics known as Heisenberg's uncertainty principle sounded the death knell on the Bohr's model. a) Zeeman and Stark effects - special effects spectroscopy When the samp le is placed in a magnetic field (Zeeman effect) or an electric field (Stark effect) the atomic spectra shows a splitting of the spectral lines as observed in an absence of external fields. These effects could not be explained within the framework of Bohr's theory and required the presence of multiple non-circular orbits corresponding to each energy level. An effort, with some degree of success, was made by Sommerfield to explain these phenomena by invoking multiple elliptical orbits corresponding to each energy level. The Bohr-Sommerfield theory was ultimately discarded along with Bohr's model for other reasons. b) Many electron atoms and hyperfine structures in atomic spectra Bohr's model specially with Sommerfield's extension explained most details of the hydrogen atom spectrum. However, it failed to explain the spectrum of many electron atoms. Any electron in such an atom would be expected to be influenced by the electric field of other electrons and by the magnetic field due to the motion of other electrons. The latter was also the reason for the presence of hyperfine structures within the spectral lines of many electron atoms. Bohr's model had no provision for the incorporation of inter electronic interactions in many electron atoms. It may be noted that in a crude form, Bohr's model could be applied to certain many electron atoms. The example below gives an idea how.

Example An atom such as lithium has two electrons in n = 1 level and one electron in the n = 2 level. These are the K and L shells, respectively. The lone electron in the L shell is attracted by the nucleus containing three protons and also repelled by the K shell electrons. The L shell electron may crudely be considered to be moving in a central field of an effective nuclear charge 'Zeff'= Z - where Z is the actual atomic number and is a shielding or screening constant due to repulsion by the K shell electrons. The ionisation energy (energy required to remove the outermost electron) of lithium is 5.363 eV. Given that the ionisation energy of the hydrogen atom is 13.6 eV, use Bohr's theory to calculate due to the K shell electrons in lithium.

Atomic models Answer According to Bohr's theory

Ionisation energy of Li = 5.363 eV where E0 = 13.6 eV and n = 2

or Zeff = But Zeff = Z - or = Z - Zeff = 3 - 1.26 = 1.74 (Note: The screening effect due to repulsion by inner shell electrons play a very important note in determining atomic properties.) Quantum Mechanical Principles - Wave Particle Duality We may rewrite Einstein's equations: E = mc2 = (mc) c = pc ...(i) (where p stands for the momentum of a photon) and Planck's equation:

...(ii) The dual nature of light can be represented by equating the two expressions of energy, the equation number (i) describing it as a particle of momentum 'p' and equation (ii) describing it as a wave of wavelength .

We get

or ...(iii) De Broglie in 1920 argued along these lines to generalise this equation for all particles in motion and put forth the principle of dual nature of matter.

Using de Broglie relation Example (a) Calculate the (i) velocity and (ii) wavelength of an electron accelarated across a potential difference of 10 V. (b) If a proton is accelerated similarly, would it have a longer or a shorter wavelength than the electron? Explain. Answer a) The kinetic energy of the electron is

i)

=1.88 106 m s-1 ii) We can find p = mev = 9.1 10-31 1.88 106 kg m s-1 = 1.71 10-24 kg m s-1 [Alternatively,

= 1.71 10-24 kg m s-1]

We now have

b) Both the proton, as well as the electron would have the same kinetic energy of 10 eV. If me and mp are masses of the electron and the proton and ve and vp are the velocities of the electron and proton respectively, then (using similar subscript notation for and p)

or,

The ratio of the momenta

But,

Assignment A positively charged particle of mass 1.67 10-27 kg and a charge of 1.6 10-19 C was accelerated across two parallel plates having a potential difference of 'x' volts. If the effective wavelength of the accelerated particle was 5 10-8 m, calculate 'x'. (Answer 33)

Atomic models

Answer The uncertainity in velocity ' ' is 1% i.e 107 1/100 m s-1 = 105 m s-1

= 10-27 105 = 10-22 kg m s-1

PLANCK'S CONSTANT 'h' You should understand that the de Broglie relationship and Heisenberg's uncertainty principle are not two unrelated concepts. They highlight two aspects of the quantum world and are related

through h as the central link. If h was zero, then = = 0

and then particles would not have any wave nature

and spread of would not be there, respectively. A non-zero value of h is thus the common basis. The fact that the value of h is very small (but not zero) is the reason why the 'classical world' is valid for larger bodies. It the value of h were larger then, even larger bodies would have shown wave nature and uncertainties in position and momentum. Atomic models Answer The uncertainity in velocity ' ' is 1% i.e 107 1/100 m s-1 = 105 m s-1

= 10-27 105 = 10-22 kg m s-1

PLANCK'S CONSTANT 'h' You should understand that the de Broglie relationship and Heisenberg's uncertainty principle are not two unrelated concepts. They highlight two aspects of the quantum world and are related through h as the

central link. If h was zero, then = = 0

and then particles would not have any

wave nature and spread of would not be there, respectively. A non-zero value of h is thus the common basis. The fact that the value of h is very

small (but not zero) is the reason why the 'classical world' is valid for larger bodies. It the value of h were larger then, even larger bodies would have shown wave nature and uncertainties in position and momentum. Implication of the Uncertainty Principle and the Dual Nature Matter - Wave Mechanics Applied to Electrons in an Atom There are two important implications of the principles elaborated above.

1. Bohr theory must be discarded: The Bohr theory considers electrons to be moving along specified orbits with specific velocity. It is inherent that the electrons in an atom possess precise momentum and position defined and determinable at all points of time. This is in complete violation of the uncertainty principle. 2. Electrons in atoms have wave like properties: In a way this is also another reason for discarding Bohr's theory, since Bohr implied that electrons are particles.

If Bohr's theory is wrong, how did it explain the hydrogen atom so nicely? This can be understood as follows:

Bohr's postulate proposed by an empirical manipulation to explain the Balmer equation and the value of Rydberg's constant also 'implied' the de Broglie relationship. Let us see how. Let us consider that electrons are matter waves. An electron bound to the nucleus cannot be a progressive wave. It must be a stationary wave. Actually we can see a parallel between Bohr's idea of stationary states and stationary matter waves of the electrons by using the de Broglie relation. If you recall the idea of stationary waves on a vibrating string you will remember that we had the wave length ' ' related to the length of the string by the relation

Where n takes integral values n = 1, 2, 3, ...etc. Extending the same idea we can say that for the electron in the nth Bohr orbit of radius rn and velocity vn: Wave length of the electron is

Implication of the Uncertainty Principle and the Dual Nature Matter - Wave Mechanics Applied to Electrons in an Atom There are two important implications of the principles elaborated above.

1. Bohr theory must be discarded: The Bohr theory considers electrons to be moving along specified orbits with specific velocity. It is inherent that the electrons in an atom possess precise momentum and position defined and determinable at all points of time. This is in complete violation of the uncertainty principle. 2. Electrons in atoms have wave like properties: In a way this is also another reason for discarding Bohr's theory, since Bohr implied that electrons are particles.

If Bohr's theory is wrong, how did it explain the hydrogen atom so nicely? This can be understood as follows:

Bohr's postulate proposed by an empirical manipulation to explain the Balmer equation and the value of Rydberg's constant also 'implied' the de Broglie relationship. Let us see how. Let us consider that electrons are matter waves. An electron bound to the nucleus cannot be a progressive wave. It must be a stationary wave. Actually we can see a parallel between Bohr's idea of stationary states and stationary matter waves of the electrons by using the de Broglie relation. If you recall the idea of stationary waves on a vibrating string you will remember that we had the wave length ' ' related to the length of the string by the relation

Where n takes integral values n = 1, 2, 3, ...etc. Extending the same idea we can say that for the electron in the nth Bohr orbit of radius rn and velocity vn:

Wave length of the electron is

Length covered by the electron wave is equal to

the circumference of the orbit =

We must have an integral multiple of equal to this circumference if the wave is to be stationary.

(using de Broglie relation) Rearranging the above equation we

have which is nothing but the Bohr postulate.

Assignment Find out the number of waves made by a Bohr electron in one complete revolution in its third orbit

.

Answer: 3