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

7.5 ACCELERATION DUE TO GRAVITY OF THE EARTH 4π 3 density and hence its mass is M E = R E ρ 3 The earth can be imagined to be a sphere made where ME is the mass of the earth RE is its radiusof a large number of concentric spherical shells and ρ is the density. On the other hand thewith the smallest one at the centre and the largest one at its surface. A point outside the 4π 3 ρr and mass of the sphere Mr of radius r isearth is obviously outside all the shells. Thus, 3 Reprint 2025-26 GRAVITATION 133 hence its distance from the centre of the earth is (RE + h ). If F (h) denoted the magnitude of the force on the point mass m , we get from G m M E Eq. (7.5) : = 3 r (7.10) R E If the mass m is situated on the surface of GM E m F (h ) = earth, then r = RE and the gravitational force on ( R E + h )2 (7.13) it is, from Eq. (7.10) The acceleration experienced by the point M E m F = G 2 (7.11) mass is F (h )/ m ≡ g (h ) and we get R E The acceleration experienced by the mass F (h ) GM E . g (h ) = = (7.14)m, which is usually denoted by the symbol g is m ( R E + h )2 related to F by Newton’s 2nd law by relation This is clearly less than the value of g on the F = mg. Thus GM E . g = surface of earth : GM F For h << R E , we can E R E2 g = = 2 (7.12) m R E expand the RHS of Eq. (7.14) : E Acceleration g is readily measurable. RE is a g (h ) = 2 GM 2 = g (1 + h / R E )−2known quantity. The measurement of G by R E (1 + h / R E ) Cavendish’s experiment (or otherwise), combined h << 1 , using binomial expression,with knowledge of g and RE enables one to For R E estimate ME from Eq. (7.12). This is the reason  2h why there is a popular statement regarding g (h ) ≅ g 1 − . (7.15)Cavendish : “Cavendish weighed the earth”.  RE  7.6 ACCELERATION DUE TO GRAVITY BELOW Equation (7.15) thus tells us that for small AND ABOVE THE SURFACE OF EARTH heights h above the value of g decreases by a factor (1 − 2h / R E ).Consider a point mass m at a height h above the Now, consider a point mass m at a depthsurface of the earth as shown in Fig. 7.8(a). The d below the surface of the earth (Fig. 7.8(b)),radius of the earth is denoted by RE . Since this so that its distance from the centre of thepoint is outside the earth, earth is ( R E − d ) as shown in the figure. The earth can be thought of as being composed of a smaller sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer shell of thickness d is zero because the result quoted in the previous section. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence according to the result quoted earlier, the force due to this smaller sphere is just as if the entire mass of the smaller sphere is concentrated at the centre. If Ms is the mass of the smaller sphere, then, Ms/ME = ( RE – d)3 / RE3 ( 7.16) Since mass of a sphere is proportional to be Fig. 7.8 (a) g at a height h above the surface of the cube of its radius. earth. Reprint 2025-26 134 PHYSICS close to the surface of earth, at distances from the surface much smaller than the radius of the earth. In such cases, the force of gravity is practically a constant equal to mg, directed towards the centre of the earth. If we consider a Ms ME point at a height h1 from the surface of the earth and another point vertically above it at a height h2 from the surface, the work done in lifting the particle of mass m from the first to the second position is denoted by W12 Fig. 7.8 (b) g at a depth d. In this case only the smaller W12 = Force × displacement = mg (h2 – h1) (7.20) sphere of radius (RE–d) contributes to g. Thus the force on the point mass is If we associate a potential energy W(h) at a point at a height h above the surface such that F (d) = G Ms m / (RE – d ) 2 (7.17) W(h) = mgh + Wo (7.21) Substituting for Ms from above , we get (where Wo = constant) ; F (d) = G ME m ( RE – d ) / RE 3 (7.18) then it is clear that and hence the acceleration due to gravity at W12 = W(h2) – W(h1) (7.22) a depth d, The work done in moving the particle is just the difference of potential energy between its F ( d ) final and initial positions.Observe that the g(d) = is m constant Wo cancels out in Eq. (7.22). Setting h = 0 in the last equation, we get W ( h = 0 ) = Wo. F (d ) GM E g ( d ) = = 3 ( R E − d ) . h = 0 means points on the surface of the earth. m R E Thus, Wo is the potential energy on the surface of the earth. R E − d = g = g (1 − d / R E ) (7.19) If we consider points at arbitrary distance R E from the surface of the earth, the result just Thus, as we go down below earth’s surface, derived is not valid since the assumption that the acceleration due gravity decreases by a factor the gravitational force mg is a constant is no (1 − d / R E ). The remarkable thing about longer valid. However, from our discussion we know that a point outside the earth, the force of acceleration due to earth’s gravity is that it is gravitation on a particle directed towards the maximum on its surface decreasing whether you centre of the earth is go up or down. G ME m F = 2 (7.23)7.7 GRAVITATIONAL POTENTIAL ENERGY r where ME = mass of earth, m = mass of theWe had discussed earlier the notion of potential particle and r its distance from the centre of theenergy as being the energy stored in the body at earth. If we now calculate the work done inits given position. If the position of the particle lifting a particle from r = r1 to r = r2 (r2 > r1) alongchanges on account of forces acting on it, then a vertical path, we get instead of Eq. (7.20) the change in its potential energy is just the amount of work done on the body by the force. r2 G M m W12 2 d rAs we had discussed earlier, forces for which the =∫r1 r work done is independent of the path are the conservative forces.  1 1  = − G M E m − (7.24) The force of gravity is a conservative force  r2 r1  and we can calculate the potential energy of a In place of Eq. (7.21), we can thus associate body arising out of this force, called the a potential energy W(r) at a distance r, such that gravitational potential energy. Consider points Reprint 2025-26 GRAVITATION 135 G M E m W (r ) =− + W1 , (7.25) r valid for r > R , so that once again W12 = W(r2) – W(r1). Setting r = infinity in the last equation, we get W ( r = infinity ) = W1 . Thus, W1 is the potential energy at infinity. One should note that only the difference of potential energy between two points has a definite meaning from Eqs. (7.22) and (7.24). One conventionally sets W1 equal to zero, so that the potential energy at a point is just the amount of work done in displacing the particle from infinity to that point. We have calculated the potential energy at a point of a particle due to gravitational forces Fig. 7.9 on it due to the earth and it is proportional to the mass of the particle. The gravitational The gravitational potential at the centre of potential due to the gravitational force of the the square r = 2 l/2 is ( )earth is defined as the potential energy of a particle of unit mass at that point. From the G m U (r ) = − 4 2 . ⊳earlier discussion, we learn that the gravitational l potential energy associated with two particles of masses m1 and m2 separated by distance by a 7.8 ESCAPE SPEED distance r is given by If a stone is thrown by hand, we see it falls back Gm 1m 2 to the earth. Of course using machines we can V = – (if we choose V = 0 as r →∞) r shoot an object with much greater speeds and It should be noted that an isolated system of with greater and greater initial speed, the object particles will have the total potential energy that scales higher and higher heights. A natural equals the sum of energies (given by the above query that arises in our mind is the following: equation) for all possible pairs of its constituent ‘can we throw an object with such high initial particles. This is an example of the application speeds that it does not fall back to the earth?’ of the superposition principle. The principle of conservation of energy helps us to answer this question. Suppose the object ⊳ Example 7.3 Find the potential energy of did reach infinity and that its speed there was a system of four particles placed at the Vf. The energy of an object is the sum of potential vertices of a square of side l. Also obtain and kinetic energy. As before W1 denotes that the potential at the centre of the square. gravitational potential energy of the object at infinity. The total energy of the projectile at infinity then is Answer Consider four masses each of mass m at the corners of a square of side l; See Fig. 7.9. 2 mV fWe have four mass pairs at distance l and two E ( ∞=) W1 + (7.26) 2 diagonal pairs at distance 2 l If the object was thrown initially with a speed Hence, Vi from a point at a distance (h+RE) from the G m 2 G m 2 centre of the earth (RE = radius of the earth), its W (r ) = − 4 − 2 l 2 l energy initially was 2 2 1 2 GmM E 2 G m  1  G m E (h + R E ) = mVi – + W1 (7.27) 5.41 = − 2 (h + R E ) 2 +  = − l  l  2  Reprint 2025-26 136 PHYSICS By the principle of energy conservation ⊳ Eqs. (7.26) and (7.27) must be equal. Hence Example 7.4 Two uniform solid spheres 2 2 of equal radii R, but mass M and 4 M have mV f mVi GmM E a centre to centre separation 6 R, as shown – = (7.28) 2 (h + R E ) 2 in Fig. 7.10. The two spheres are held fixed. The R.H.S. is a positive quantity with a A projectile of mass m is projected from the minimum value zero hence so must be the L.H.S. surface of the sphere of mass M directly Thus, an object can reach infinity as long as Vi towards the centre of the second sphere. is such that Obtain an expression for the minimum speed v of the projectile so that it reaches mVi 2 GmM E the surface of the second sphere. – ≥ 0 (7.29) 2 (h + R E ) The minimum value of Vi corresponds to the case when the L.H.S. of Eq. (7.29) equals zero. Thus, the minimum speed required for an object to reach infinity (i.e. escape from the earth) Fig. 7.10corresponds to 1 2 GmM E Answer The projectile is acted upon by two m = (7.30) mutually opposing gravitational forces of the two ( Vi )min 2 h + R E spheres. The neutral point N (see Fig. 7.10) is defined as the position where the two forces If the object is thrown from the surface of cancel each other exactly. If ON = r, we have the earth, h = 0, and we get G M m 4 G M m = 2GM E r 2 (6 R −r )2 (Vi )min = (7.31) (6R – r)2 = 4r2 R E 6R – r = ±2r r = 2R or – 6R. 2 The neutral point r = – 6R does not concern Using the relation g = GM E / R E , we get us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a speed (Vi )min = 2 gR E (7.32) which would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the surface Using the value of g and RE, numerically of M is (Vi)min≈11.2 km/s. This is called the escape 1 2 G M m 4 G M mspeed, sometimes loosely called the escape E i = m v − − . velocity. 2 R 5 R Equation (7.32) applies equally well to an At the neutral point N, the speed approaches object thrown from the surface of the moon with zero. The mechanical energy at N is purely g replaced by the acceleration due to Moon’s potential. gravity on its surface and rE replaced by the G M m 4 G M m − EN = − .radius of the moon. Both are smaller than their 2 R 4 R values on earth and the escape speed for the From the principle of conservation of moon turns out to be 2.3 km/s, about five times mechanical energy smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the 1 2 GM 4GM GM GMsurface of the moon having velocities larger than v − − = − − this will escape the gravitational pull of the 2 R 5 R 2R R moon. or Reprint 2025-26 GRAVITATION 137 + h) with speed V. Its 2 2 G M  4 1  traverses a distance 2π(RE v = − time period T therefore is R  5 2  2π( R E + h ) 2π( R E + h )3 / 2 T = = (7.37)  3 G M 1/2 V ⊳ G M E v =  5 R  on substitution of value of V from Eq. (7.35). A point to note is that the speed of the projectile Squaring both sides of Eq. (7.37), we get is zero at N, but is nonzero when it strikes the 2 T = k ( RE + h)3 (where k = 4 π2 / GME) (7.38)heavier sphere 4 M. The calculation of this speed is left as an exercise to the students. which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a 7.9 EARTH SATELLITES satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (7.38). Earth satellites are objects which revolve around Hence, for such satellites, T is To, where the earth. Their motion is very similar to the motion of planets around the Sun and hence T 0 = 2π R E / g (7.39) Kepler’s laws of planetary motion are equally If we substitute the numerical values applicable to them. In particular, their orbits g ≃ 9.8 m s-2 and RE = 6400 km., we get around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near 6.4 × 10 6 T 0 = 2π s circular orbit with a time period of approximately 9.8

2.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?