Structure of Atom

Introduction

The concept of the atom as the fundamental building block of matter has its origins in ancient Indian and Greek philosophy around 400 B.C. Philosophers like Kanada in India and Democritus in Greece proposed that matter is composed of indivisible particles called atoms (from the Greek word 'a-tomio' meaning 'uncuttable'). They believed that continuous division of matter would ultimately yield these smallest particles which cannot be further divided. This early atomic concept laid the foundation for modern atomic theory. Over centuries, scientific investigations have revealed that atoms themselves are composed of smaller subatomic particles such as electrons, protons, and neutrons. The differences in the internal structure of atoms of different elements explain the rich diversity of chemical behavior observed in nature. This chapter explores the discovery of these subatomic particles, various atomic models proposed to explain atomic structure, and the development of the quantum mechanical model which accurately describes the behavior of electrons in atoms. **Table on page 1 (1×2)** | The rich diversity of chemical behaviour of differen elements can be traced to the differences in the intern Objectives structure of atoms of these elements. After studying this unit you will be able to • know about the discovery of The existence of atoms has been proposed since the tim electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) wh their characteristics; were of the view that atoms are the fundamental buildin • describe Thomson, Rutherford blocks of matter. According to them, the continue and Bohr atomic models; subdivisions of matter would ultimately yield atom which would not be further divisible. The word ‘atom • understand the important features has been derived from the Greek word ‘a-tomio’ whic of the quantum mechanical model means ‘uncut-able’ or ‘non-divisible’. These earlier idea of atom; were mere speculations and there was no way to tes • u n d e r s t a n d n a t u r e o f them experimentally. These ideas remained dormant fo electromagnetic radiation and a very long time and were revived again by scientists i Planck’s quantum theory; the nineteenth century. • explain the photoelectric effect The atomic theory of matter was first propose and describe features of atomic on a firm scientific basis by John Dalton, a Britis spectra; school teacher in 1808. His theory, called Dalton’ • state the de Broglie relation and atomic theory, regarded the atom as the ultimat Heisenberg uncertainty principle; particle of matter (Unit 1). Dalton’s atomic theory wa able to explain the law of conservation of mass, law o • define an atomic orbital in terms constant composition and law of multiple proportio of quantum numbers; very successfully. However, it failed to explain the result • state aufbau principle, Pauli of many experiments, for example, it was known tha exclusion principle and Hund’s substances like glass or ebonite when rubbed with sil rule of maximum multiplicity; and or fur get electrically charged. • write the electronic confgi urations In this unit we start with the experimental observation of atoms. made by scientists towards the end of nineteenth an | | | --- | --- | | | t al e o g d s ’ h s t r n d h s e s f n s t k s d | **Table on page 12 (1×2)** | (intensity of radiation) from a black bod body radiation mentioned above was given and its spectral distribution depends onl by Max Planck in 1900. Let us first try to on its temperature. At a given temperature understand this phenomenon, which is given intensity of radiation emitted increase below: with the increase of wavelength, reaches Hot objects emit electromagnetic maximum value at a given wavelength an radiations over a wide range of wavelengths. then starts decreasing with further increase o At high temperatures, an appreciable wavelength, as shown in Fig. 2.8. Also, as th proportion of radiation is in the visible temperature increases, maxima of the curv region of the spectrum. As the temperature shifts to short wavelength. Several attempt is raised, a higher proportion of short were made to predict the intensity of radiatio wavelength (blue light) is generated. For as a function of wavelength. example, when an iron rod is heated in a But the results of the above experimen furnace, it fri st turns to dull red and then could not be explained satisfactorily o progressively becomes more and more red the basis of the wave theory of light. Ma as the temperature increases. As this is Planck arrived at a satisfactory relationshi heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. This means that red radiation is most intense at a particular temperature and the blue radiation is more intense at another temperature. This means intensities of radiations of different wavelengths emitted by hot body depend upon its temperature. By late 1850’s it was known that objects made of different material and kept at different temperatures emit different amount of radiation. Also, when the surface of an object is irradiated with light (electromagnetic radiation), a part of radiant energy is generally reflected as such, a part Fig. 2.8 Wavelength-intensity relationship is absorbed and a part of it is transmitted. The reason for incomplete absorption is that ordinary objects are as a rule imperfect absorbers of radiation. An ideal body, which emits and absorbs radiations of all frequencies uniformly, is called a black body and the radiation emitted by such a body is called black body radiation. In practice, no such | | | --- | --- | | body exists. Carbon black approximates fairly closely to black body. A good physical approximation to a black body is a cavity with a | | **Table on page 14 (1×7)** | Metal | Li | Na | K | Mg | Cu | Ag | | --- | --- | --- | --- | --- | --- | --- | | W /eV 0 | 2.42 | 2.3 | 2.25 | 3.7 | 4.8 | 4.3 | **Table on page 15 (2×4)** | | Problem 2.6 Calculate energy of one mole of photons of radiation whose frequency is 5 ×1014 Hz. Solution Energy (E) of one photon is given by the | | Solution The energy (E) of a 300 nm photon is given by = 6.626 × 10–19 J The energy of one mole of photons = 6.626 ×10–19 J × 6.022 × 1023 mol–1 = 3.99 × 105 J mol–1 The minimum energy needed to remove one mole of electrons from sodium = (3.99 –1.68) 105 J mol–1 = 2.31 × 105 J mol–1 The minimum energy for one electron This corresponds to the wavelength hc = E 6.626 10 34J s 3.0 10 8m s 1 = 3.84 10 19J = 517 nm (This corresponds to green light) Problem 2.9 The threshold frequency ν for a metal is 0 7.0 ×1014 s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency ν =1.0 ×1015 s–1 hits the metal. Solution According to Einstein’s equation Kinetic energy = ½ m v2=h(ν – ν ) e 0 = (6.626 × 10–34 J s) (1.0 × 1015 s–1 – 7.0 ×1014 s–1) = (6.626 × 10–34 J s) (10.0 × 1014 s–1 – 7.0 ×1014 s–1) = (6.626 × 10–34 J s) × (3.0 × 1014 s–1) = 1.988 × 10–19 J | | --- | --- | --- | --- | | | expression E = hν h = 6.626 ×10–34 J s ν = 5×1014 s–1 (given) E = (6.626 ×10–34 J s) × (5 ×1014 s–1) = 3.313 ×10–19 J Energy of one mole of photons = (3.313 ×10–19 J) × (6.022 × 1023 mol–1) = 199.51 kJ mol–1 Problem 2.7 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 = hν = hc/λ 6.6261034 J s310 8 m s1 = 400109m = 4.969 × 10–19 J Number of photons emitted 100 J s1  2 .0121020s1 4 .9691019J Problem 2.8 When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 ×105 J mol–1. What is the minimum energy needed to remove | | = 6.626 × 10–19 J The energy of one mole of photons = 6.626 ×10–19 J × 6.022 × 1023 mol–1 = 3.99 × 105 J mol–1 The minimum energy needed to remove one mole of electrons from sodium = (3.99 –1.68) 105 J mol–1 = 2.31 × 105 J mol–1 The minimum energy for one electron This corresponds to the wavelength hc = E 6.626 10 34J s 3.0 10 8m s 1 = 3.84 10 19J = 517 nm (This corresponds to green light) Problem 2.9 The threshold frequency ν for a metal is 0 7.0 ×1014 s–1. Calculate the kinetic energy of an electron emitted when radiation of frequency ν =1.0 ×1015 s–1 hits the metal. Solution According to Einstein’s equation Kinetic energy = ½ m v2=h(ν – ν ) e 0 = (6.626 × 10–34 J s) (1.0 × 1015 s–1 – 7.0 ×1014 s–1) = (6.626 × 10–34 J s) (10.0 × 1014 s–1 – 7.0 ×1014 s–1) = (6.626 × 10–34 J s) × (3.0 × 1014 s–1) | | | an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted? | | | **Table on page 17 (5×6)** | (a) | | | | | | | --- | --- | --- | --- | --- | --- | | (b) Fig. 2.10 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any othe element) can be passed through a prism and separated into certain discrete wavelengths. Thus an emissio spectrum, which is a photographic recording of the separated wavelengths is called as line spectrum. An sample of reasonable size contains an enormous number of atoms. Although a single atom can be in onl one excited state at a time, the collection of atoms contains all possible excited states. The light emitted a these atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption. When whit light is passed through unexcited atomic hydrogen and then through a slit and prism, the transmitted ligh is lacking in intensity at the same wavelengths as are emitted in (a) The recorded absorption spectrum i also a line spectrum and the photographic negative of the emission spectrum. that if spectral lines are expressed in terms The value 109,677 cm–1 is called th of wavenumber ( ), then the visible lines of Rydberg constant for hydrogen. The fri st fvi the hydrogen spectrum obey the following series of lines that correspond to n = 1, 2, 3 1 formula: 4, 5 are known as Lyman, Balmer, Paschen Bracket and Pfund series, respectively (2.8) Table 2.3 shows these series of transitions i the hydrogen spectrum. Fig. 2.11 (page, 46 where n is an integer equal to or greater than shows the Lyman, Balmer and Paschen serie 3 (i.e., n = 3,4,5,....) of transitions for hydrogen atom. The series of lines described by this formula Of all the elements, hydrogen atom ha are called the Balmer series. The Balmer the simplest line spectrum. Line spectru series of lines are the only lines in the hydrogen spectrum which appear in the visible region Table 2.3 The Spectral Lines for Atomi of the electromagnetic spectrum. The Swedish Hydrogen spectroscopist, Johannes Rydberg, noted that all series of lines in the hydrogen Series n n Spectral Region spectrum could be described by the following 1 2 expression : Lyman 1 2,3.... Ultraviolet Balmer 2 3,4.... Visible (2.9) Paschen 3 4,5.... Infrared | | | | | | | | 0 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or an | | | | | | | | Series | n n 1 2 | Spectral Region | | | | | Lyman Balmer Paschen | 1 2,3... 2 3,4... 3 4,5... | . Ultraviolet . Visible . Infrared | | | where n =1,2........ 1 n = n + 1, n + 2...... 2 1 1 | | Brackett Pfund | 4 5,6... 5 6,7... | | | **Table on page 18 (7×10)** | | | | | | | | | | | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | are arranged concentrically around th nucleus. ii) The energy of an electron in the orbi does not change with time. However the electron will move from a lowe stationary state to a higher stationar state when required amount of energ is absorbed by the electron or energy i emitted when electron moves from highe stationary state to lower stationary stat (equation 2.16). The energy change doe not take place in a continuous manner Angular Momentum Just as linear momentum is the product Fig. 2.11 Transitions of the electron in the of mass (m) and linear velocity (v), angular hydrogen atom (The diagram shows the Lyman, Balmer and Paschen series of momentum is the product of moment of transitions) inertia (I) and angular velocity (ω). For an electron of mass m , moving in a circular e becomes more and more complex for heavier path of radius r around the nucleus, atom. There are, however, certain features angular momentum = I × ω which are common to all line spectra, i.e., Since I = m r2, and ω = v/r where v is the (i) line spectrum of element is unique and e linear velocity, (ii) there is regularity in the line spectrum of ∴angular momentum = m r2 × v/r = m vr each element. The questions which arise are: e e What are the reasons for these similarities? iii) The frequency of radiation absorbed o Is it something to do with the electronic emitted when transition occurs betwee structure of atoms? These are the questions two stationary states that differ i need to be answered. We shall find later that energy by ∆E, is given by: the answers to these questions provide the key in understanding electronic structure of (2.10 these elements. Where E and E are the energies of th 2.4 Bohr’s Model for Hydrogen 1 2 lower and higher allowed energy state Atom respectively. This expression is commonl Neils Bohr (1913) was the first to explain known as Bohr’s frequency rule. quantitatively the general features of the iv) The angular momentum of an electro structure of hydrogen atom and its spectrum. is quantised. In a given stationary stat | | | | | | | | | | | | | | | | | | | Angular Momentum Just as linear momentum is the product of mass (m) and linear velocity (v), angular momentum is the product of moment of inertia (I) and angular velocity (ω). For an electron of mass m , moving in a circular e path of radius r around the nucleus, angular momentum = I × ω Since I = m r2, and ω = v/r where v is the e linear velocity, ∴angular momentum = m r2 × v/r = m vr e e | | | | | | | | | | | | r n n ) e s y n e | **Table on page 19 (4×3)** | Where m is the mass of electron, v is the e velocity and r is the radius of the orbit in which electron is moving. Thus an electron can move only in those orbits for which its angular momentum is integral multiple of h/2π. That means angular momentum is quantised. Radiation is emitted or obsorbed only when transition of electron | Niels Bohr (1885–1962) Niels Bohr, a Danish physicist received his Ph.D. from the University of Copenhagen in 1911. He then spent a year with J.J. Thomson and Ernest Rutherford in England. In 1913, he returned to Copenhagen where he remained for the rest of his life. In 1920 he was named Director of the Institute of theoretical Physics. After first World War, Bohr worked energetically for peaceful uses of atomic energy. He received the fri st Atoms for Peace award in 1957. Bohr was awarded the Nobel Prize in Physics in 1922. | | | --- | --- | --- | | takes place from one quantised value of he returned to Copenhagen where he remained for angular momentum to another. Therefore, the rest of his life. In 1920 he was named Director of the Institute of theoretical Physics. After first Maxwell’s electromagnetic theory does not World War, Bohr worked energetically for peaceful apply here that is why only certain fxi ed orbits uses of atomic energy. He received the fri st Atoms are allowed. for Peace award in 1957. Bohr was awarded the The details regarding the derivation of Nobel Prize in Physics in 1922. energies of the stationary states used by Bohr, are quite complicated and will be discussed in higher classes. However, according to Bohr’s Fig. 2.11 depicts the energies of differen theory for hydrogen atom: stationary states or energy levels of hydroge a) The stationary states for electron are atom. This representation is called an energ level diagram. numbered n = 1,2,3.......... These integral numbers (Section 2.6.2) are known as When the electron is free from the inful enc Principal quantum numbers. of nucleus, the energy is taken as zero. Th b) The radii of the stationary states are electron in this situation is associated with th expressed as: stationary state of Principal Quantum numbe = n = ∞ and is called as ionized hydrogen atom r = n2 a (2.12) n 0 When the electron is attracted by the nucleu where a = 52.9 pm. Thus the radius of and is present in orbit n, the energy is emitte 0 the fri st stationary state, called the Bohr and its energy is lowered. That is the reaso orbit, is 52.9 pm. Normally the electron in the hydrogen atom is found in this What does the negative electronic orbit (that is n=1). As n increases the energy (E ) for hydrogen atom mean? value of r will increase. In other words n the electron will be present away from The energy of the electron in a hydrogen the nucleus. atom has a negative sign for all possible orbits (eq. 2.13). What does this negative c) The most important property associated sign convey? This negative sign means with the electron, is the energy of its that the energy of the electron in the stationary state. It is given by the atom is lower than the energy of a free expression. electron at rest. A free electron at rest  1  is an electron that is infinitely far away E n R H  n2 n = 1,2,3.... (2.13) from the nucleus and is assigned the energy value of zero. Mathematically, this where R is called Rydberg constant and H corresponds to setting n equal to infinity its value is 2.18×10–18 J. The energy of the in the equation (2.13) so that E =0. As the lowest state, also called as the ground state, is ∞ electron gets closer to the nucleus (as n 1 decreases), E becomes larger in absolute E 1 = –2.18×10–18 ( 12 ) = –2.18×10–18 J. The value and mon re and more negative. The | he returned to Copenhagen where he remained for the rest of his life. In 1920 he was named Director of the Institute of theoretical Physics. After first World War, Bohr worked energetically for peaceful uses of atomic energy. He received the fri st Atoms for Peace award in 1957. Bohr was awarded the Nobel Prize in Physics in 1922. | | | | | t n y e e e r . s d n | | | What does the negative electronic energy (E ) for hydrogen atom mean? n The energy of the electron in a hydrogen atom has a negative sign for all possible orbits (eq. 2.13). What does this negative sign convey? 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 equation (2.13) so that E =0. As the ∞ electron gets closer to the nucleus (as n decreases), E becomes larger in absolute n value and more and more negative. The | | | energy of the stationary state for n = 2, will 1 be : E = –2.18×10–18J ( ) = –0.545×10–18 2 22 J. | | |

• Atoms are the fundamental building blocks of matter.
• The word 'atom' is derived from Greek 'a-tomio' meaning indivisible.
• Ancient philosophers proposed that matter is composed of atoms.
• Modern science has discovered subatomic particles within atoms.
• Differences in atomic structure explain chemical diversity.
• This chapter covers discovery, models, and quantum theory of atoms.
• 📌 Atom: The smallest unit of matter that retains the properties of an element.
• 📌 Subatomic particles: Particles smaller than an atom, such as electrons, protons, and neutrons.

Discovery of Electron

The discovery of the electron was a landmark event in atomic theory. In the late 19th century, experiments with cathode rays led to the identification of electrons as negatively charged particles present in atoms. A cathode ray tube consists of two electrodes sealed in a glass tube from which air has been evacuated. When a high voltage is applied across the electrodes, a stream of particles called cathode rays is emitted from the cathode (negative electrode) and travels towards the anode (positive electrode). These rays were found to be deflected by electric and magnetic fields, indicating that they are charged particles. J.J. Thomson conducted experiments using cathode ray tubes and measured the charge-to-mass ratio (e/m) of these particles, establishing that they are much lighter than atoms and carry a negative charge. This discovery proved that atoms are not indivisible but contain smaller charged particles. Millikan's oil drop experiment later measured the charge of an electron, allowing calculation of its mass. The electron has a charge of –1.602176 × 10⁻¹⁹ coulombs and a mass of 9.1094 × 10⁻³¹ kg.

• Cathode rays are streams of negatively charged particles called electrons.
• J.J. Thomson measured charge-to-mass ratio (e/m) of electrons.
• Electrons are much lighter than atoms and carry negative charge.
• Millikan's oil drop experiment measured the charge of an electron.
• Electron charge = –1.602176 × 10⁻¹⁹ C; mass = 9.1094 × 10⁻³¹ kg.
• Discovery showed atoms are divisible into subatomic particles.
• 📌 Electron: Negatively charged subatomic particle found in atoms.
• 📌 Cathode ray tube: Device used to study cathode rays and discover electrons.
• 📌 Charge-to-mass ratio (e/m): Ratio of electric charge to mass of a particle.