11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 273 Reprint 2025-26 Physics Chapter Eleven DUAL NATURE OF RADIATION AND MATTER 11.1 INTRODUCTION The Maxwell’s equations of electromagnetism and Hertz experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light. Towards the same period at the end of 19th century, experimental investigations on conduction of electricity (electric discharge) through gases at low pressure in a discharge tube led to many historic discoveries. The discovery of X-rays by Roentgen in 1895, and of electron by J. J. Thomson in 1897, were important milestones in the understanding of atomic structure. It was found that at sufficiently low pressure of about 0.001 mm of mercury column, a discharge took place between the two electrodes on applying the electric field to the gas in the discharge tube. A fluorescent glow appeared on the glass opposite to cathode. The colour of glow of the glass depended on the type of glass, it being yellowish-green for soda glass. The cause of this fluorescence was attributed to the radiation which appeared to be coming from the cathode. These cathode rays were discovered, in 1870, by William Crookes who later, in 1879, suggested that these rays consisted of streams of fast moving negatively charged particles. The British physicist J. J. Thomson (1856-1940) confirmed this hypothesis. By applying mutually perpendicular electric and magnetic fields across the discharge 274 tube, J. J. Thomson was the first to determine experimentally the speed Reprint 2025-26 Dual Nature of Radiation and Matter and the specific charge [charge to mass ratio (e/m)] of the cathode ray particles. They were found to travel with speeds ranging from about 0.1 to 0.2 times the speed of light (3 ×108 m/s). The presently accepted value of e/m is 1.76 × 1011 C/kg. Further, the value of e/m was found to be independent of the nature of the material/metal used as the cathode (emitter), or the gas introduced in the discharge tube. This observation suggested the universality of the cathode ray particles. Around the same time, in 1887, it was found that certain metals, when irradiated by ultraviolet light, emitted negatively charged particles having small speeds. Also, certain metals when heated to a high temperature were found to emit negatively charged particles. The value of e/m of these particles was found to be the same as that for cathode ray particles. These observations thus established that all these particles, although produced under different conditions, were identical in nature. J. J. Thomson, in 1897, named these particles as electrons, and suggested that they were fundamental, universal constituents of matter. For his epoch-making discovery of electron, through his theoretical and experimental investigations on conduction of electricity by gasses, he was awarded the Nobel Prize in Physics in 1906. In 1913, the American physicist R. A. Millikan (1868-1953) performed the pioneering oil-drop experiment for the precise measurement of the charge on an electron. He found that the charge on an oil-droplet was always an integral multiple of an elementary charge, 1.602 × 10–19 C. Millikan’s experiment established that electric charge is quantised. From the values of charge (e) and specific charge (e/m), the mass (m) of the electron could be determined.

11.11 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). 289 Reprint 2025-26 Physics Chapter Twelve ATOMS 12.1 INTRODUCTION By the nineteenth century, enough evidence had accumulated in favour of atomic hypothesis of matter. In 1897, the experiments on electric discharge through gases carried out by the English physicist J. J. Thomson (1856 – 1940) revealed that atoms of different elements contain negatively charged constituents (electrons) that are identical for all atoms. However, atoms on a whole are electrically neutral. Therefore, an atom must also contain some positive charge to neutralise the negative charge of the electrons. But what is the arrangement of the positive charge and the electrons inside the atom? In other words, what is the structure of an atom? The first model of atom was proposed by J. J. Thomson in 1898. According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. This model was picturesquely called plum pudding model of the atom. However subsequent studies on atoms, as described in this chapter, showed that the distribution of the electrons and positive charges are very different from that proposed in this model. We know that condensed matter (solids and liquids) and dense gases at all temperatures emit electromagnetic radiation in which a continuous distribution of several wavelengths is present, though with different 290 intensities. This radiation is considered to be due to oscillations of atoms Reprint 2025-26 Atoms and molecules, governed by the interaction of each atom or molecule with its neighbours. In contrast, light emitted from rarefied gases heated in a flame, or excited electrically in a glow tube such as the familiar neon sign or mercury vapour light has only certain discrete wavelengths. The spectrum appears as a series of bright lines. In such gases, the average spacing between atoms is large. Hence, the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules. In the early nineteenth century it was also established that each element is associated with a characteristic spectrum of radiation, for example, hydrogen always gives a set of lines with fixed relative position between the lines. ErnstErnstErnstErnstErnst RutherfordRutherfordRutherfordRutherfordRutherford (1871(1871(1871(1871(1871 ––––– This fact suggested an intimate relationship between the 1937)1937)1937)1937)1937) New Zealand born, internal structure of an atom and the spectrum of British physicist who did radiation emitted by it. In 1885, Johann Jakob Balmer pioneering work on ERNST (1825 – 1898) obtained a simple empirical formula which radioactive radiation. He gave the wavelengths of a group of lines emitted by atomic discovered alpha-rays and hydrogen. Since hydrogen is simplest of the elements beta-rays. Along with known, we shall consider its spectrum in detail in this Federick Soddy, he created chapter. the modern theory of Ernst Rutherford (1871–1937), a former research radioactivity. He studied student of J. J. Thomson, was engaged in experiments on the ‘emanation’ of thorium α-particles emitted by some radioactive elements. In 1906, and discovered a new noble RUTHERFORDhe proposed a classic experiment of scattering of these gas, an isotope of radon, α-particles by atoms to investigate the atomic structure. now known as thoron. By This experiment was later performed around 1911 by Hans scattering alpha-rays from the metal foils, he (1871Geiger (1882–1945) and Ernst Marsden (1889–1970, who discovered the atomic –was 20 year-old student and had not yet earned his nucleus and proposed the bachelor’s degree). The details are discussed in Section planetary model of the 12.2. The explanation of the results led to the birth of atom. He also estimated theRutherford’s planetary model of atom (also called the 1937) approximate size of the nuclear model of the atom). According to this the entire nucleus. positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus with electrons revolving around the nucleus just as planets revolve around the sun. Rutherford’s nuclear model was a major step towards how we see the atom today. However, it could not explain why atoms emit light of only discrete wavelengths. How could an atom as simple as hydrogen, consisting of a single electron and a single proton, emit a complex spectrum of specific wavelengths? In the classical picture of an atom, the electron revolves round the nucleus much like the way a planet revolves round the sun. However, we shall see that there are some serious difficulties in accepting such a model. 12.212.212.212.212.2 ALPHALPHALPHALPHA-PARLPHA ARARARTICLEAR TICLETICLETICLETICLE SSSCASSCACACATTERINGCATTERINGTTERINGTTERINGTTERING ANDANDANDANDAND RUTHERFORDUTHERFORDUTHERFORDUTHERFORD’SUTHERFORD NNNUCLEARNNUCLEARUCLEARUCLEARUCLEAR MODELODELODELODELODEL OFOFOFOFOF ATOMTOMTOMTOMTOM At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden performed some experiments. In one of their experiments, as shown in 291 Reprint 2025-26 Physics Fig. 12.1, they directed a beam of 5.5 MeV a-particles emitted from a 21483Bi radioactive source at a thin metal foil made of gold. Figure 12.2 shows a schematic diagram of this experiment. Alpha-particles emitted by a 21483Bi radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10–7 m. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. These flashes may be viewed through a microscope and the FIGURE 12.1 Geiger-Marsden scattering experiment. distribution of the number of scattered The entire apparatus is placed in a vacuum chamber particles may be studied as a function (not shown in this figure). of angle of scattering. FIGURE 12.2 Schematic arrangement of the Geiger-Marsden experiment. A typical graph of the total number of a-particles scattered at different angles, in a given interval of time, is shown in Fig. 12.3. The dots in this figure represent the data points and the solid curve is the theoretical prediction based on the assumption that the target atom has a small, dense, positively charged nucleus. Many of the a-particles pass through the foil. It means that they do not suffer any collisions. Only about 0.14% of the incident a-particles scatter by more than 1°; and about 1 in 8000 deflect by more than 90°. Rutherford argued that, to deflect the a-particle 292 backwards, it must experience a large repulsive force. This force could Reprint 2025-26 Atoms be provided if the greater part of the mass of the atom and its positive charge were concentrated tightly at its centre. Then the incoming α-particle could get very close to the positive charge without penetrating it, and such a close encounter would result in a large deflection. This agreement supported the hypothesis of the nuclear atom. This is why Rutherford is credited with the discovery of the nucleus. In Rutherford’s nuclear model of the atom, the entire positive charge and most of the mass of the atom are concentrated in the nucleus with the electrons some distance away. The electrons would be moving in orbits FIGUREFIGUREFIGUREFIGUREFIGURE 12.312.312.312.312.3 Experimental data points (shown byabout the nucleus just as the planets dots) on scattering of α-particles by a thin foil at do around the sun. Rutherford’s different angles obtained by Geiger and Marsden experiments suggested the size of using the setup shown in Figs. 12.1 and the nucleus to be about 10–15 m to 12.2. Rutherford’s nuclear model predicts the solid 10–14 m. From kinetic theory, the size curve which is seen to be in good agreement with of an atom was known to be 10–10 m, experiment. about 10,000 to 100,000 times larger than the size of the nucleus. Thus, the electrons would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the size of the nucleus itself. Thus, most of an atom is empty space. With the atom being largely empty space, it is easy to see why most α-particles go right through a thin metal foil. However, when α-particle happens to come near a nucleus, the intense electric field there scatters it through a large angle. The atomic electrons, being so light, do not appreciably affect the α-particles. The scattering data shown in Fig. 12.3 can be analysed by employing Rutherford’s nuclear model of the atom. As the gold foil is very thin, it can be assumed that α-particles will suffer not more than one scattering during their passage through it. Therefore, computation of the trajectory of an alpha-particle scattered by a single nucleus is enough. Alpha- particles are nuclei of helium atoms and, therefore, carry two units, 2e, of positive charge and have the mass of the helium atom. The charge of the gold nucleus is Ze, where Z is the atomic number of the atom; for gold Z = 79. Since the nucleus of gold is about 50 times heavier than an α-particle, it is reasonable to assume that it remains stationary throughout the scattering process. Under these assumptions, the trajectory of an alpha-particle can be computed employing Newton’s second law of motion and the Coulomb’s law for electrostatic force of repulsion between the alpha-particle and the positively charged nucleus. 293 Reprint 2025-26 Physics The magnitude of this force is 1 (2e )( Ze ) F = 2 (12.1) 4 πε0 r where r is the distance between the a-particle and the nucleus. The force is directed along the line joining the a-particle and the nucleus. The magnitude and direction of the force on an a-particle continuously changes as it approaches the nucleus and recedes away from it. 12.2.1 Alpha-particle trajectory The trajectory traced by an a-particle depends on the impact parameter, b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the a-particle from the centre of the nucleus (Fig. 12.4). A given beam of a-particles has a distribution of impact parameters b, so that the beam is scattered in various directions with different probabilities (Fig. 12.4). (In a beam, all particles have nearly same kinetic energy.) It is seen that an a-particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the a-particle rebounds back (q @ p). For a large impact parameter, the a-particle goes nearly undeviated and has a small deflection (q @ 0). FIGURE 12.4 Trajectory of a-particles in the The fact that only a small fraction of the coulomb field of a target nucleus. The impact number of incident particles rebound back parameter, b and scattering angle q indicates that the number of a-particles are also depicted. undergoing head on collision is small. This, in turn, implies that the mass and positive charge of the atom is concentrated in a small volume. Rutherford scattering therefore, is a powerful way to determine an upper limit to the size of the nucleus. Example 12.1 In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10–15 m) is analogous to the sun about which the electron move in orbit (radius » 10–10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is? The radius of earth’s orbit is about 1.5 ´ 1011 m. The radius of sun is taken as 7 ´ 108 m. Solution The ratio of the radius of electron’s orbit to the radius of nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s orbit is 105 times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 105 times larger than the radius of the sun, the radius of the earth’s orbit would be 105 ´ 7 ´ 108 m = 12.1 7 ´ 1013 m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty EXAMPLE 294 space than our solar system does. Reprint 2025-26 Atoms Example 12.2 In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV a-particle before it comes momentarily to rest and reverses its direction? Solution The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an a-particle and a gold nucleus is conserved. The system’s initial mechanical energy is Ei, before the particle and nucleus interact, and it is equal to its mechanical energy Ef when the a-particle momentarily stops. The initial energy Ei is just the kinetic energy K of the incoming a- particle. The final energy Ef is just the electric potential energy U of the system. The potential energy U can be calculated from Eq. (12.1). Let d be the centre-to-centre distance between the a-particle and the gold nucleus when the a-particle is at its stopping point. Then we can write the conservation of energy Ei = Ef as 1 (2e )( Ze ) 2 Ze 2 K = = 4 πε0 d 4 πε0 d Thus the distance of closest approach d is given by 2 Ze 2 d = 4 πε0 K The maximum kinetic energy found in a-particles of natural origin is 7.7 MeV or 1.2 × 10–12 J. Since 1/4pe0 = 9.0 × 109 N m2/C2. Therefore with e = 1.6 × 10–19 C, we have, (2)(9.0 × 10 9 Nm 2 / C 2 )(1.6 × 10 –19 C )2 Z d = − 12 1.2 × 10 J = 3.84 × 10–16 Z m The atomic number of foil material gold is Z = 79, so that d (Au) = 3.0 × 10–14 m = 30 fm. (1 fm (i.e. fermi) = 10–15 m.) The radius of gold nucleus is, therefore, less than 3.0 × 10–14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the EXAMPLE distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the a-particle reverses its motion without ever actually touching the gold nucleus. 12.2 12.2.2 Electron orbits The Rutherford nuclear model of the atom which involves classical concepts, pictures the atom as an electrically neutral sphere consisting of a very small, massive and positively charged nucleus at the centre surrounded by the revolving electrons in their respective dynamically stable orbits. The electrostatic force of attraction, Fe between the revolving electrons and the nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus, for a dynamically stable orbit in a hydrogen atom Fe = Fc 1 e 2 mv 2 = (12.2) 295 r 4 πε0 r 2 Reprint 2025-26 Physics Thus the relation between the orbit radius and the electron velocity is e 2 r = 2 (12.3) 4 πε0mv The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are 1 2 e 2 e 2 K = mv = and U = − 2 8 πε0r 4 πε0r (The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus the total energy E of the electron in a hydrogen atom is e 2 e 2 E = K + U = − 8 πε0r 4 πε0r e 2 = − (12.4) 8 πε0r The total energy of the electron is negative. This implies the fact that the electron is bound to the nucleus. If E were positive, an electron will not follow a closed orbit around the nucleus. Example 12.3 It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom. Solution Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10–19 J = –2.2 ×10–18 J. Thus from Eq. (12.4), we have e 2 − 18 E = − = −2.2 × 10 J 8 πε0r This gives the orbital radius e 2 (9 × 10 9 N m 2 /C 2 )(1.6 × 10 −19 C)2 r = − = − −18 8 πε0 E (2)(–2.2 × 10 J) = 5.3 × 10–11 m. 12.3 The velocity of the revolving electron can be computed from Eq. (12.3) with m = 9.1 × 10–31 kg, e 6 v = = 2.2 × 10 m/s. EXAMPLE 4 πε0 mr 12.3 ATOMIC SPECTRA As mentioned in Section 12.1, each element has a characteristic spectrum of radiation, which it emits. When an atomic gas or vapour is excited at low pressure, usually by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wavelengths 296 only. A spectrum of this kind is termed as emission line spectrum and it Reprint 2025-26 Atoms consists of bright lines on a dark background. The spectrum emitted by atomic hydrogen is shown in Fig. 12.5. Study of emission line spectra of a material can therefore serve as a type of “fingerprint” for identification of the gas. When white light passes through a gas and we analyse the transmitted light using a spectrometer we find some dark lines in the FIGURE 12.5 Emission lines in the spectrum of hydrogen. spectrum. These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas. This is called the absorption spectrum of the material of the gas.