12.1 Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

12.1 (a) No different from (b) Thomson’s model; Rutherford’s model (c) Rutherford’s model (d) Thomson’s model; Rutherford’s model (e) Both the models

2.2 Atomic Models measuring charge ‘e’. In chamber, Observations obtained from the experiments the forces acting on oil drop are: gravitational, electrostatic due tomentioned in the previous sections have electrical field and a viscous dragsuggested that Dalton’s indivisible atom is force when the oil drop is moving. composed of sub-atomic particles carrying positive and negative charges. The major problems before the scientists after the • to explain the formation of different discovery of sub-atomic particles were: kinds of molecules by the combination of different atoms and,• to account for the stability of atom, • to understand the origin and nature of• to compare the behaviour of elements the characteristics of electromagnetic in terms of both physical and chemical radiation absorbed or emitted by atoms. properties, Reprint 2025-26 structure of atom 33 Table 2.1 Properties of Fundamental Particles Name Symbol Absolute Relative Mass/kg Mass/u Approx. charge/C charge mass/u Electron e – 1.602176×10–19 –1 9.109382×10–31 0.00054 0 Proton p + 1.602176×10–19 +1 1.6726216×10–27 1.00727 1 Neutron n 0 0 1.674927×10–27 1.00867 1 Different atomic models were proposed to explain the distributions of these charged In the later half of the nineteenth century particles in an atom. Although some of these different kinds of rays were discovered, models were not able to explain the stability besides those mentioned earlier. Wilhalm of atoms, two of these models, one proposed Röentgen (1845-1923) in 1895 showed by J.J. Thomson and the other proposed by that when electrons strike a material in Ernest Rutherford are discussed below. the cathode ray tubes, produce rays which can cause fluorescence in the fluorescent2.2.1 Thomson Model of Atom materials placed outside the cathode ray J. J. Thomson, in 1898, proposed that an tubes. Since Röentgen did not know the atom possesses a spherical shape (radius nature of the radiation, he named themapproximately 10–10 m) in which the positive X-rays and the name is still carried on. It wascharge is uniformly distributed. The electrons noticed that X-rays are produced effectivelyare embedded into it in such a manner as to when electrons strike the dense metal anode,give the most stable electrostatic arrangement (Fig. 2.4). Many different names are given called targets. These are not deflected by the to this model, for example, plum pudding, electric and magnetic fields and have a very raisin pudding or watermelon. This model 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 (∼0.1 nm) and possess electro-magnetic character (Section 2.3.1). Henri Becqueral (1852-1908) observed that there are certain elements which emit radiation on their own and named this Fig.2.4 Thomson model of atom phenomenon as radioactivity and the can be visualised as a pudding or watermelon elements known as radioactive elements. of positive charge with plums or seeds This field was developed by Marie Curie, (electrons) embedded into it. An important Piere Curie, Rutherford and Fredrick Soddy. feature of this model is that the mass of the It was observed that three kinds of rays i.e., atom is assumed to be uniformly distributed α, β- and γ-rays are emitted. Rutherford over the atom. Although this model was able found that α-rays consists of high energy to explain the overall neutrality of the atom, particles carrying two units of positive charge but was not consistent with the results of later and four unit of atomic mass. He concluded experiments. Thomson was awarded Nobel that α- particles are helium nuclei as when α- Prize for physics in 1906, for his theoretical particles combined with two electrons yielded and experimental investigations on the helium gas. β-rays are negatively charged conduction of electricity by gases. Reprint 2025-26 34 chemistry represented in Fig. 2.5. A stream of high particles similar to electrons. The γ-rays energy α–particles from a radioactive source are high energy radiations like X-rays, are was directed at a thin foil (thickness ∼ 100 nm) neutral in nature and do not consist of of gold metal. The thin gold foil had a circular particles. As regards penetrating power, fluorescent zinc sulphide screen around it. α-particles are the least, followed by β-rays Whenever α–particles struck the screen, a (100 times that of α–particles) and γ-rays tiny flash of light was produced at that point. (1000 times of that α-particles). The results of scattering experiment were quite unexpected. According to Thomson model of atom, the mass of each gold atom2.2.2 Rutherford’s Nuclear Model of Atom in the foil should have been spread evenly Rutherford and his students (Hans Geiger over the entire atom, and α–particles had and Ernest Marsden) bombarded very thin enough energy to pass directly through such a gold foil with α–particles. Rutherford’s famous uniform distribution of mass. It was expected –particle scattering experiment is that the particles would slow down and change directions only by a small angles as they passed through the foil. It was observed that: (i) most of the α–particles passed through the gold foil undeflected. (ii) a small fraction of the α–particles was deflected by small angles. (iii) a very few α–particles (∼1 in 20,000) bounced back, that is, were deflected by A. Rutherford’s scattering experiment nearly 180°. On the basis of the 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 α–particles passed through the foil undeflected. (ii) A few positively charged α–particles were deflected. 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. The positive charge has to be concentrated in a very small volume that repelled and deflected the positively charged α–particles. B. Schematic molecular view of the gold foil (iii) Calculations by Rutherford showed that the volume occupied by the nucleusFig. 2.5 Schematic view of Rutherford’s is negligibly small as compared to the scattering experiment. When a beam total volume of the atom. The radius of of alpha () particles is “shot” at a thin gold foil, most of them pass through the atom is about 10–10 m, while that of without much effect. Some, however, nucleus is 10–15 m. One can appreciate are deflected. this difference in size by realising that if Reprint 2025-26 structure of atom 35 a cricket ball represents a nucleus, then The total number of nucleons is termed as the radius of atom would be about 5 km. mass number (A) of the atom. On the basis of above observations and mass number (A) = number of protons (Z ) conclusions, Rutherford proposed the nuclear + number of model of atom. According to this model: neutrons (n) (2.4) (i) The positive charge and most of the mass 2.2.4 Isobars and Isotopes of the atom was densely concentrated in The composition of any atom can be extremely small region. This very small represented by using the normal element portion of the atom was called nucleus symbol (X) with super-script on the left hand by Rutherford. side as the atomic mass number (A) and (ii) The nucleus is surrounded by electrons subscript (Z) on the left hand side as the that move around the nucleus with a atomic number (i.e., AZ X). very high speed in circular paths called Isobars are the atoms with same mass orbits. Thus, Rutherford’s model of atom number but different atomic number for resembles the solar system in which the example, 14C6 and 14N.7 On the other hand, nucleus plays the role of sun and the atoms with identical atomic number but electrons that of revolving planets. different atomic mass number are known(iii) Electrons and the nucleus are held as Isotopes. In other words (according to together by electrostatic forces of equation 2.4), it is evident that difference attraction. between the isotopes is due to the presence 2.2.3 Atomic Number and Mass Number of different number of neutrons present in the nucleus. For example, considering ofThe presence of positive charge on the nucleus is due to the protons in the nucleus. As hydrogen atom again, 99.985% of hydrogen established earlier, the charge on the proton atoms contain only one proton. This isotope is is equal but opposite to that of electron. The called protium (11H). Rest of the percentage of number of protons present in the nucleus is hydrogen atom contains two other isotopes, equal to atomic number (Z ). For example, the the one containing 1 proton and 1 neutron number of protons in the hydrogen nucleus is called deuterium (12D, 0.015%) and the is 1, in sodium atom it is 11, therefore their other one possessing 1 proton and 2 neutrons atomic numbers are 1 and 11 respectively. is called tritium (13T ). The latter isotope is In order to keep the electrical neutrality, found in trace amounts on the earth. Other the number of electrons in an atom is equal examples of commonly occuring isotopes are: to the number of protons (atomic number, carbon atoms containing 6, 7 and 8 neutrons Z ). For example, number of electrons in 12 13 14 besides 6 protons ( 6 C, 6 C, 6 C ); chlorinehydrogen atom and sodium atom are 1 and atoms containing 18 and 20 neutrons besides 11 respectively. 35 37 17 protons ( 17 Cl, 17 Cl ). Atomic number (Z) = number of protons in Lastly an important point to mention the nucleus of an atom regarding isotopes is that chemical properties = number of electrons of atoms are controlled by the number of in a nuetral atom (2.3) electrons, which are determined by the number While the positive charge of the nucleus of protons in the nucleus. Number of neutrons is due to protons, the mass of the nucleus, present in the nucleus have very little effect due to protons and neutrons. As discussed on the chemical properties of an element. earlier protons and neutrons present in the Therefore, all the isotopes of a given element nucleus are collectively known as nucleons. show same chemical behaviour. Reprint 2025-26 36 chemistry of the massive sun and the electrons being Problem 2.1 similar to the lighter planets. When classical Calculate the number of protons, 80 mechanics* is applied to the solar system, it neutrons and electrons in 35Br . shows that the planets describe well-defined Solution orbits around the sun. The gravitational force between the planets is given by the expression In this case, 8035Br , Z = 35, A = 80, species  m 1m 2  2  where m1 and m2 are the masses, is neutral  G. r Number of protons = number of electrons r is the distance of separation of the masses = Z = 35 and G is the gravitational constant. The theory Number of neutrons = 80 – 35 = 45, can also calculate precisely the planetary (equation 2.4) orbits and these are in agreement with the Problem 2.2 experimental measurements. The number of electrons, protons and The similarity between the solar system neutrons in a species are equal to 18, 16 and nuclear model suggests that electrons and 16 respectively. Assign the proper should move around the nucleus in well symbol to the species. defined orbits. Further, the coulomb force Solution (kq1q2/r2 where q1 and q2 are the charges, r is the distance of separation of the charges The atomic number is equal to and k is the proportionality constant) between number of protons = 16. The element is electron and the nucleus is mathematically sulphur (S). similar to the gravitational force. However, Atomic mass number = number of when a body is moving in an orbit, it protons + number of neutrons undergoes acceleration even if it is moving = 16 + 16 = 32 with a constant speed in an orbit because Species is not neutral as the number of of changing direction. So an electron in the protons is not equal to electrons. It is nuclear model describing planet like orbits anion (negatively charged) with charge is under acceleration. According to the equal to excess electrons = 18 – 16 = 2. electromagnetic theory of Maxwell, charged Symbol is . particles when accelerated should emit Note : Before using the notation AZ X, electromagnetic radiation (This feature does find out whether the species is a neutral not exist for planets since they are uncharged). atom, a cation or an anion. If it is a Therefore, an electron in an orbit will emit neutral atom, equation (2.3) is valid, i.e., radiation, the energy carried by radiation number of protons = number of electrons comes from electronic motion. The orbit will = atomic number. If the species is an thus continue to shrink. Calculations show ion, determine whether the number of that it should take an electron only 10–8 s protons are larger (cation, positive ion) to spiral into the nucleus. But this does or smaller (anion, negative ion) than the number of electrons. Number of neutrons not happen. Thus, the Rutherford model is always given by A–Z, whether the cannot explain the stability of an atom. species is neutral or ion. If the motion of an electron is described on the basis of the classical mechanics and 2.2.5 Drawbacks of Rutherford Model electromagnetic theory, you may ask that As you have learnt above, Rutherford nuclear since the motion of electrons in orbits is model of an atom is like a small scale solar leading to the instability of the atom, then system with the nucleus playing the role why not consider electrons as stationary * Classical mechanics is a theoretical science based on Newton’s laws of motion. It specifies the laws of motion of macroscopic objects. Reprint 2025-26 structure of atom 37 around the nucleus. If the electrons were was developed in the early 1870’s by James stationary, electrostatic attraction between Clerk Maxwell, which was experimentally the dense nucleus and the electrons would confirmed later by Heinrich Hertz. Here, we pull the electrons toward the nucleus to will learn some facts about electromagnetic form a miniature version of Thomson’s model radiations. of atom. James Maxwell (1870) was the first to Another serious drawback of the give a comprehensive explanation about the Rutherford model is that it says nothing interaction between the charged bodies and about distribution of the electrons around the the behaviour of electrical and magnetic nucleus and the energies of these electrons. fields on macroscopic level. He suggested