Preview:

It was believed that atoms were the smallest particles which could not be further divided. But it was established by experiments like, electrical discharge through gas that, atoms were further composed of sub-atomic particles like: electrons. We shall discuss this experiment in detail. Next, we shall discuss J.J. Thomson’s experiment to find the electrical charge of an electron to its mass followed by Millikan’s oil drop experiment to determine the charge on an electron.

We shall then discuss briefly about the discovery of protons and neutrons. The smallest and lightest positive ion obtained from hydrogen was called, proton. Electrically neutral particles having a mass slightly greater than that of protons were called, neutrons. This lead JJ Thompson to propose the model of an atom called plum pudding or raisin pudding or watermelon model in which, the mass of the atom is assumed to be uniformly distributed with embedded electrons in them like seeds of the watermelon.

Subsequently, we shall look at Rutherford’s model according to which, an atom is made of a tiny positively charged nucleus, at its centre with electrons revolving around it in circular orbits. As a modified version of Rutherford’s model of an atom, we shall study Bohr’s model of an atom in which electrons of an atom move around the nucleus in certain fixed, closed, circular paths called orbits with positively charged nucleus at the centre.

With further modifications and for simplification for Bohr’s model of the atom, the quantum mechanical model was proposed, which was based on, Dual behaviour of matter (electrons having both wave-like and particle-properties) and Hesenberg’s uncertainty principle (according to which, it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron). According to the quantum model, energy of electrons can only have certain specific values (called, quanta) when electrons are bound to the nucleus in atoms; following which, we shall describe the four quantum numbers required to describe the position of an electron accurately in an atom.

Then, we shall define atomic orbital as the three dimensional space around the nucleus of an atom where there is high probability of finding electrons.

You will see that different orbitals have different intriguing shapes.

Next, we shall study the concept of distribution of electrons into these orbitals in a certain order, which we call, Electronic configuration of an atom.

It is found that, an atom is most stable when its electrons fill the orbitals exactly to half its capacity or full. We shall study about this attribute of an atom in some detail, with which we shall come to the end of the chapter.

2.0 Introduction:

According to Greek philosophers, the continued subdivisions of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ has been derived from the Greek word ‘a-tomio’ which means ‘uncutable’ or ‘indivisible’.

The major problems before scientists due to this assumption were:

• to account for the stability of atom after the discovery of sub-atomic particles,
• to account for the difference in behaviour of one element from other in terms of both physical and chemical properties,
• to explain the formation of different kinds of molecules by the combination of different atoms and,
• to understand the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms.

Questions for the section:

1. What were the flaws in the hypothesis that atoms were indivisible?

2.1 Sub-Atomic Particles

Though Dalton’s atomic theory was able to explain many laws like, the law of conservation of mass, law of constant composition and law of multiple proportions very successfully, it failed to explain the results of many experiments. For example, substances like glass or ebonite when rubbed with silk or fur generating electricity.

Thus, many different kinds of sub-atomic particles were discovered. In this section, we will talk about only two particles, namely: Electron and Proton.

2.1.1 Discovery of Electron

An insight into the structure of atom was obtained from results obtained from experiments on electrical discharge through gases. Before we discuss these results we need to keep in mind, a basic rule regarding the behaviour of charged particles: “Like charges repel each other and unlike charges attract each other”.

Concept box: Cathode rays: A beam of electrons emitted from the cathode of a high-vacuum tube.

• Evidence for existence of electrons came from experiments on cathode ray discharge tube. It is depicted in Figure 2.1.
• A cathode ray tube is made of glass containing two thin pieces of metal, called electrodes, sealed in it.
• The electrical discharge through the gases could only be observed at very low pressures and at very high voltages.
• The pressure of different gases can be adjusted by evacuation and with sufficiently high voltage applied across the electrodes, current flows through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles.
• The flow of current from cathode to anode was further facilitated by making a hole in the anode and coating the tube behind anode with phosphorescent material zinc sulphide.
• When the rays, after passing through anode, strike the zinc sulphide coating, a bright spot on the coating is developed (same thing happens in a television set) [as shown in Figure 2.1(b)].

The results of these experiments are summarised below;

(i) The cathode rays start from cathode and move towards the anode.

(ii) These rays themselves are not visible but their behaviour can be observed with the help of certain kind of materials (fluorescent or phosphorescent) which glow when hit by these rays. Television picture tubes are cathode ray tubes and television pictures result due to fluorescence on the television screen coated with certain fluorescent or phosphorescent materials.

(iii) In the absence of electrical or magnetic field, these rays travel in straight lines (Figure 2.2).

(iv) In the presence of electrical or magnetic field, the behaviour of cathode rays was similar to that expected from negatively charged particles, suggesting that the cathode rays consist of negatively charged particles, which was called electrons.

(v) The characteristics of cathode rays (electrons) do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.

Thus, it was concluded that electrons are basic constituent of all the atoms.

2.1.2 Charge to Mass Ratio of Electron

British physicist, J.J. Thomson measured the ratio of electrical charge (e) to the mass of electron (m_e) by using cathode ray tube and applying electrical and magnetic field perpendicular to each other, as well as to the path of electrons (Figure 2.2).

Thomson argued that, the amount of deviation of the particles from their path in the presence of electrical or magnetic field depends upon:

(i) The magnitude of the negative charge on the particle— greater the magnitude of the charge on the particle, greater is the interaction with the electric or magnetic field and thus greater is the deflection.

(ii) The mass of the particle — lighter the particle, greater the deflection.

(iii) The strength of the electrical or magnetic field — the deflection of electrons from its original path increases with the increase in the voltage across the electrodes, or the strength of the magnetic field.

When only electric field is applied, the electrons deviate from their path and hit the cathode ray tube at point A (figure 2.2). Similarly, when only magnetic field is applied, electron strikes the cathode ray tube at point C. By balancing the electrical and magnetic field strength, it is possible to bring back the electron to the path followed as in the absence of electric or magnetic field in which case they hit the screen at point B.

Note box: By finding accurate measurements of the amount of deflection of electrons by varying the electric field strength or magnetic field strength, Thomson was able to determine the value of $e/m_e$ as: $$ \frac {e}{m_e} = 1.758820 \times 10^{11} C \space kg^{-1}….(2.1)$$ Where, “$m_e$” is the mass of the electron in kg and “e” is the magnitude of the charge on the electron in coulomb (C). Since electrons are negatively charged, the charge on electron is –e.

2.1.3 Charge on the Electron

R.A. Millikan devised a method known as Oil drop experiment to determine the charge on the electrons as described:

Millikan’s Oil Drop Method

Figure 2.3: The Millikan oil drop apparatus for measuring charge ‘e’. (In chamber, the forces acting on oil drop are: gravitational, electrostatic due to electrical field and a viscous drag force when the oil drop is moving).

• In this method, oil droplets in the form of mist, produced by the atomiser, were allowed to enter through a tiny hole in the upper plate of electrical condenser.
• The downward motion of these droplets was viewed through the telescope, equipped with a micrometer eye piece.
• By measuring the rate of fall of these droplets, Millikan was able to measure the mass of oil droplets.
• Then, the air inside the chamber was ionized by passing a beam of X-rays through it.
• The gaseous ions could now provide electrical charge on these oil droplets through collisions.
• The fall of these charged oil droplets could be retarded, accelerated or made stationary by varying the charge on the droplets or, by varying the polarity or, the strength of the voltage applied to the plate.
• By measuring the effects of electrical field strength on the motion of oil droplets, Millikan concluded that, the magnitude of electrical charge, q, on the droplets is always an integral multiple of the electrical charge, e, that is, q = n e, where n = 1, 2, 3… .

Inference:

• Millikan found the charge on the electron to be: – 1.6 × $10^{–19}$ C. Now, the accepted value of electrical charge is – 1.6022 × $10^{–19}$ C.
• The mass of the electron ($m_e$) was determined by combining this result with Thomson’s value of $e/m_e$ ratio as:

$$ m_e = \frac {e}{e/m_e} = \frac {1.6022 \times 10^{-19} C }{1.758820 \times 10^{11} C \space kg^{-1} } $$

$$= 9.1094 \times 10^{-31} kg …(2.2)$$

2.1.4 Discovery of Protons and Neutrons

Electrical discharge carried out in the modified cathode ray tube led to the discovery of particles carrying positive charge, also known as canal rays (because, these rays pass through holes in cathode).

The characteristics of these positively charged particles are listed below:

(i) Unlike cathode rays, the positively charged particles depend upon the nature of gas present in the cathode ray tube. These cathode rays are simply the positively charged gaseous ions.

(ii) The charge to mass ratio of the particles is found to depend on the gas from which these originate.

(iii) Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.

(iv) The behaviour of these particles in the magnetic or electrical field is opposite to that observed for electron or cathode rays. That is, the deflection of positively charged particles from its original path decreases with the increase in the voltage across the electrodes, or the strength of the magnetic field.

The smallest and lightest positive ion was obtained from hydrogen and was called proton.

Discovery of neutrons:

• Later, James Chadwick discovered an electrically neutral particle as one of the constituents of an atom by bombarding a thin sheet of beryllium with α-particles.
• On bombarding, electrically neutral particles having a mass slightly greater than that of protons were emitted. He named these particles as neutrons.

$$ \ce { ^{9}_{4}Be + ^{4}_{2} He -> ^{12}_{6}C + ^{1}_{0}n }$$

The important properties of these fundamental particles are given in Table 2.1.

Questions from section 2.1:

1. Describe with relevant diagram, the cathode ray discharge tube experiment. What were the conclusions drawn from it?

2. What are the factors upon which the amount of deviation of the particles from their path (in the presence of electrical or magnetic field) depends, in a cathode ray tube experiment?

3. Decribe Millikan’s oil drop experiment and state the inferences drawn from it.

4. List the properties of positively charged particles as a consequence of cathode ray tube experiment.

2.2 Atomic Models

Different atomic models were proposed to explain the distributions of these charged particles (electrons, protons and neutrons) in an atom. Two such models, proposed by J. J. Thomson and Ernest Rutherford are discussed below.

2.2.1 Thomson Model of Atom

• J. J. Thomson, in 1898, proposed that, an atom has a spherical shape (radius approximately 10^–10 m) in which the positive charge is uniformly distributed.
• The mass of the atom is assumed to be uniformly distributed over the atom.
• The electrons are embedded into it in such a manner so as to provide the most stable electrostatic arrangement (Figure 2.4).
• This model can be visualised as a pudding or watermelon of positive charge with plums or seeds (electrons) embedded into it. Hence, it was given many different names like, plum pudding, raisin pudding or watermelon.

• Although this model was able to explain the overall neutrality of the atom, but was not consistent with the results of later experiments.

Note box: Thomson was awarded Nobel Prize for physics in 1906, for his theoretical and experimental investigations on the conduction of electricity by gases.

Discovery of X-rays:

• Wilhalm Röentgen showed that, when electrons strike a barium platinocyanide covered screen in the cathode ray tubes, produce rays which can cause fluorescence of the screen even when placed outside the cathode ray tubes.
• Since Röentgen did not know the nature of the radiation, he named them X-rays and the name is still carried on.
• It was noticed that, X-rays are produced effectively when electrons strike the dense metal anode, called targets.
• These rays were not deflected by the electric and magnetic fields and have a very high penetrating power through the matter and that is the reason that these rays are used to study the interior of the objects.
• These rays are of very short wavelengths (of range, 0.1 nm) and possess electro-magnetic character.

Radioactivity

• Henri Becqueral observed that there are certain elements which emit radiation on their own and named this phenomenon as radioactivity. The elements known as radioactive elements. This field was developed by Marie Curie, Piere Curie, Rutherford and Fredrick Soddy.
• It was observed that three kinds of rays, a, b and g-rays are emitted.
• Rutherford suggested that a-rays consist of high energy particles carrying two units of positive charge and four unit of atomic mass. He concluded that, α-particles are helium nuclei and as a-particles combined with two electrons, yielded helium gas. Further, b-rays are negatively charged particles like electrons. The g-rays are high energy radiations, neutral in nature and do not consist of particles. With regard to penetrating power, α-particles are the least, followed by b-rays (100 times that of α–particles) and g-rays (1000 times of that α-particles); [g > b > a].

2.2.2 Rutherford’s Nuclear Model of Atom

Rutherford’s a–particle scattering experiment:

• Rutherford and his students (Hans Geiger and Ernest Marsden) bombarded very thin gold foil with α–particles (represented in Figure 2.5). For this purpose, a stream of high energy α–particles from a radioactive source was directed at a thin foil (with thickness of about 100 nm) of gold metal.

Figure 2.5: Schematic view of Rutherford’s scattering experiment. When a beam of alpha (α) particles is “shot” at a thin gold foil, most of them pass through without much effect. Some, however, are deflected.

• The thin gold foil had a circular fluorescent zinc sulphide screen around it.
• Whenever α–particles struck the screen, a tiny flash of light was produced at that point.

Observations:

• According to Thomson model of atom, the mass of each gold atom in the foil should have been spread evenly over the entire atom. Hence, α – particles had enough energy to pass directly through such a uniform distribution of mass.
• Thus, it was expected that, the particles would slow down and change directions only by small angles as they passed through the foil.
• But, it was observed that:

(i) Most of the α–particles passed through the gold foil un-deflected.

(ii) A small fraction of the α–particles was deflected by small angles.

(iii) A very few α–particles (about 1 in 20,000) bounced back, that is, were deflected by nearly 180°.

Inferences:

Based on the experimental observations, Rutherford drew the following conclusions regarding the structure of atom:

(i) Most of the space in the atom is empty as most of the a–particles passed through the foil un-deflected.

(ii) Since a few positively charged α–particles were deflected, he suggested that, the deflection must be due to enormous repulsive force, showing that the positive charge of the atom is not spread throughout the atom as Thomson had presumed; instead, the positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged a–particles.

(iii) Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom. The radius of the atom is about $10^{–10}$ m, while that of nucleus is $10^{–15}$ m.

On the basis of above observations and conclusions, Rutherford proposed the Nuclear Model of Atom according to which,

(i) The positive charge and most of the mass of the atom was densely concentrated in extremely small region. This very small portion of the atom was called nucleus by Rutherford.

(ii) The nucleus is surrounded by electrons that move around the nucleus with a very high speed in circular paths called orbits. (Thus, Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of sun and the electrons that of revolving planets!)

(iii) Electrons and the nucleus are held together by electrostatic forces of attraction.

Drawbacks of Rutherford’s model:

a) When a charged particle moves in an orbit, it undergoes acceleration. According to electromagnetic theory by Maxwell, charged particles when accelerated should emit electrons. Therefore, an electron in an orbit will emit radiation. Thus, lose energy and fall into the nucleus. Hence, Rutherford’s model could not explain the stability of an atom.

b) Rutherford’s model of an atom does not explain line emission spectrum.

2.2.3 Atomic Number and Mass Number

The charge on the proton is equal and opposite to that of an electron. Also, in order to keep the electrical neutrality, the number of electrons in an atom is equal to the number of protons. While the positive charge of the nucleus is due to protons, the mass of the nucleus is due to protons and neutrons.

• Atomic number (Z): The number of protons present in the nucleus is equal to atomic number.

Atomic number (Z) = Number of protons in the nucleus of an atom = Number of electrons in a neutral atom …………….       (2.3)

• Nucleons: The protons and neutrons present in the nucleus are collectively known as nucleons.

• Mass number (A): The total number of nucleons is termed as mass number (A) of the atom.

Mass number (A) = Number of protons (Z) + Number of neutrons (n)………….        (2.4)

Concept box: Composition of any atom can be represented by using the normal element symbol (X) with super-script on the left hand side as the atomic mass number (A) and subscript (Z) on the left hand side as the atomic number that is, $ ^{A}_{Z} X

For example, $ ^{238}_{92}U$, the atomic number (Z) is 92 and the mass number (A) is 238. Thus, number of neutrons, (A – Z) = 238 – 92 = 146.

Note box: Mosely found a linear relationship between square-root of frequencies of the X-ray given by a cathode ray tube and atomic number of the element as: $$ \sqrt ϑ=Z (a-b) $$ Where, is the frequency, “Z” is atomic number and; “a” and “b” are constants.

2.2.4 Isobars and Isotopes

Definition box: 1. Isobars: Atoms with same mass number but different atomic number are called, Isobars. For example: $ ^{14}_6C$ and $ ^{14}_{7}N$. 2. Isotopes: Atoms with identical atomic number but different atomic mass number are known as Isotopes. For example: $^{14}_{6}C , ^{15}_{7}N , ^{16}_8O$

Note box: 1. Considering an atom of hydrogen, 99.985% of hydrogen atoms contain only one proton. This isotope is called, protium ($^1_1 H$). Rest of the percentage of hydrogen atom contains two other isotopes, the one containing 1 proton and 1 neutron is called, deuterium ($^2_1 D$)., 0.015%) and the other one possessing 1 proton and 2 neutrons is called, tritium ($^3_1 T$). 2. Chemical properties of atoms are controlled by the number of electrons, which are determined by the number of protons in the nucleus. Number of neutrons present in the nucleus thus, have very little effect on the chemical properties of an element. Therefore, all the isotopes of a given element show same chemical behaviour.

3. Isotones: Atoms of different elements which contain the same number of neutrons are called, Isotones.

Examples: $^{14}_6C, ^{15}_7 N , ^{16}_8 O$            

4. Isoelectronic species: Species that have the same number of electrons are called, isoelectronic species.

Examples: $O^{2-}, F^- , Ne , Na^+, Mg^{2+}, Al^{3+}$

Example: Calculate the number of protons, neutrons and electrons in $^{80}_{35}Br$. 

Solution:

In this case,  $^{80}_{35}Br, \; Z = 35, \; A = 80,$ species is neutral.

Number of protons = number of electrons = Z = 35 

Number of neutrons = 80 – 35 = 45

Example: The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.

Solution: 

The atomic number is equal to number of protons = 16. The element is sulphur (S).

Atomic mass number = number of protons + number of neutrons   $16 + 16 = 32$

Species is not neutral as the number of protons is not equal to electrons. It is an anion (negatively charged) with charge equal to excess electrons:  $18 – 16 = 2$

Symbol is:   $^{32}_{16}S^{2-}$

Note:  Before using the notation $^{A}_{Z}X$, find out whether the species is a neutral atom, a cation or an anion. If it is a neutral atom, equation (2.3) is valid, i.e., number of protons = number of electrons = atomic number.  If the species is an ion, determine whether the number of protons are larger (cation, positive ion) or smaller (anion, negative ion) than the number of electrons.  Number of neutrons is always given by $A – Z$, whether the species is neutral or ion.

Questions from section 2.2:

1. Describe JJ Thomson’s model of an atom.

2. Write a note on:

a) Discovery of x-rays

b) Radioactivity

3. Describe Rutherford’s model of an atom.

4. Define the terms:

a) Atomic number

b) Mass number

c) Nucleons

d) Isobars

e) Isotopes

f) Isotones

g) Isoelectronic species

5. Give the expression for Mosley’s relation, which relates frequency of x-ray and atomic number of the element.

2.3 Developments leading to the Bohr’s Model of Atom

Story box: Niels Bohr (1885–1962) Niels Bohr, a Danish physicist received his Ph.D. from the University of Copenhagen in 1911. In 1920 he was named Director of the Institute of theoretical Physics. After the First World War, Bohr worked energetically for peaceful uses of atomic energy. He received the first Atoms for Peace award in 1957. Bohr was awarded the Nobel Prize in Physics in 1922.

Two developments played a major role in the formulation of Bohr’s model of atom. These were:

(i) Dual character of the electromagnetic radiation—which means that radiations possess both wave like and particle like properties.

(ii) Experimental results of atomic spectra— which can be explained only by assuming quantized electronic energy levels in atoms (shall be discussed in Section 2.4).

2.3.1 Wave Nature of Electromagnetic Radiation

Maxwell suggested that, when an electrically charged particle moves with acceleration, alternating electrical and magnetic fields are produced and transmitted. These fields are transmitted in the form of waves called electromagnetic waves or electromagnetic radiation.

Properties of Electromagnetic radiation:

a) The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and; both are perpendicular to the direction of propagation of the wave (as shown in Figure 2.6).

Figure 2.6 The electric and magnetic field components of an electromagnetic wave. These components have the same wavelength, frequency, speed and amplitude, but they vibrate in two mutually perpendicular planes.

b) Unlike sound waves or water waves, electromagnetic waves do not require medium and can move in vacuum.

(c) There are many types of electromagnetic radiations, which differ from one another in wavelength (or frequency). These constitute what is called electromagnetic spectrum (Figure 2.7).

Fig. 2.7 (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible region is only a small part of the entire spectrum.

Different regions of the spectrum are identified by different names. Some examples are: radio frequency region around 106 Hz, used for broadcasting; microwave region around 1010 Hz used for radar; infrared region around 1013 Hz used for heating; ultraviolet region around 1016Hz a component of sun’s radiation. The small portion around 1015 Hz, is what is ordinarily called visible light. It is only this part which our eyes can see (or detect). Special instruments are required to detect non-visible radiation.

(d) Different kinds of units are used to represent electromagnetic radiation.

(e) These radiations are characterised by the properties: frequency (n) and wavelength (l).

Note box: The SI unit for frequency (n) is hertz (Hz, $s^{–1}$), after Heinrich Hertz. Frequency is defined as the number of waves that pass a given point in one second. Wavelength has the unit of length. The SI unit of wavelength is thus, meter (m). In vacuum, all types of electromagnetic radiations, regardless of wavelength, travel at the same speed, that is, 3.0 × $10^8$ m $s^{–1}$ (2.997925 × $10^8$ m $s^{–1}$, to be precise). This is called speed of light and is given the symbol ‘c‘. The frequency (n), wavelength (l) and velocity of light (c) are related by the equation 2.5.  ………………  (2.5) The other commonly used quantity in spectroscopy is the wave number (n). Wave number  is defined as the number of wavelengths per unit length. Its units are reciprocal of wavelength unit, that is, $m^{–1}$. However, commonly used unit is $cm^{–1}$ (not SI unit).

Example: 

The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to? 

Solution:

The wavelength, $\lambda$, is equal to $\frac{c}{\nu}$, where $c$ is the speed of electromagnetic radiation in vacuum and $\nu$ is the frequency. Substituting the given values, we have 

$$\lambda = \frac{c}{\nu}$$ 

$$= \frac{3.00 \times 10^8 \; m \, s^{-1}}{1368 \; kHz}$$ 

$$= \frac{3.00 \times 10^8 \; m \, s^{-1}}{1368 \times 10^3 \; s^{-1}}$$ 

$$= 219.3 \, m$$ 

This is a characteristic radiowave wavelength.

Example: The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz). (1 nm = 10$^{-9}$ m) 

Solution: 

Using equation 2.5, frequency of violet light 

$$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \; m \, s^{-1}}{400 \times 10^{-9} \; m}$$ 

$$= 7.50 \times 10^{14} \; Hz$$ 

Frequency of red light 

$$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \; m \, s^{-1}}{750 \times 10^{-9} \; m} = 4.00 \times 10^{14} \; Hz$$ 

The range of visible spectrum is from $4.0 \times 10^{14}$ to $7.5 \times 10^{14}$ Hz in terms of frequency units.

Example: Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å. 

Solution:

(a) Calculation of wavenumber ($\bar{\nu}$) 

$$\lambda = 5800 \, \text{Å} = 5800 \times 10^{-8} \, cm$$ 

$$= 5800 \times 10^{-10} \, m$$

$$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5800 \times 10^{-10} \, m}$$ 

$$= 1.724 \times 10^6 \, m^{-1}$$ 

$$= 1.724 \times 10^4 \, cm^{-1}$$ 

(b) Calculation of the frequency ($\nu$) 

$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, m \, s^{-1}}{5800 \times 10^{-10} \, m} = 5.172 \times 10^{14} \, s^{-1}$$

2.3.2 Particle Nature of Electromagnetic Radiation: Planck’s Quantum Theory

Phenomenon of the black body radiation:

• It was given by Max Planck in 1900.
• When solids are heated, they emit radiation over a wide range of wavelengths. For example, when an iron rod is heated in a furnace, it first turns to dull red and then progressively gets redder as the temperature increases. As this is heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high.
• In terms of frequency, it means that the frequency of emitted radiation goes from a lower frequency to a higher frequency as the temperature increases. (The red colour lies in the lower frequency region while blue colour belongs to the higher frequency region of the electromagnetic spectrum).
• The ideal body, which emits and absorbs radiations of all frequencies, is called a black body and the radiation emitted by such a body is called black body radiation.
• The exact frequency distribution (that is, intensity versus frequency curve of the radiation) of the emitted radiation from a black body depends only on its temperature.

At a given temperature, intensity of radiation emitted increases with increase of wavelength, reaches a maximum value at a given wavelength and then starts decreasing with further increase of wavelength, as shown in Figure 2.8.

The above experimental results cannot be explained satisfactorily on the basis of the wave theory of light. Thus, Planck suggested that, atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner. Planck gave the name, quantum to the smallest quantity of energy that can be emitted or absorbed in the form of electromagnetic radiation.

The energy (E) of a quantum of radiation is proportional to its frequency (n) and is expressed by equation:

The proportionality constant, ‘h’ is known as Planck’s constant and has the value 6.626×10^–34 J s.

Story box: Max Planck (1858 – 1947) Max Planck, a German physicist, received his Ph.D in theoretical physics from the University of Munich. He was appointed Director of the Institute of Theoretical Physics at the University of Berlin. Planck was awarded the Nobel Prize in Physics in 1918 for his quantum theory.

Photoelectric Effect

The Photoelectric Effect was discovered by H. Hertz through his experiment.

Photoelectric Effect is the phenomenon in which, electrons (or electric current) were ejected when certain metals (for example potassium, rubidium, caesium) were exposed to a beam of light as shown in Figure 2.9.

Figure 2.9 Equipment for studying the photoelectric effect. Light of a particular frequency strikes a clean metal surface inside a vacuum chamber. Electrons are ejected from the metal and are counted by a detector that measures their kinetic energy.

The results observed in this experiment were:

(i) The electrons are ejected from the metal surface as soon as the beam of light strikes the surface, that is, there is no time lag between the striking of light beam and the ejection of electrons from the metal surface.

(ii) The number of electrons ejected is proportional to the intensity or brightness of light.

(iii) For each metal, there is a characteristic minimum frequency of light wave, $n_0$ (also known as threshold frequency) below which, photoelectric effect is not observed.

At a frequency $ n > n_0$, the ejected electrons come out with certain kinetic energy. The kinetic energy of ejected electrons increase with increase in frequency of the light used.

For example, red light [  = (4.3 to 4.6) × $10^{14}$ Hz] of any brightness (intensity) may shine on a piece of potassium metal for hours but no photoelectrons are ejected. But, as soon as even a very weak yellow light (  = 5.1–5.2 × $10^{14}$ Hz) shines on the potassium metal, the photoelectric effect is observed because the threshold frequency ($n_0$) for potassium metal is 5.0× $10^{14}$ Hz.

Einstein’s Theory of Photoelectric Effect

Story box: Albert Einstein, a German born American physicist, is regarded by many as one of the two great physicists the world has known (the other is Isaac Newton). His three research papers (on special relativity, Brownian motion and the photoelectric effect) which he published in 1905, while he was employed as a technical assistant in a Swiss patent office in Berne have profoundly influenced the development of physics. He received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect.

Einstein explained the photoelectric effect based on Planck’s quantum theory of electromagnetic radiation. According to which,

• A photon of sufficient energy when strikes an electron in the atom of the metal, the photon will transfer its energy instantaneously to the electron during the collision and the electron is ejected without any time lag or delay.
• Greater the energy possessed by the photon, greater will be transfer of energy to the electron and greater the kinetic energy of the ejected electron. In other words, kinetic energy of the ejected electron is proportional to the frequency of the electromagnetic radiation (since electromagnetic radiations are considered to be made of a stream of photons).
• Since the striking photon has energy equal to h  and the minimum energy required to eject the electron is $hϑ_0$ (also called, work function, $W_0$; refer Table 2.2), the difference in energy $(hϑ – hϑ_0)$  is the amount of energy transferred as the kinetic energy of the photoelectron.

• According to conservation of energy principle, the kinetic energy of the ejected electron is given by the equation:

$$hϑ = hϑ_0 + \frac {1}{2}m_ev^2…(2.7)$$

Where, $m_e$ is the mass of the electron and v is the velocity associated with the ejected electron.

• A more intense beam of light consists of larger number of photons. Consequently, the number of electrons ejected is also larger as compared to that ejected when a beam of weaker intensity of light is employed.

Example: Calculate energy of one mole of photons of radiation whose frequency is $5 × 10^{14}$ Hz. 

Solution: 

Energy (E) of one photon is given by the expression 

$$E = h\nu$$ 

$$h = 6.626 \times 10^{-34} \, J \, s$$ 

$$\nu = 5 \times 10^{14} \, s^{-1} \; (given)$$ 

$$E = (6.626 \times 10^{-34} \, J \, s) \times (5 \times 10^{14} \, s^{-1})$$ 

$$= 3.313 \times 10^{-19} \, J$$ 

Energy of one mole of photons 

$$= (3.313 \times 10^{-19} \, J) \times (6.022 \times 10^{23} \, mol^{-1})$$ 

$$= 199.51 \, kJ \, mol^{-1}$$

Example: A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Solution: 

Power of the bulb = 100 watt 

$$= 100 \, J \, s^{-1}$$ 

Energy of one photon $E = h\nu = \frac{hc}{\lambda}$ 

$$= \frac{6.626 \times 10^{-34} \, J \, s \times 3 \times 10^8 \, m \, s^{-1}}{400 \times 10^{-9} \, m}$$ 

$$= 4.969 \times 10^{-19} \, J$$ 

Number of photons emitted 

$$\frac{100 \, J \, s^{-1}}{4.969 \times 10^{-19} \, J} = 2.012 \times 10^{20} \, s^{-1}$$

Example: When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68 \times 10^5 \, J \, mol^{-1}$. What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted? 

Solution: 

The energy (E) of a 300 nm photon is given by 

$$h\nu = \frac{hc}{\lambda}$$ 

$$= \frac{6.626 \times 10^{-34} \, J \, s \times 3.0 \times 10^8 \, m \, s^{-1}}{300 \times 10^{-9} \, m}$$ 

$$= 6.626 \times 10^{-19} \, J$$ 

The energy of one mole of photons 

$$= 6.626 \times 10^{-19} \, J \times 6.022 \times 10^{23} \, mol^{-1}$$ 

$$= 3.99 \times 10^5 \, J \, mol^{-1}$$ 

The minimum energy needed to remove one mole of electrons from sodium 

$$= (3.99 – 1.68) \times 10^5 \, J \, mol^{-1}$$ 

$$= 2.31 \times 10^5 \, J \, mol^{-1}$$ 

The minimum energy for one electron

$$= \frac{2.31 \times 10^5 \, J \, mol^{-1}}{6.022 \times 10^{23} \, electrons \, mol^{-1}}$$ 

$$= 3.84 \times 10^{-19} \, J$$ 

This corresponds to the wavelength 

$$\therefore \lambda = \frac{hc}{E}$$ 

$$= \frac{6.626 \times 10^{-34} \, J \, s \times 3.0 \times 10^8 \, m \, s^{-1}}{3.84 \times 10^{-19} \, J}$$ 

$$= 517 \, nm$$ 

(This corresponds to green light)

Example:  The threshold frequency $\nu_0$ for a metal is $7.0 \times 10^{14} \, s^{-1}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $\nu = 1.0 \times 10^{15} \, s^{-1}$ hits the metal. 

Solution: 

According to Einstein’s equation 

$$Kinetic \, energy = \tfrac{1}{2} m_e v^2 = h(\nu – \nu_0)$$ 

$$= (6.626 \times 10^{-34} \, J \, s) (1.0 \times 10^{15} \, s^{-1} – 7.0 \times 10^{14} \, s^{-1})$$ 

$$= (6.626 \times 10^{-34} \, J \, s) (10.0 \times 10^{14} \, s^{-1} – 7.0 \times 10^{14} \, s^{-1})$$ 

$$= (6.626 \times 10^{-34} \, J \, s) \times (3.0 \times 10^{14} \, s^{-1})$$ 

$$= 1.988 \times 10^{-19} \, J$$

2.3.3 Evidence for the quantized Electronic Energy Levels: Atomic spectra

Refraction of light: The bending of light when it passes from one medium to another (having different densities) is called refraction of light.

Spectrum: Ordinary white light consists of waves with all the wavelengths in the visible range. Thus, when a ray of white light emerges out of the prism, it is spread out into a series of coloured bands called, spectrum.

It is observed that, when the ray of white light is passed through a prism, the wave with shorter wavelength bends more than the one with a longer wavelength. For example, the light of red colour which has longest wavelength is deviated the least while the violet light, which has shortest wavelength, is deviated the most.

The spectrum of white light, that we can see, ranges from violet at 7.50 × $10^{14}$ Hz to red at 4×$10^{14}$ Hz.  Such a spectrum is called continuous spectrum. It is called a continuous because, violet merges into blue, blue into green and so on. A similar spectrum is produced when a rainbow forms in the sky.

Emission and Absorption Spectra

Emission spectrum: The spectrum of radiation emitted by a substance that has absorbed energy is called an emission spectrum.

To produce an emission spectrum, energy is supplied to a sample by heating it or irradiating it. The wavelength (or frequency) of the radiation emitted as the sample gives up the absorbed energy, is then recorded.

Absorption spectrum: A spectrum of electromagnetic radiation transmitted through a substance, showing dark lines or bands due to absorption at specific wavelengths is called, an absorption spectrum.

To produce an absorption spectrum, a continuum of radiation is passed through a sample which absorbs radiation of certain wavelengths. The missing wavelength which corresponds to the radiation absorbed by the matter, leave dark spaces in the bright continuous spectrum.An absorption spectrum appears like the photographic negative of an emission spectrum.

Spectroscopy: The study of emission or absorption spectra is referred to as spectroscopy.

Fig. 2.10 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any other element) can be passed through a prism and separated into certain discrete wavelengths. Thus an emission spectrum, which is a photographic recording of the separated wavelengths is called as line spectrum. Any sample of reasonable size contains an enormous number of atoms. Although a single atom can be in only one excited state at a time, the collection of atoms contains all possible excited states. The light emitted as these atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption. When white light is passed through unexcited atomic hydrogen and then through a slit and prism, the transmitted light is lacking in intensity at the same wavelengths as are emitted in (a) The recorded absorption spectrum is also a line spectrum and the photographic negative of the emission spectrum.

Line emission spectra:

• The emission spectra of atoms in the gas phase do not show a continuous spread of wavelength from red to violet, instead, they emit light only at specific wavelengths with dark spaces between them. Such spectra are called line spectra or atomic spectra because the emitted radiation is identified by the appearance of bright lines in the spectra (as shown in Figure 2.10 above).
• Each element has a unique line emission spectrum. The characteristic lines in atomic spectra can thus be used in chemical analysis to identify unknown atoms.

Note box: German chemist, Robert Bunsen (1811-1899) was one of the first investigators to use line spectra to identify elements. Elements like rubidium (Rb), caesium (Cs) thallium (Tl), indium (In), gallium (Ga) and scandium (Sc) were discovered when their minerals were analysed by spectroscopic methods. The element, helium (He) was discovered in the sun by spectroscopic method.

Line Spectrum of Hydrogen

When an electric discharge is passed through gaseous hydrogen, the $H_2$ molecules dissociate and the energetically excited hydrogen atoms emit electromagnetic radiation of discrete frequencies.

• Balmer showed on the basis of experimental observations that if spectral lines are expressed in terms of wavenumber (n), then the visible lines of the hydrogen spectrum obey the following formula:

$$ \bar v = 109677 \bigg( \frac {1}{2^2}- \frac {1}{n^2} \bigg)cm^{-1} …..(2.11)$$

Where, “n” is an integer equal to or greater than 3 (that is, n = 3, 4, 5,….)

The series of lines obtained by this formula (2.11) are called the Balmer series. The Balmer series of lines are the only lines in the hydrogen spectrum which appear in the visible region of the electromagnetic spectrum.

• The Swedish spectroscopist, Johannes Rydberg, noted that, all series of lines in the hydrogen spectrum could be described by the following expression:

$$ \bar v = 109677 \bigg( \frac {1}{n_1^2}- \frac {1}{n_2^2} \bigg)cm^{-1} …..(2.12)$$

Where, $n_1 $= 1, 2……..

$$ n_2 = n_1 + 1, n_1 + 2…… $$

The value 109,677 $cm^{–1}$ is called the Rydberg constant for hydrogen. The first five series of lines that correspond to $n_1$ = 1, 2, 3, 4, 5 are known as Lyman, Balmer, Paschen, Bracket and Pfund series, respectively.

Table 2.3 shows these series of transitions in the hydrogen spectrum.

Fig. 2.11 depicts the energies of different stationary states or energy levels of hydrogen atom. Hence, it is called, Energy level diagram.

Figure 2.11 Transitions of the electron in the hydrogen atom (The diagram shows the Lyman, Balmer and Paschen series of transitions).

Example: What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?

Solution:

Since $n_1 = 5$ and $n_f = 2$, this transition gives rise to a spectral line in the visible region of the Balmer series. From equation (2.17)

$$\Delta E = 2.18 \times 10^{-18} \, J \left[ \frac{1}{5^2} – \frac{1}{2^2} \right]$$

$$= -4.58 \times 10^{-19} \, J$$

It is an emission energy

The frequency of the photon (taking energy in terms of magnitude) is given by

$$\nu = \frac{\Delta E}{h}$$

$$= \frac{4.58 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, J \, s}$$

$$= 6.91 \times 10^{14} \, Hz$$

$$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8 \, m \, s^{-1}}{6.91 \times 10^{14} \, Hz} = 434 \, nm$$

Example: Calculate the energy associated with the first orbit of $He^+$. What is the radius of this orbit? 

Solution:

$$E_n = -\frac{(2.18 \times 10^{-18} \, J) Z^2}{n^2} \; atom^{-1}$$ 

For He⁺, $n = 1$, $Z = 2$ 

$$E_1 = -\frac{(2.18 \times 10^{-18} \, J)(2^2)}{1^2} = – 8.72 \times 10^{-18} \, J$$ 

The radius of the orbit is given by equation (2.15) 

$$r_n = \frac{(0.0529 \, nm) n^2}{Z}$$ 

Since $n = 1$, and $Z = 2$ 

$$r_n = \frac{(0.0529 \, nm)1^2}{2} = 0.02645 \, nm$$

Features common to all line spectra:

(i) Line spectrum for an element is unique

(ii) There is regularity in the line spectrum of each element.

Questions from section 2.3:

1. What were the developments that lead to Bohr’s model of the atom.

2. List the properties of electromagnetic radiation.

3. Define:

a) Frequency

b) Wave number

4. Describe:

a) Black body radiation

b) Einstein’s Theory of Photoelectric Effect

c) Spectrum

d) Continuous spectrum

e) Emission spectrum

f) Absorption spectrum

g) Spectroscopy

5. What are the features of a line spectra?

2.4.1 Explanation of Line Spectrum of Hydrogen

Radiation (energy) is absorbed when the electron moves from the orbit of smaller Principal quantum number to the orbit of higher Principal quantum number, whereas the radiation (energy) is emitted if the electron moves from higher orbit to lower orbit.

The energy gap between the two orbits is given by equation:

$$\Delta E = E_f – E_i \qquad (2.16)$$ 

Combining equations (2.13) and (2.16) 

$\Delta E = \left(-\frac{R_H}{n_f^2}\right) – \left(-\frac{R_H}{n_i^2}\right) \quad$ (where $n_i$ and $n_f$ stand for initial orbit and final orbits)

$\Delta E = R_H \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) = 2.18 \times 10^{-18} \, J \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) \quad (2.17)$

The frequency ($\nu$) associated with the absorption and emission of the photon can be evaluated by using equation (2.18) 

$$\nu = \frac{\Delta E}{h} = \frac{R_H}{h} \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)$$ 

$$= \frac{2.18 \times 10^{-18} \, J}{6.626 \times 10^{-34} \, J \, s} \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) \quad (2.18)$$

$$= 3.29 \times 10^{15} \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) \, Hz \quad (2.19)$$ 

and in terms of wavenumbers ($\bar{\nu}$) 

$$\bar{\nu} = \frac{\nu}{c} = \frac{R_H}{hc} \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) \quad (2.20)$$ 

$$= \frac{3.29 \times 10^{15} \, s^{-1}}{3 \times 10^8 \, m \, s^{-1}} \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right)$$ 

$$= 1.09677 \times 10^7 \left(\frac{1}{n_i^2} – \frac{1}{n_f^2}\right) \, m^{-1} \quad (2.21)$$

In case of absorption spectrum, $ n_f > n_i $ and the term in the parenthesis is positive and energy is absorbed. On the other hand in case of emission spectrum $ n_i > n_f$, ΔE is negative and energy is released.  

Bohr’s Model of Atom:

A modification to Rutherford’s model of atom was suggested by Neils Bohr in order to explain the stability of an atom and the emission of spectral lines. This model is based on the following assumptions:

a) Electrons in an atom move around the nucleus in certain fixed, closed, circular paths called orbits, without emitting any radiations.

b) These orbits are associated with definite energies and are also called energy levels or stationary states.

c) Electrons gain and lose energy by jumping from one allowed orbit to another. This loss or gain in energy is equal to the difference in energies of the two orbits, given by: $ΔE = E_2 – E_1 = h ν$ (where, “ν” is frequency and “h” is Planck’s constant). The frequency of radiation emitted or absorbed will thus be,

$$  v = \frac {\Delta E}{h}=\frac {E_2-E_1}{h} $$

d) The angular momentum of an electron in a given stationary state can be expressed as in equation:

$$ m_e v r = n.\frac {h}{2 \pi}   \hspace{10mm} n=1,2,3,…$$

e) Electrons can revolve only in these stationary states in which their angular momentum is equal to an integral multiple of constant, h/2π.

So, mvr = nh/2π

Where, “m” is mass of electron; “v” is velocity; “r” is radius of the orbit; “n” is an integer (= 1, 2, 3..) called Principal quantum number; “h” is Planck’s constant.

Bohr’s model for hydrogen atom:

• The stationary states for each electron are numbered, n = 1, 2, 3… These are known as Principal quantum numbers.
• The radii ($r_n$) of stationary states are expressed as:

$$ r_n= n^2a_0$$

Where, $a_0$= 52.9 pm (called Bohr radius) and since electron of hydrogen is generally in its first orbit, n=1. As “n” increases, the value of “r” will also increase.

• The most important property associated with the electron is the energy of its stationary state. It is given by:

$$E_n = – R_H \bigg( \frac {1}{n^2} \bigg) \hspace{10mm} n=1,2,3,… $$

• This theory for Hydrogen atom can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, $He^+, Li^{2+}, Be^{3+}$ and so on. The energies of the stationary states associated with these ions are given by the expression:

 $$ E_n = -2.18 \times 10^{-18} \bigg( \frac {Z^2}{n^2} \bigg) J $$

and radii by the expression

$$ r_n = \frac {52.9(n^2)}{Z}pm $$

Where, “Z” is atomic number and has values, 2 and 3 for helium and lithium atoms respectively.

• It is also possible to calculate the velocities of electrons moving in these orbits. It is found that, velocity of electron increases with increase of positive charge on the nucleus and decreases with increase of principal quantum number.

Q and A:

What does the negative electronic energy ($E_n$) for hydrogen atom mean?

The energy of the electron in a hydrogen atom has a negative sign for all possible orbits in the equation: . $$E_n = – R_H \bigg( \frac {1}{n^2} \bigg) \hspace{10mm} n=1,2,3,… $$

This negative sign means that, the energy of the electron in the atom is lower than the energy of a free electron at rest.

A free electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero. Mathematically, this corresponds to setting “n” equal to infinity in the above equation so that, $E_∞$ =0

As the electron gets closer to the nucleus (as n decreases),$ E_n$ becomes larger in absolute value and more and more negative. The most negative energy value is given by n=1 which corresponds to the most stable orbit. We call this the ground state.

Example: What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom? 

Solution: 

Since $n_1 = 5$ and $n_f = 2$, this transition gives rise to a spectral line in the visible region of the Balmer series. From equation (2.17) 

$$\Delta E = 2.18 \times 10^{-18} \, J \left[\frac{1}{5^2} – \frac{1}{2^2}\right]$$ 

$$= -4.58 \times 10^{-19} \, J$$

It is an emission energy 

The frequency of the photon (taking energy in terms of magnitude) is given by 

$$\nu = \frac{\Delta E}{h}$$ 

$$= \frac{4.58 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, J \, s}$$ 

$$= 6.91 \times 10^{14} \, Hz$$ 

$$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8 \, m \, s^{-1}}{6.91 \times 10^{14} \, Hz} = 434 \, nm$$

Example:  Calculate the energy associated with the first orbit of $He^+$. What is the radius of this orbit? 

Solution:

$$E_n = -\frac{(2.18 \times 10^{-18} \, J)Z^2}{n^2} \; atom^{-1}$$ 

For $He^+$, $n = 1$, $Z = 2$ 

$$E_1 = -\frac{(2.18 \times 10^{-18} \, J)(2^2)}{1^2} = -8.72 \times 10^{-18} \, J$$ 

The radius of the orbit is given by equation (2.15) 

$$r_n = \frac{(0.0529 \, nm)n^2}{Z}$$ 

Since $n = 1$, and $Z = 2$ 

$$r_n = \frac{(0.0529 \, nm)(1^2)}{2} = 0.02645 \, nm$$

2.4.2 Limitations of Bohr’s Model

Although Bohr’s model of the hydrogen atom was an improvement over Rutherford’s nuclear model, as it could account for the stability and line spectra of hydrogen atom and hydrogen like ions (for example, $He^+, Li^{2+}, Be^{3+}$, and so on), it was too simple to account for the following:

• It fails to account for the finer details like, double (which is two closely spaced lines) of the hydrogen atom spectrum observed by using sophisticated spectroscopic techniques.
• This model is also unable to explain the spectrum of atoms other than hydrogen, for example, helium atom which possesses only two electrons.
• Further, Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of magnetic field (Zeeman effect) or in an electric field (Stark effect).
• It could not explain the ability of atoms to form molecules by chemical bonds.

2.5 Towards Quantum Mechanical Model of the Atom

In view of the shortcoming of the Bohr’s model, attempts were made to develop a more suitable and general model for atoms. Two important developments which contributed significantly in the formulation of such a model were:

1. Dual behaviour of matter,

2. Heisenberg uncertainty principle.

Questions from section 2.4:

1. Describe Bohr’s Model of an atom.

2. Describe Bohr’s Model for hydrogen atom. Also state its limitations.

2.5.1 Dual Behaviour of Electromagnetic Radiation or Light – De Broglie equation

Story box: Louis de Broglie (1892 – 1987) Louis de Broglie, a French physicist, studied history as an undergraduate in the early 1910’s. His interest turned to science as a result of his assignment to radio communications in World War I. He received his Dr. Sc. from the University of Paris in 1924. He was professor of theoretical physics at the University of Paris from 1932 until his retirement in 1962. He was awarded the Nobel Prize in Physics in 1929.

According to Louis de Broglie, if light waves can have particle-like behaviour, then the particles of matter can show properties of wave as well under suitable conditions. Based on this principle, a relationship between wavelength of the wave and momentum of the particle has been derived as:

$$ \lambda = \frac {h}{mν} = \frac {h}{p} $$

Where, m is the mass of the particle, v its velocity and p its momentum.

Fact box: de Broglie’s prediction was confirmed experimentally when it was found that, an electron beam undergoes diffraction, a phenomenon characteristic of waves. This fact has been put to use in making an electron microscope, which is based on the wavelike behaviour of electrons just as an ordinary microscope utilises the wave nature of light. The wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected. The wavelengths associated with electrons and other subatomic particles (with very small mass) can however be detected experimentally.

Example: What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m $s^{-1}$? 

Solution: 

According to de Broglie equation (2.22) 

$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \, J \, s}{(0.1 \, kg)(10 \, m \, s^{-1})}$$ 

$$= 6.626 \times 10^{-34} \, m \quad (J = kg \, m^2 \, s^{-2})$$

Example: The mass of an electron is $9.1 \times 10^{-31} \, kg$. If its K.E. is $3.0 \times 10^{-25} \, J$, calculate its wavelength.

Solution: 

Since K.E. = $\frac{1}{2} mv^2$ 

$$v = \left(\frac{2 K.E.}{m}\right)^{1/2} = \left(\frac{2 \times 3.0 \times 10^{-25} \, kg \, m^2 s^{-2}}{9.1 \times 10^{-31} \, kg}\right)^{1/2}$$ 

$$= 812 \, m \, s^{-1}$$ 

$$\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \, J \, s}{(9.1 \times 10^{-31} \, kg)(812 \, m \, s^{-1})}$$ 

$$= 8967 \times 10^{-10} \, m = 896.7 \, nm$$

Example: Calculate the mass of a photon with wavelength 3.6 Å. 

Solution: 

$$\lambda = 3.6 \, \text{Å} = 3.6 \times 10^{-10} \, m$$ 

Velocity of photon = velocity of light 

$$m = \frac{h}{\lambda v} = \frac{6.626 \times 10^{-34} \, J \, s}{(3.6 \times 10^{-10} \, m)(3 \times 10^8 \, m \, s^{-1})}$$ 

$$= 6.135 \times 10^{-29} \, kg$$

2.5.2 Heisenberg’s Uncertainty Principle

Story box: Werner Heisenberg (1901 – 1976) received his Ph.D. in physics from the University of Munich in 1923. He then spent a year working with Max Born at Gottingen and three years with Niels Bohr in Copenhagen. He was professor of physics at the University of Leipzig from 1927 to 1941. During World War II, Heisenberg was in charge of German research on the atomic bomb. After the war he was named director of Max Planck Institute for physics in Gottingen. He was also accomplished mountain climber. Heisenberg was awarded the Nobel Prize in Physics in 1932. It was given by German physicist, Werner Heisenberg.

Definition box: It states that, it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.

Mathematically, it can be given as:

$$ \Delta x \times \Delta p_x \geq \frac {h}{4 \pi} $$

$$ or \space \Delta x \times \Delta (mv_x) \geq \frac {h}{4 \pi }$$

$$ or \space \Delta x \times \Delta v_x \geq  \frac {h}{4 \pi m }$$

Where, Δx is the uncertainty in position and $Δp_x$ (or, $Δv_x$) is the uncertainty in momentum (or velocity) of the particle.

Thus, if the position of the electron is known with high degree of accuracy (Δx is small), then the velocity of the electron will be uncertain [Δ($v_x$) is large]. On the other hand, if the velocity of the electron is known precisely (Δ $(v_x$) is small), then the position of the electron will be uncertain (Δx will be large).

Significance of Uncertainty Principle

• It rules out the existence of definite paths or trajectories of electrons and other similar particles—since for a sub-atomic object such as an electron, it is not possible simultaneously to determine the position and velocity at any given instant to an arbitrary degree of precision, it is not possible to talk of the trajectory of an electron.
• The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. This can be seen from the following examples. If uncertainty principle is applied to an object of mass, say about a milligram ($10^{–6}$kg), then,

$$ \Delta v .\Delta x = \frac  {h}{4 \pi m } $$

$$ = \frac {6.626 \times 10^{-34} Js}{4 \times 3.1416 \times 10^{-6} kg} \approx 10^{-28}m^2s^{-1}$$

The value of ΔvΔx obtained is extremely small and is insignificant. Therefore, we can conclude that, in dealing with milligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence.

In the case of a microscopic object like an electron on the other hand, Δv. Δx obtained is much larger and such uncertainties are of real consequence. For example, for an electron whose mass is 9.11× $10^{–31}$ kg, according to Heisenberg uncertainty principle,

$$ \Delta v .\Delta x = \frac  {h}{4 \pi m } $$

$$ = \frac {6.626 \times 10^{-34} Js}{4 \times 3.1416 \times 9.11 \times 10^{-31} kg} \approx 10^{-4}m^2s^{-1}$$

It therefore means that, if one tries to find the exact location of the electron, say to an uncertainty of only $10^{–8}$ m, then the uncertainty, Δv in velocity would be:

$$ \frac {10^{-4}m^2s^{-1}}{10^{-8}m}  \approx 10^4ms^{-1}$$

Example: A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity? 

Solution:

$$\Delta x \, \Delta p = \frac{h}{4 \pi} \quad \text{or} \quad \Delta x \, m \, \Delta v = \frac{h}{4 \pi}$$ 

$$\Delta v = \frac{h}{4 \pi \, \Delta x \, m}$$ 

$$\Delta v = \frac{6.626 \times 10^{-34} \, J \, s}{4 \times 3.14 \times 0.1 \times 10^{-10} \, m \times 9.11 \times 10^{-31} \, kg}$$ 

$$= 0.579 \times 10^7 \, m \, s^{-1} \; (1J = 1 \, kg \, m^2 \, s^{-2})$$ 

$$= 5.79 \times 10^6 \, m \, s^{-1}$$ 

Example: A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.

Solution:

The uncertainty in the speed is 2%, i.e., 

$$45 \times \frac{2}{100} = 0.9 \, m \, s^{-1}$$ 

Using the equation (2.22) 

$$\Delta x = \frac{h}{4 \pi m \Delta v}$$ 

$$= \frac{6.626 \times 10^{-34} \, J \, s}{4 \times 3.14 \times 40 g \times 10^{-3} \, kg \, g^{-1} (0.9 \, m \, s^{-1})}$$ 

$$= 1.46 \times 10^{-33} \, m$$ 

This is nearly $\sim 10^{18}$ times smaller than the diameter of a typical atomic nucleus. 

As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Reasons for the Failure of the Bohr Model

• The wave character of the electron is not considered in Bohr model. Hence, it ignores the duel character of matter.
• Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. But this is not possible according to Heisenberg uncertainty principle.

Questions from section 2.5:

1. State Heisenberg’s Uncertainty Principle, give the mathematical expressions depicting the same and explain briefly, its significance.

2. State the reasons for failure of Bohr’s model of an atom.            

2.6 Quantum Mechanical Model of Atom

Classical mechanics fails when applied to subatomic particles like, electrons since it ignores the duel behaviour of matter. Quantum mechanics is the branch of science that takes into account, the dual behaviour of matter.

Definition box: Quantum mechanics is a theoretical science that deals with the study of the motions of the microscopic objects that have both observable wave like and particle like properties. It specifies the laws of motion that these objects obey.

Quantum mechanics was developed independently in 1926 by Werner Heisenberg and Erwin Schrödinger.

Fact box: The fundamental equation of quantum mechanics was developed by Schrödinger and it won him the Nobel Prize in Physics in 1933.

For a system, such as an atom or a molecule whose energy does not change with time, the Schrödinger equation is written as:

$$ H \psi = E \psi $$

Where, ıH is a mathematical operator called Hamiltonian.

Schrödinger arrived at this operator from the expression for the total energy of the system. The total energy of the system takes into account the kinetic energies of all the sub-atomic particles (electrons, nuclei), attractive potential between the electrons and nuclei and repulsive potential among the electrons and nuclei individually. Solution of this equation gives E and .

Introduction to Hydrogen Atom and the Schrödinger Equation

• The wave function is a mathematical function whose value depends upon the coordinates of the electron in the atom and does not carry any physical meaning.
• When Schrödinger equation is solved for hydrogen atom, the solution gives the possible energy levels the electron can occupy and the corresponding wave function(s): ( ) of the electron associated with each energy level.
• When an electron is in any energy state, the wave function corresponding to that energy state contains all information about the electron. Such wave functions of hydrogen or hydrogen like species with one electron are called atomic orbitals and these wave functions pertaining to one-electron species are called, one-electron systems.
• The probability of finding an electron at a point within an atom is proportional to the $| Ψ |^2$ at that point.
• Application of Schrödinger equation to multi-electron atoms presents a difficulty: the Schrödinger equation cannot be solved exactly for a multi-electron atom. The principal difference lies in the consequence of increased nuclear charge. Because of this all the orbitals are somewhat contracted.
• Further, unlike orbitals of hydrogen or hydrogen like species, whose energies depend only on the quantum number n, the energies of the orbitals in multi-electron atoms depend on quantum numbers n and l.

Important Features of the Quantum Mechanical Model of Atom

Quantum mechanical model of atom is the picture of the structure of the atom, which emerges from the application of the Schrödinger equation to atoms. The following are the important features of the quantum mechanical model of atom:

1. The energy of electrons in atoms is quantized. That is, it can only have certain specific values when electrons are bound to the nucleus in atoms.

2. The quantized electronic energy levels are due to the wave like properties of electrons. Thus, the solutions of Schrödinger wave equation can be applied to them.

3. Both the exact position and exact velocity of an electron in an atom cannot be determined simultaneously (Heisenberg uncertainty principle). The path of an electron in an atom therefore, can never be determined or known accurately. Hence, probability can be applied to find the electron at different points in an atom.

4. An atomic orbital is defined by the wave function,  for an electron in an atom. So, whenever an electron is described by a wave function, we say that the electron occupies that orbital. Since many such wave functions are possible for an electron, there are many atomic orbitals in an atom. In each orbital, the electron has a definite energy. An orbital cannot contain more than two electrons.

In a multi-electron atom, the electrons are filled in various orbitals in the order of increasing energy. For each electron of a multi-electron atom, there exists an orbital wave function, characteristic of the orbital it occupies. All the information about the electron in an atom is stored in its orbital wave function,  which we can extract.

5. The probability of finding an electron at a point within an atom is proportional to the square of the orbital wave function that is$| Ψ |^2$ at that point. $| Ψ |^2$ is known as probability density and is always positive. From the value of $| Ψ |^2$ at different points within an atom, it is possible to predict the region around the nucleus where electron will most probably be found.

Orbitals and Quantum Numbers:

Quantum Numbers: The set of four numbers which give the complete description of an electron in an atom—energy, orbital occupied, size, shape, orientation of the orbital, the direction of electron from the nucleus and the main energy level to which the electrons belongs—is called Quantum Numbers. They are:

1. Principal quantum number (n):

• The principal quantum number ‘n’ is a positive integer with value of n = 1,2,3….
• The principal quantum number determines the size and the energy of the orbital.
• The principal quantum number also identifies the shell (like, K, L, M and N).
• With the increase in the value of ‘n’, the number of allowed orbital increases.
• The maximum number of electrons that the shell can accommodate is given by: $‘n^2’$.

Shell	Name	Value of n	Maximum number of electrons
I  shell	K  shell	n=1	2 x $1^2$=2
II  shell	L  shell	n=2	2 x $2^2$=8
III  shell	M shell	n=3	2 x $3^2$=18
IV  shell	N  shell	n=4	2 x $4^2$= 32

Q and A:

Why do the farthest shells have higher energy levels?

Size of an orbital also increases with increase of principal quantum number ‘n’. In other words, the electron will be located away from the nucleus. Since energy is required in shifting away the negatively charged electron from the positively charged nucleus, the energy of the shell and orbital will increase with increase of n.

2. Azimuthal quantum number (l):

• Azimuthal quantum number, ‘l’ is also known as orbital angular momentum or subsidiary quantum number.
• It defines the three dimensional shape of the orbital.
• For a given value of n, “l” can have “n” values ranging from “0” to “n – 1”, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ………. (n–1).

For example, when n = 1, value of l is 0.

For n = 2, the possible value of l can be: 0 and 1.

For n = 3, the possible l values are 0, 1 and 2.

• These values for “l” correspond to the subshell or sub-level.
• The number of subshells in any shell is equal to the principal quantum number, “n”. For example, in the first shell (n = 1), there is only one sub-shell which corresponds to l = 0. There are two sub-shells (l = 0, 1) in the second shell (n = 2), three (l = 0, 1, 2) in third shell (n = 3) and so on. Each sub-shell is assigned an azimuthal quantum number.
• Sub-shells corresponding to different values of “l” are represented by the following symbols: s, p, d, f, g and h (for l= 0, 1, 2, 3, 4 and 5 respectively).

Table 2.4 shows the permissible values of ‘l’ for a given principal quantum number and the corresponding sub-shell notation.

3. Magnetic quantum number (m):

• Magnetic orbital quantum number, ‘m_l’ gives information about the spatial orientation of the orbital with respect to standard set of co-ordinate axis.
• For any sub-shell with an ‘l’ value, “2l+1” values of “m_l” are possible.
• “m_l” can take the values from “-l” to “+l” including zero, denoted by: m_l = – l, – (l–1), – (l–2)… 0, 1… (l –2), (l–1), l.
• The total number of values of m_l which is found as: (2l + 1) gives the total number of orbitals in the subshell.

For example, for l = 0 (so, s-orbital), the only permitted value of m_l = 0, [obtained as: 2(0) + 1 = 1, that is, one s-orbital].

For l = 1 (so, p-orbital), m_l can be –1, 0 and +1 [obtained as: 2(1) + 1 = 3, three p-orbitals].

For l = 2 (so, d-orbital), m_l = –2, –1, 0, +1 and +2 [obtained as: 2(2) + 1 = 5, five d-orbitals].

The following chart summarizes the relation between the sub-shell and the number of orbitals associated with it:

4. Spin quantum number $(m_s):$

• Spin quantum number describes the spin of the electron on its own axis.
• It may spin either clockwise or anticlockwise.
• It can take the values: +1/2 and -1/2 which are symbolically represented as:  .
• Each orbital can have two electrons and these two electrons should have opposite spins.

To sum up, the four quantum numbers provide the following information:

i) “n” defines the shell, the size of the orbital and also to a large extent, the energy of the orbital.

ii) “L” identifies the subshell and determines the shape of the orbital. There are n subshells in the n^th shell. There are (2l+1) orbitals of each type in a subshell, that is, one s-orbital (l = 0), three p orbitals (l = 1) and five d orbitals (l = 2) per subshell. To some extent l also determines the energy of the orbital in a multi-electron atom.

iii) “m_l” designates the orientation of the orbital. For a given value of l, m_l has (2l+1) values, the same as the number of orbitals per subshell. It means that, the number of orbitals is equal to the number of ways in which they are oriented.

iv) “m_s” refers to orientation of the spin of the electron.

Example: What is the total number of orbitals associated with the principal quantum number $n = 3$?

Solution:  

For $n = 3$, the possible values of $l$ are 0, 1 and 2. Thus there is one 3s orbital ($n = 3, l = 0$ and $m_l = 0$); there are three 3p orbitals ($n = 3, l = 1$ and $m_l = -1, 0, +1$); there are five 3d orbitals ($n = 3, l = 2$ and $m_l = -2, -1, 0, +1, +2$). 

Therefore, the total number of orbitals is $1 + 3 + 5 = 9$ 

The same value can also be obtained by using the relation; number of orbitals $= n^2$, i.e. $3^2 = 9$. 

Example: Using s, p, d, f notations, describe the orbital with the following quantum numbers 

(a) $n = 2, l = 1$, (b) $n = 4, l = 0$, (c) $n = 5, l = 3$, (d) $n = 3, l = 2$ 

Solution: 

Orbitals:

Definition box: Orbital: An orbital is the three-dimensional space around the nucleus of an atom where there is high probability of finding electrons.

Probability density: According to German physicist, Max Born, $| Ψ |^2$ at any point in an atom gives the value of probability density at that point. Probability density is defined as the probability per unit volume.

The product of $| Ψ |^2$ and a small volume (called a volume element) yields the probability of finding the electron in that volume (the reason for specifying a small volume element is that $| Ψ |^2$ varies from one region to another in space but its value can be assumed to be constant within a small volume element).

The total probability of finding the electron in a given volume can then be calculated by the sum of all the products of $| Ψ |^2$  and the corresponding volume elements. It is thus possible to get the probability distribution of an electron in an orbital.

2.6.2 Shapes of Atomic Orbitals

For 1s orbital, the probability density is maximum at the nucleus and it decreases sharply as we move away from it.

On the other hand, for 2s orbital the probability density first decreases sharply to zero and again starts increasing. After reaching small maxima it decreases again and approaches zero as the value of r increases further. The region where this probability density function reduces to zero is called nodal surfaces or simply nodes. In general, it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n. In other words, number of nodes for 2s orbital is one, two for 3s and so on.

Fig. 2.12 The plots of (a) the orbital wave function $\Psi (r)$ (b) the variation of probability density $\Psi^2 (r)$ as a function of distance r of the electron from the nucleus for 1s and 2s orbitals.

Boundary surface diagrams of constant probability density for different orbitals give a fairly good representation of the shapes of the orbitals. In this representation, a boundary surface or contour surface is drawn in space for an orbital on which the value of probability density $| Ψ |^2$ is constant. In principle many such boundary surfaces may be possible. However, for a given orbital, only that boundary surface diagram of constant probability density is taken to be good representation of the shape of the orbital which encloses a region or volume in which the probability of finding the electron is very high.

Boundary surface diagrams of s, p and d orbitals:

s-orbital:

If l = 0, the orbital is known as s-orbital and is spherical in shape.

p-orbital:

The p-orbitals are dumb-bell shaped and are of same energy. Hence, they are said to be three-fold degenerate.

d-orbital:

The five d-orbitals are designated as $d_{xy}, d_{yz}, d_{xz},d_{x^2-y^2}$ and $d_{z^2}$. The first four orbitals have double dumbbell shapes and $d_{z^2}$ has a dumbbell shape along the z-axis and doughnut shape in the x-y plane.

f-orbital:

They have complex shapes.

Types of nodes:

a) Angular nodes: Angular nodes are typically flat plane (at fixed angles), as in case of $p_z$ orbital, xy-plane is a nodal plane, in case of $d_{xy}$ orbital, there are two nodal planes passing through the origin and bisecting the xy plane containing z-axis. These are called angular nodes. The number of angular nodes in an orbital is given by: “l” (quantum number).

b) Radial nodes: Radial nodes are spherical nodes (at fixed radius) that occur as the principle quantum number increases. Total number of radial nodes is given by: (n – l – 1).

The total number of nodes of an orbital is the sum of angular and radial nodes.

2.6.3 Energies of Orbitals

• The energy of an electron in a hydrogen atom is determined solely by the principal quantum number. Thus, the energy of the orbitals increases as follows:

$$1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < \qquad (2.23)$$

Figure 2.14 Energy level diagrams for the few electronic shells of (a) hydrogen atom and (b) multi-electronic atoms. Note that orbitals for the same value of principal quantum number, have the same energies even for different azimuthal quantum number for hydrogen atom. In case of multi-electron atoms, orbitals with same principal quantum number possess different energies for different azimuthal quantum numbers.

• Although the shapes of 2s and 2p orbitals are different, an electron has the same energy when it is in the 2s orbital as and in 2p orbital. Such orbitals having the same energy are called degenerate.
• The 1s orbital in a hydrogen atom, as said earlier, corresponds to the most stable condition and is called the ground state and an electron residing in this orbital is most strongly held by the nucleus. An electron in the 2s, 2p or higher orbitals in a hydrogen atom is said to be in excited state.

Why does the energy of an electron in a multi-electron atom not only depend on the principal quantum number?

1. Difference in azimuthal quantum number—The energy of an electron in a multi-electron atom, unlike that of the hydrogen atom, depends also on its azimuthal quantum number (subshell) because, for a given principal quantum number, azimuthal quantum numbers (s, p, d, f..) have different energies.
2. Interactive forces in hydrogen atom—The only electrical interaction present in hydrogen atom is the attraction between the negatively charged electron and the positively charged nucleus.
3. Interactive forces in multi-electron atoms—In multi-electron atoms, besides the presence of attraction between the electron and nucleus, there is mutual repulsion between electrons in the atom. The stability of an electron in multi-electron atom is because total attractive interactions are more than the repulsive interactions. On the other hand, the attractive forces between electrons increase with increase of positive charge (Ze) on the nucleus. Due to the presence of electrons in the inner shells, the electron in the outer shell will not experience the full positive charge of the nucleus (Ze). The effect will be lowered due to the partial screening of positive charge on the nucleus by the inner shell electrons. This is known as the shielding of the outer shell electrons from the nucleus by the inner shell electrons, and the net positive charge experienced by the outer electrons due to the nucleus is known as effective nuclear charge ($Z_{eff}$e).
4. Impact of shielding effect—Despite the shielding of the outer electrons from the nucleus by the inner shell electrons, the attractive force experienced by the outer shell electrons increases with increase of nuclear charge. It follows that, the energy of interaction between, the nucleus and electron (that is orbital energy) decreases (that is more negative) with the increase of atomic number (Z) since the energy required to draw the outer-shell electrons would be less due to increase in positive charge.
5. Effect of size and shape of orbital—Both the attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron is present. For example, electrons present in spherical shaped, s orbital shields the outer electrons from the nucleus more effectively as compared to electrons present in p orbital. Similarly, electrons present in p orbitals shield the outer electrons from the nucleus more than the electrons present in d orbitals, even though all these orbitals are present in the same shell. Further within a shell, due to spherical shape of s orbital, the s orbital electron spends more time close to the nucleus in comparison to p orbital electron which spends more time in the vicinity of nucleus in comparison to d orbital electron. In other words, for a given shell (principal quantum number), the $Z_{eff}$ experienced by the electron decreases with increase of azimuthal quantum number (l). So, the s orbital electron will be more tightly bound to the nucleus than p orbital electron and p orbital electron will be more tightly bound than the d orbital electron. Consequently, the energy of electrons in s orbital will be lower (more negative) than that of p orbital electron which will have less energy than that of d orbital electron and so on.
6. Effect of shielding on energy of orbital—Since the extent of shielding from the nucleus is different for electrons in different orbitals, it leads to the splitting of energy levels within the same shell (or same principal quantum number), that is, energy of electron in an orbital, as mentioned earlier, depends upon the values of n and l. It follows a simple rule: The lower the value of (n + l) for an orbital, the lower is its energy. Also, if two orbitals have the same value of (n + l), the orbital with lower value of “n” will have the lower energy. Thus, different subshells of a particular shell have different energies in case of multi–electrons atoms. However, in hydrogen atom, these have the same energy.
7. Effect of atomic number on energy of orbital—Lastly, energies of the orbitals in the same subshell decrease with increase in the atomic number ($Z_{eff}$). For example, energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on, that is, $E_{2s}(H) > E_{2s}(Li) > E_{2s}(Na) > E_{2s}(K)$.

2.6.4 Filling of Orbitals in Atom

Definition box: Electron configuration is the distribution of electrons into orbitals of an atom.

The following principles are necessary to write the electronic configuration of elements:

Aufbau’s Principle

• The word, ‘aufbau’ in German means: ‘building up’. The building up of orbitals means, the filling up of orbitals with electrons.
• According to the principle, “In the ground state of an atom, the orbitals are filled in order of their increasing energies”.
• In other words, electrons first occupy the lowest energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled.
• The order in which the energies of the orbitals increase and hence the order in which the orbitals are filled is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s…

Starting from the top, the direction of the arrows gives the order of filling of orbitals, starting from right top and ending at bottom left:

• To find exact order of increasing energy, the (n+l) rule is used (as depicted in the table below):

According to this rule,

• The subshell with lower (n+l) value has a lower energy.
• If two subshells have the same (n+l) value, then the subshell with a lower “n” value has a lower energy. Thus, 2p and 3s have same (n+l) value but 2p having lower “n” value is filled first. Similarly, between 3p and 4s, the orbital, 3p with lower “n” is filled first. Between 3d and 4s, 4s orbital with lower (n+l) value is preferred.  Thus, the order of energy is got as: 1s< 2s< 2p< 3s< 3p< 4s< 3d< 4p< 5s< 4d< 5p< 4f< 5d< 6p< 7s.

Pauli Exclusion Principle

• It was given by the Austrian scientist, Wolfgang Pauli.
• According to this principle, “No two electrons in an atom can have the same set of four quantum numbers”.
• Pauli’s exclusion principle can also be stated as: “Only two electrons may exist in the same orbital and these electrons must have opposite spin.”
• The two statements can be summarised as: two electrons can have the same value of three quantum numbers n, l and m_l, but must have the opposite spin quantum number. For example, the helium atom has the configuration represented as: ↑↓1$s^2$.

Note box: Configurations like, $1s^2$ is read as: “1 s 2” and not as: “1 s squared”.

Hund’s Rule of Maximum Multiplicity

This rule deals with the filling of electrons into the orbitals belonging to the same subshell (that is, orbitals of equal energy, called degenerate orbitals).

It states that, “pairing of electrons in the orbitals belonging to the same subshell (p, d or f) does not take place until each orbital belonging to that subshell has got one electron each that is, it is singly occupied”.

For example,

“s” subshell contains only one orbital. So, the first and second electron gets paired with the entry of second electron.

“p” subshell contains three orbitals, $p_x, p_y$ and $p_z$. After accommodating one electron in each, only the 4^th electron gets paired.

“d” subshell contains five orbitals. Only the 6^th electron gets paired.

“f” has seven orbitals. Only the 8^th electron gets paired.

Note box: It has been observed that half-filled and fully filled degenerate set of orbitals acquire extra stability due to their symmetry.

Ways of representing the electronic configuration of atoms:

$s^ap^bd^c$ …… notation:

• In this type of notation, the subshell is represented by the respective letter symbol and the number of electrons present in the subshell is depicted, as the super script, like a, b, c.
• The similar subshells that exist for different shells are differentiated by writing the principal quantum number before the respective subshell.

Second type:

• In the second notation, each orbital of the subshell is represented by a box and the electron is represented by an arrow ( ) a positive spin or an arrow ( ) a negative spin.
• The advantage of second notation over the first is that, it represents all the four quantum numbers.

The electronic configurations of the known elements (as determined by spectroscopic methods) are tabulated in Table 2.6.

Table 2.6 Electronic Configurations of the Elements

Element    Z	1s	2s     2p	3s     3p     3d	4s     4p     4d     4f	5s     5p     5d     5f	6s     6p     6d     7s
H       1 He     2	1 2
Li        3 Be      4 B        5 C        6 N        7 O        8 F         9 Ne      10	2 2 2 2 2 2 2 2	1 2 2      1 2      2 2      3 2      4 2      5 2      6
Na      11 Mg     12 Al       13 Si       14 P        15 S        16 Cl       17 Ar      18 K        19 Ca      20 Sc       21 Ti       22 V        23 Cr*    24 Mn    25 Fe      26 Co      27 Ni      28 Cu*     29 Zn      30	2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2	2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6	1 2 2     1 2     2 2     3 2     4 2     5 2     6 2     6 2     6 2     6     1 2     6     2 2     6     3 2     6     5 2     6     5 2     6     6 2     6     7 2     6     8 2     6     10 2     6     10	1 2 2 2 2 1 2 2 2 2 1 2
Ga      31 Ge      32 As      33 Se      34 Br      35 Kr      36 Rb      37 Sr      38 Y       39 Zr      40 Nb*     41 Mo*     42 Tc      43 Ru*     44 Rh*     45 Pd*     46 Ag*     47 Cd      48 In      49 Sn      50 Sb      51 Te      52 I        53 Xe      54	2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2	2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6 2      6	2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10	2      1 2      2 2      3 2      4 2      5 2      6 2      6 2      6 2      6      1 2      6      2 2      6      4 2      6      5 2      6      7 2      6      8 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10 2      6      10	1 2 2 2 1 1 2 1 1 1 2     1 2     2 2     3 2     4 2     5 2     6

Structure of Atom

Element    Z	1s	2s     2p	3s     3p     3d	4s     4p     4d     4f	5s     5p     5d     5f	6s     6p     6d     7s	7s
Cs     55 Ba     56 La*    57 Ce*    58 Pr     59 Nd     60 Pm     61 Sm     62 Eu     63 Gd*    64 Tb     65 Dy     66 Ho     67 Er     68 Tm     69 Yb     70 Lu     71 Hf     72 Ta     73 W      74 Re     75 Os     76 Ir*    77 Pt**   78 Au**   79 Hg     80 Tl     81 Pb     82 Bi     83 Po     84 At     85 Rn     86	2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2	2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6	2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10	2     6     10 2     6     10 2     6     10 2     6     10     2 2     6     10     3 2     6     10     4 2     6     10     5 2     6     10     6 2     6     10     7 2     6     10     7 2     6     10     9 2     6     10     10 2     6     10     11 2     6     10     12 2     6     10     13 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14	2     6 2     6 2     6     1 6     2 2     6 2     6 2     6 2     6 2     6 2     6     1 2     6 2     6 2     6 2     6 2     6 2     6 2     6     1 2     6     2 2     6     3 2     6     4 2     6     5 2     6     6 2     6     7 2     6     9 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10	1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2     1 2     2 2     3 2     4 2     5 2     6
Fr     87 Ra     88 Ac     89 Th     90 Pa     91 U      92 Np     93 Pu     94 Am     95 Cm     96 Bk     97 Cf     98 Es     99 Fm     100 Md     101 No     102 Lr     103 Rf     104 Db     105 Sg     106 Bh     107 Hs     108 Mt     109 Ds     110 Rg**   111	2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2	2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6 2     6	2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10 2     6     10	2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14	2     6     10 2     6     10 2     6     10 2     6     10 2     6     10     2 2     6     10     3 2     6     10     4 2     6     10     6 2     6     10     7 2     6     10     7 2     6     10     8 2     6     10     10 2     6     10     11 2     6     10     12 2     6     10     13 2     6     10     14 2     6     10     10 2     6     10     11 2     6     10     12 2     6     10     13 2     6     10     14 2     6     10     14 2     6     10     14 2     6     10     14	2     6 2     6 2     6     1 2     6     2 2     6     1 2     6     1 2     6     1 2     6 2     6 2     6     1 2     6     1 2     6 2     6 2     6 2     6 2     6 2     6     1 2     6     2 2     6     3 2     6     4 2     6     5 2     6     6 2     6     7 2     6     8 2     6     10	1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1

• For example, the hydrogen atom has only one electron which goes in the orbital with the lowest energy, namely 1s. The electronic configuration of the hydrogen atom is $1s^1$ meaning that it has one electron in the 1s orbital. There are two electrons in in helium (He) which can occupy the 1s orbital. Thus, its configuration is: $1s^2$. As mentioned above, the two electrons differ from each other with opposite spin, as can be seen from the orbital diagram. Similarly, the third electron of lithium (Li) is not allowed in the 1s orbital because of Pauli’s exclusion principle. It, therefore, takes the next available choice, the 2s orbital. The electronic configuration of Li is $1s^22s^1$. The 2s orbital can accommodate one more electron. The configuration of beryllium (Be) atom is, therefore, $1s^2 2s^2$.   

• In the next six elements—boron (B, $1s^22s^22p^1$), carbon (C, $1s^22s^22p^2$), nitrogen (N, $1s^22s^22p^3$), oxygen (O, $1s^22s^22p^4$), fluorine (F, $1s^22s^22p^5$) and neon (Ne, $1s^22s^22p^6$), the 2p orbitals get progressively filled, with electrons in the neon atom completely occupying the p-orbital. The orbital picture of these elements can be represented as follows:

• The electronic configuration of the elements from sodium (Na, $1s^22s^22p^63s^1$) to argon (Ar, $1s^22s^22p^63s^23p^6$) follows exactly the same pattern as the elements from lithium to neon with the difference that, the 3s and 3p orbitals are getting filled now.
• Hence, the process of writing the electronic configurations for further elements (after neon) can be simplified if we represent the total number of electrons in the first two shells: s and p by the name of element neon (Ne) instead of writing, $1s^22s^22p^6$. The electronic configuration of the elements from sodium to argon can be written as (Na: $[Ne]3s^1)$ to (Ar: [Ne] $3s^23p^6$).

Concept box: Core electrons: The electrons in the completely filled shells are known as core electrons Valence electrons: The electrons that are added to the electronic shell with the highest principal quantum number are called valence electrons. For example, the electrons in Ne are the core electrons and the electrons from Na to Ar are the valence electrons. In potassium (K) and calcium (Ca), the 4s orbital, being lower in energy than the 3d orbitals, is occupied by one and two electrons respectively.

• A new pattern is followed beginning with scandium (Sc). The 3d orbital, being lower in energy than the 4p orbital, is filled first. Consequently, in the next ten elements, scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu) and zinc (Zn), the five 3d orbitals are progressively occupied.
• Chromium and copper have five and ten electrons in 3d orbitals rather than four and nine as their position would have indicated with two-electrons in the 4s orbital. The reason is that, fully filled orbitals and half-filled orbitals have extra stability (that is, lower energy). Thus, $p^3, p^6, d^5, d^{10}, f^7, f^{14}$and so on. Configurations, which are either half-filled or fully filled, are more stable. Chromium and copper therefore adopt the $d^5$ and $d^{10}$ configurations respectively.
• With the saturation of the 3d orbitals, the filling of the 4p orbital starts at gallium (Ga) and is complete at krypton (Kr). In the next eighteen elements from rubidium (Rb) to xenon (Xe), the pattern of filling the 5s, 4d and 5p orbitals are similar to that of 4s, 3d and 4p orbitals as discussed above. Caesium (Cs) and the barium (Ba), fill the 6s orbitals with one and two electrons, respectively. Then from lanthanum (La) to mercury (Hg), the filling up of electrons takes place in 4f and 5d orbitals. After this, 6p orbital gets filled, then 7s and finally 5f and 6d orbitals.

Note box: The elements after uranium (U) are all short-lived and all of them are produced artificially.

Importance of electronic configuration:

Electronic distribution (or configuration) helps us to understand and explain chemical behaviour. For instance, it helps explain how two or more atoms combine to form molecules, the metallic nature of some elements, reactivity of elements like halogens; inert nature of helium and argon and so on.

2.6.6 Stability of Completely Filled and Half Filled Subshells

The ground state electronic configuration of the atom of an element always corresponds to the state of the lowest total electronic energy.

However, in certain elements such as Cu, or Cr, where the two subshells (4s and 3d) differ slightly in their energies, an electron shifts from a subshell of lower energy (4s) to a subshell of higher energy (3d), only when such a shift can cause all orbitals of the subshell of higher energy getting either completely filled or half filled. The valence electronic configurations of Cr and Cu, therefore, are $3d^5 4s^1$ and $3d^{10} 4s^1$ respectively and not $3d^4 4s^2$ and $3d^9 4s^2$ because, there is extra stability associated with these electronic configurations.

Causes of Stability of Completely Filled and Half Filled Sub-shells:

The completely filled and completely half-filled sub-shells are stable due to the following reasons:

1. Symmetrical distribution of electrons: Symmetry leads to stability. The completely filled or half -filled subshells have symmetrical distribution of electrons in them and are therefore more stable. Electrons in the same subshell (here 3d) have equal energy but differ in spatial distribution. Consequently, their shielding of one another is relatively small and the electrons are more strongly attracted by the nucleus.

2. Exchange Energy: The stabilizing effect arises whenever two or more electrons with the same spin are present in the degenerate orbitals of a subshell. These electrons tend to exchange their positions and the energy released due to this exchange is called exchange energy. The number of exchanges that can take place is maximum when the subshell is either half-filled or completely filled (refer Figure 2.16 below). As a result, the exchange energy is maximum and so is the stability.

Note box: Exchange energy is on the basis of Hund’s rule that electrons which enter orbitals of equal energy have parallel spins as far as possible. In other words, the extra stability of half-filled and completely filled subshell is due to: (i) relatively small shielding (ii) smaller coulombic repulsion energy and (iii) larger exchange energy.

Questions from section 2.6:

1. Define:

a) Quantum mechanics

b) Wave function

2. Explain the features of the Quantum Mechanical Model of Atom.

3. Describe:

a) Principal quantum number

b) Azimuthal quantum number

c) Magnetic quantum number

d) Spin quantum number

e) Boundary surface diagrams

4. Define:

a) Orbital

b) Probability density

5. Trace the Boundary surface diagrams for s, p and d orbitals.

6. Name the two types of nodes.

7. What are degenerate orbitals?

8. What is ground state and excited state with regard to orbitals?

9. Explain why the energy of an electron in a multi-electron atom not only depend on the principal quantum number.

10. Define the term: Electronic configuration and explain Aufbau’s Principle of building up of electrons.

11. Explain Pauli’s exclusion principle with an example.

12. Explain Hund’s Rule of Maximum Multiplicity.

13. Explain the two ways of expressing electronic configurations in brief.

14. What are core electrons and valence electrons?

15. Explain the reasons for stability conferred by half-filled and completely filled orbital.