Structure of Atom PDF

26 CHEMISTRY UNIT 2 STRUCTURE OF ATOM The rich diversity of chemical be...

26 CHEMISTRY UNIT 2 STRUCTURE OF ATOM The rich diversity of chemical behaviour of different elements can be traced to the differ ences in the internal 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 time electron, proton and neutron and of early Indian and Greek philosophers (400 B.C.) who their characteristics; were of the view that atoms are the fundamental building describe Thomson, Rutherford blocks of matter. According to them, the continued and Bohr atomic models; subdivisions of matter would ultimately yield atoms which understand the important would not be further divisible. The word ‘atom’ has been features of the quantum derived from the Greek word ‘a-tomio’ which means ‘uncut- mechanical model of atom; able’ or ‘non-divisible’. These earlier ideas were mere speculations and there was no way to test them understand nature of experimentally. These ideas remained dormant for a very electromagnetic radiation and Planck’s quantum theory; long time and were revived again by scientists in the nineteenth century. explain the photoelectric effect The atomic theory of matter was first proposed on a and describe features of atomic firm scientific basis by John Dalton, a British school spectra; teacher in 1808. His theory, called Dalton’s atomic state the de Broglie relation and theory, regarded the atom as the ultimate particle of Heisenberg uncertainty principle; matter (Unit 1). define an atomic orbital in terms In this unit we start with the experimental of quantum numbers; observations made by scientists towards the end of state aufbau principle, Pauli nineteenth and beginning of twentieth century. These exclusion principle and Hund’s established that atoms can be further divided into sub- rule of maximum multiplicity; atomic particles, i.e., electrons, protons and neutrons— a concept very different from that of Dalton. The major write the electronic configurations problems before the scientists at that time were: of atoms. to account for the stability of atom after the discovery of sub-atomic particles, to compare the behaviour of one element from other in terms of both physical and chemical properties, 2015-16 STRUCTURE OF ATOM 27 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. 2.1 SUB-ATOMIC PARTICLES Dalton’s atomic theory was able to explain the law of conservation of mass, law of Fig. 2.1(a) A cathode ray discharge tube constant composition and law of multiple proportion very successfully. However, it failed stream of particles moving in the tube from to explain the results of many experiments, the negative electrode (cathode) to the positive for example, it was known that substances electrode (anode). These were called cathode like glass or ebonite when rubbed with silk or rays or cathode ray particles. The flow of fur generate electricity. Many different kinds current from cathode to anode was further of sub-atomic particles were discovered in the checked by making a hole in the anode and twentieth century. However, in this section coating the tube behind anode with we will talk about only two particles, namely phosphorescent material zinc sulphide. When electron and proton. these rays, after passing through anode, strike the zinc sulphide coating, a bright spot on 2.1.1 Discovery of Electron the coating is developed(same thing happens In 1830, Michael Faraday showed that if in a television set) [Fig. 2.1(b)]. electricity is passed through a solution of an electrolyte, chemical reactions occurred at the electrodes, which resulted in the liberation and deposition of matter at the electrodes. He formulated certain laws which you will study in class XII. These results suggested the particulate nature of electricity. An insight into the structure of atom was obtained from the experiments on electrical discharge through gases. Before we discuss these results we need to keep in mind a basic Fig. 2.1(b) A cathode ray discharge tube with rule regarding the behaviour of charged perforated anode particles : “Like charges repel each other and The results of these experiments are unlike charges attract each other”. summarised below. In mid 1850s many scientists mainly (i) The cathode rays start from cathode and Faraday began to study electrical discharge move towards the anode. in partially evacuated tubes, known as cathode ray discharge tubes. It is depicted (ii) These rays themselves are not visible but in Fig. 2.1. A cathode ray tube is made of glass their behaviour can be observed with the containing two thin pieces of metal, called help of certain kind of materials electrodes, sealed in it. The electrical (fluorescent or phosphorescent) which discharge through the gases could be glow when hit by them. Television observed only at very low pressures and at picture tubes are cathode ray tubes and very high voltages. The pressure of different television pictures result due to gases could be adjusted by evacuation. When fluorescence on the television screen sufficiently high voltage is applied across the coated with certain fluorescent or electrodes, current starts flowing through a phosphorescent materials. 2015-16 28 CHEMISTRY (iii) In the absence of electrical or magnetic (ii) the mass of the particle — lighter the field, these rays travel in straight lines particle, greater the deflection. (Fig. 2.2). (iii) the strength of the electrical or magnetic (iv) In the presence of electrical or magnetic field — the deflection of electrons from field, the behaviour of cathode rays are its original path increases with the similar to that expected from negatively increase in the voltage across the charged particles, suggesting that the electrodes, or the strength of the cathode rays consist of negatively magnetic field. charged particles, called electrons. When only electric field is applied, the (v) The characteristics of cathode rays electrons deviate from their path and hit the (electrons) do not depend upon the cathode ray tube at point A. Similarly when material of electrodes and the nature of only magnetic field is applied, electron strikes the gas present in the cathode ray tube. the cathode ray tube at point C. By carefully Thus, we can conclude that electrons are balancing the electrical and magnetic field basic constituent of all the atoms. strength, it is possible to bring back the electron to the path followed as in the absence 2.1.2 Charge to Mass Ratio of Electron of electric or magnetic field and they hit the In 1897, British physicist J.J. Thomson screen at point B. By carrying out accurate measured the ratio of electrical charge (e) to measurements on the amount of deflections the mass of electron (me ) by using cathode observed by the electrons on the electric field ray tube and applying electrical and magnetic strength or magnetic field strength, Thomson field perpendicular to each other as well as to was able to determine the value of e/me as: the path of electrons (Fig. 2.2). Thomson argued that the amount of deviation of the e me = 1.758820 × 10 C kg (2.1) 11 –1 particles from their path in the presence of electrical or magnetic field depends upon: Where me is the mass of the electron in kg (i) the magnitude of the negative charge on and e is the magnitude of the charge on the the particle, greater the magnitude of the charge on the particle, greater is the electron in coulomb (C). Since electrons interaction with the electric or magnetic are negatively charged, the charge on electron field and thus greater is the deflection. is –e. Fig. 2.2 The apparatus to deter mine the charge to the mass ratio of electron C:\Chemistry XI\Unit-2\Unit-2(2)-Lay-3(reprint).pmd 27.7.6, 16.10.6 (Reprint) 2015-16 STRUCTURE OF ATOM 29 2.1.3 Charge on the Electron Millikan’s Oil Drop Method R.A. Millikan (1868-1953) devised a method In this method, oil droplets in the form of known as oil drop experiment (1906-14), to mist, pr oduced by the atomiser, were allowed determine the charge on the electrons. He to enter thr ough a tiny hole in the upper plate found that the charge on the electron to be of electrical condenser. The downward motion – 1.6 × 10–19 C. The present accepted value of of these dr oplets was viewed through the electrical charge is – 1.6022 × 10–19 C. The telescope, equipped with a micrometer eye mass of the electron (me) was determined by piece. By measuring the rate of fall of these combining these results with Thomson’s value droplets, Millikan was able to measure the of e/me ratio. mass of oil dr oplets.The air inside the chamber was ionized by passing a beam of e 1.6022 × 10–19 C X-rays through it. The electrical charge on me = = these oil dr oplets was acquired by collisions e/ m e 1.758820 × 1011C kg –1 with gaseous ions. The fall of these charged = 9.1094×10–31 kg (2.2) oil droplets can be retar ded, accelerated or made stationary depending upon the charge 2.1.4 Discovery of Protons and Neutrons on the droplets and the polarity and strength Electrical discharge carried out in the of the voltage applied to the plate. By carefully modified cathode ray tube led to the discovery measuring the ef fects of electrical field of particles carrying positive charge, also strength on the motion of oil dr oplets, known as canal rays. The characteristics of Millikan concluded that the magnitude of these positively charged particles are listed electrical charge, q, on the dr oplets is always an integral multiple of the electrical charge, below. e, that is, q = n e, where n = 1, 2, 3.... (i) unlike cathode rays, the positively charged particles depend upon the nature of gas present in the cathode ray tube. These 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 Fig. 2.3 The Millikan oil dr op apparatus for measuring charge ‘e’. In chamber, the rays. forces acting on oil drop ar e : The smallest and lightest positive ion was gravitational, electrostatic due to obtained from hydrogen and was called electrical field and a viscous drag force proton. This positively charged particle was when the oil drop is moving. characterised in 1919. Later, a need was felt properties of these fundamental particles are for the presence of electrically neutral particle given in Table 2.1. as one of the constituent of atom. These particles were discovered by Chadwick (1932) 2.2 ATOMIC MODELS by bombarding a thin sheet of beryllium by Observations obtained from the experiments α-particles. When electrically neutral particles mentioned in the previous sections have having a mass slightly greater than that of suggested that Dalton’s indivisible atom is the protons was emitted. He named these composed of sub-atomic particles carrying particles as neutr ons. The important positive and negative charges. Different 2015-16 30 CHEMISTRY Table 2.1 Properties of Fundamental Particles atomic models were proposed to explain the distributions of these charged particles in an In the later half of the nineteenth century atom. Although some of these models were different kinds of rays were discovered, not able to explain the stability of atoms, two besides those mentioned earlier. Wilhalm of these models, proposed by J. J. Thomson Röentgen (1845-1923) in 1895 showed and Ernest Rutherford are discussed below. that when electrons strike a material in 2.2.1 Thomson Model of Atom the cathode ray tubes, produce rays which can cause fluorescence in the J. J. Thomson, in 1898, proposed that an fluorescent materials placed outside the atom possesses a spherical shape (radius cathode ray tubes. Since Röentgen did approximately 10–10 m) in which the positive not know the nature of the radiation, he charge is uniformly distributed. The electrons named them X-rays and the name is still are embedded into it in such a manner as to carried on. It was noticed that X-rays are give the most stable electrostatic arrangement (Fig. 2.4). Many different names are given to produced effectively when electrons this model, for example, plum pudding, strike the dense metal anode, called raisin pudding or watermelon. This model targets. These are 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 (∼0.1 nm) and possess electro-magnetic character (Section 2.3.1). Henri Becqueral (1852-1908) Fig.2.4 Thomson model of atom observed that there are certain elements can be visualised as a pudding or watermelon which emit radiation on their own and of positive charge with plums or seeds named this phenomenon as (electrons) embedded into it. An important radioactivity and the elements known feature of this model is that the mass of the as radioactive elements. This field was atom is assumed to be uniformly distributed developed by Marie Curie, Piere Curie, over the atom. Although this model was able Rutherford and Fredrick Soddy. It was to explain the overall neutrality of the atom, observed that three kinds of rays i.e., α, but was not consistent with the results of later β- and γ-rays are emitted. Rutherford experiments. Thomson was awarded Nobel found that α-rays consists of high energy Prize for physics in 1906, for his theoretical particles carrying two units of positive and experimental investigations on the charge and four unit of atomic mass. He conduction of electricity by gases. 2015-16 STRUCTURE OF ATOM 31 concluded that α- particles are helium represented in Fig. 2.5. A stream of high nuclei as when α- particles combined energy α–particles from a radioactive source with two electrons yielded helium gas. was directed at a thin foil (thickness ∼ 100 β-rays are negatively charged particles nm) of gold metal. The thin gold foil had a similar to electrons. The γ-rays are high circular fluorescent zinc sulphide screen energy radiations like X-rays, are neutral around it. Whenever α–particles struck the in nature and do not consist of particles. screen, a tiny flash of light was produced at As regards penetrating power, α-particles that point. are the least, followed by β-rays (100 The results of scattering experiment were times that of α–particles) and γ-rays quite unexpected. According to Thomson (1000 times of that α-particles). model of atom, the mass of each gold atom in the foil should have been spread evenly over 2.2.2 Rutherford’s Nuclear Model of Atom the entire atom, and α– particles had enough energy to pass directly through such a Rutherford and his students (Hans Geiger and uniform distribution of mass. It was expected Ernest Marsden) bombarded very thin gold that the particles would slow down and foil with α–particles. Rutherford’s famous change directions only by a small angles as α –particle scattering experiment is 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 nearly 180°. A. Rutherford’s scattering experiment 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 Fig.2.5 Schematic view of Rutherford’s scattering the volume occupied by the nucleus is experiment. When a beam of alpha (α) negligibly small as compared to the total particles is “shot” at a thin gold foil, most volume of the atom. The radius of the of them pass through without much effect. atom is about 10–10 m, while that of Some, however, are deflected. nucleus is 10–15 m. One can appreciate 2015-16 32 CHEMISTRY this difference in size by realising that if earlier protons and neutrons present in the a cricket ball represents a nucleus, then nucleus are collectively known as nucleons. the radius of atom would be about 5 km. The total number of nucleons is termed as On the basis of above observations and mass number (A) of the atom. conclusions, Rutherfor d proposed the mass number (A) = number of protons (Z) nuclear model of atom (after the discovery of + number of protons). 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 The composition of any atom can be in 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 14 14 resembles the solar system in which the example, 6 C and 7 N. 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 as Isotopes. In other words (according to (iii) Electrons and the nucleus are held equation 2.4), it is evident that difference together by electrostatic forces of between the isotopes is due to the presence attraction. of different number of neutrons present in 2.2.3 Atomic Number and Mass Number the nucleus. For example, considering of The presence of positive charge on the hydrogen atom again, 99.985% of hydrogen nucleus is due to the protons in the nucleus. atoms contain only one proton. This isotope 1 As established earlier, the charge on the is called protium( 1H). Rest of the percentage proton is equal but opposite to that of of hydrogen atom contains two other isotopes, electron. The number of protons present in the one containing 1 proton and 1 neutron 2 the nucleus is equal to atomic number (Z ). is called deuterium ( 1 D, 0.015%) and the For example, the number of protons in the other one possessing 1 proton and 2 neutrons 3 hydrogen nucleus is 1, in sodium atom it is is called tritium ( 1 T ). The latter isotope is 11, therefore their atomic numbers are 1 and found in trace amounts on the earth. Other 11 respectively. In order to keep the electrical examples of commonly occuring isotopes are: neutrality, the number of electrons in an carbon atoms containing 6, 7 and 8 neutrons atom is equal to the number of protons besides 6 protons ( 12 13 14 6 C, 6 C, 6 C ); chlorine (atomic number, Z ). For example, number of atoms containing 18 and 20 neutrons besides electrons in hydrogen atom and sodium atom 17 protons ( 17 35 37 Cl, 17 Cl ). are 1 and 11 respectively. Lastly an important point to mention Atomic number (Z) = number of protons in regarding isotopes is that chemical properties the nucleus of an atom of atoms are controlled by the number of electrons, which are determined by the = number of electrons number of protons in the nucleus. Number of in a nuetral atom (2.3) neutrons present in the nucleus have very While the positive charge of the nucleus little effect on the chemical properties of an is due to protons, the mass of the nucleus, element. Therefore, all the isotopes of a given due to protons and neutrons. As discussed element show same chemical behaviour. 2015-16 STRUCTURE OF ATOM 33 Problem 2.1 playing the role of the massive sun and the electrons being similar to the lighter planets. Calculate the number of protons, Further, the coulomb force (kq1q2/r2 where q1 neutrons and electrons in 80 35 Br. and q2 are the charges, r is the distance of Solution separation of the charges and k is the In this case, 80 35 Br , Z = 35, A = 80, species proportionality constant) between electron and is neutral the nucleus is mathematically similar to the Number of protons = number of electrons  m1m 2  = Z = 35 gravitational force  G. 2  where m1 and  r  Number of neutrons = 80 – 35 = 45, m 2 are the masses, r is the distance of (equation 2.4) separation of the masses and G is the gravitational constant. When classical Problem 2.2 mechanics* is applied to the solar system, The number of electrons, protons and it shows that the planets describe well-defined neutrons in a species are equal to 18, orbits around the sun. The theory can also 16 and 16 respectively. Assign the proper calculate precisely the planetary orbits and symbol to the species. these are in agreement with the experimental Solution measurements. The similarity between the The atomic number is equal to solar system and nuclear model suggests number of protons = 16. The element is that electrons should move around the nucleus sulphur (S). in well defined orbits. However, when a body is moving in an orbit, it undergoes acceleration Atomic mass number = number of protons + number of neutrons (even if the body is moving with a constant speed in an orbit, it must accelerate because = 16 + 16 = 32 of changing direction). So an electron in the Species is not neutral as the number of nuclear model describing planet like orbits is protons is not equal to electrons. It is under acceleration. According to the anion (negatively charged) with charge electromagnetic theory of Maxwell, charged equal to excess electrons = 18 – 16 = 2. particles when accelerated should emit 32 2– Symbol is 16 S. electromagnetic radiation (This feature does Note : Before using the notation X , findA Z not exist for planets since they are uncharged). out whether the species is a neutral Therefore, an electron in an orbit will emit atom, a cation or an anion. If it is a radiation, the energy carried by radiation neutral atom, equation (2.3) is valid, i.e., comes from electronic motion. The orbit will number of protons = number of electrons thus continue to shrink. Calculations show = atomic number. If the species is an ion, that it should take an electron only 10–8 s to deter mine whether the number of spiral into the nucleus. But this does not protons are larger (cation, positive ion) happen. Thus, the Rutherford model or smaller (anion, negative ion) than the cannot explain the stability of an atom. number of electrons. Number of neutrons If the motion of an electron is described on the basis of the classical mechanics and is always given by A–Z, whether the electromagnetic theory, you may ask that species is neutral or ion. since the motion of electrons in orbits is leading to the instability of the atom, then 2.2.5 Drawbacks of Rutherford Model why not consider electrons as stationary Rutherford nuclear model of an atom is like a around the nucleus. If the electrons were small scale solar system with the nucleus stationary, electrostatic attraction between * Classical mechanics is a theoretical science based on Newton’s laws of motion. It specifies the laws of motion of macroscopic objects. 2015-16 34 CHEMISTRY the dense nucleus and the electrons would 19th century when wave nature of light was pull the electrons toward the nucleus to form established. a miniature version of Thomson’s model of Maxwell was again the first to reveal that atom. light waves are associated with oscillating Another serious drawback of the electric and magnetic character (Fig. 2.6). Rutherford model is that it says nothing Although electromagnetic wave motion is about the electronic structure of atoms i.e., complex in nature, we will consider here only how the electrons are distributed around the a few simple properties. nucleus and what are the energies of these (i) The oscillating electric and magnetic electrons. fields produced by oscillating charged 2.3 DEVELOPMENTS LEADING TO THE particles are perpendicular to each other BOHR’S MODEL OF ATOM and both are perpendicular to the direction of propagation of the wave. Historically, results observed from the studies Simplified picture of electromagnetic of interactions of radiations with matter have wave is shown in Fig. 2.6. provided immense information regarding the structure of atoms and molecules. Neils Bohr utilised these results to improve upon the model proposed by Rutherf o rd. 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, and (ii) Experimental results regarding atomic spectra which can be explained only by Fig.2.6 The electric and magnetic field assuming quantized (Section 2.4) components of an electromagnetic wave. These components have the same electronic energy levels in atoms. wavelength, fr equency, speed and 2.3.1 Wave Nature of Electromagnetic amplitude, but they vibrate in two Radiation mutually perpendicular planes. James Maxwell (1870) was the first to give a (ii) Unlike sound waves or water waves, comprehensive explanation about the electromagnetic waves do not require interaction between the charged bodies and medium and can move in vacuum. the behaviour of electrical and magnetic fields (iii) It is now well established that there are on macroscopic level. He suggested that when many types of electromagnetic electrically charged particle moves under radiations, which differ from one another accelaration, alternating electrical and in wavelength (or frequency). These magnetic fields are produced and constitute what is called transmitted. These fields are transmitted in electromagnetic spectrum (Fig. 2.7). the forms of waves called electromagnetic Different regions of the spectrum are waves or electromagnetic radiation. identified by different names. Some Light is the form of radiation known from examples are: radio frequency region early days and speculation about its nature around 106 Hz, used for broadcasting; dates back to remote ancient times. In earlier microwave region around 1010 Hz used days (Newton) light was supposed to be made for radar; infrared region around 1013 Hz of particles (corpuscules). It was only in the used for heating; ultraviolet region 2015-16 STRUCTURE OF ATOM 35 around 1016Hz a component of sun’s at the same speed, i.e., 3.0 × 10 8 m s–1 radiation. The small portion around 1015 (2.997925 × 108 m s –1, to be precise). This is Hz, is what is ordinarily called visible called speed of light and is given the symbol light. It is only this part which our eyes ‘c‘. The frequency (ν ), wavelength (λ) and velocity can see (or detect). Special instruments of light (c) are related by the equation (2.5). a re required to detect non-visible c=ν λ (2.5) radiation. The other commonly used quantity (iv) Different kinds of units are used to specially in spectroscopy, is the wavenumber represent electromagnetic radiation. (ν ). It is defined as the number of wavelengths These radiations are characterised by the properties, namely, frequency ( ν ) and per unit length. Its units are reciprocal of wavelength unit, i.e., m–1. However commonly wavelength (λ). used unit is cm–1 (not SI unit). The SI unit for frequency (ν ) is hertz (Hz, s–1), after Heinrich Hertz. It is defined as Problem 2.3 the number of waves that pass a given point The Vividh Bharati station of All India in one second. Radio, Delhi, broadcasts on a frequency Wavelength should have the units of of 1,368 kHz (kilo hertz). Calculate the length and as you know that the SI units of wavelength of the electromagnetic length is meter (m). Since electromagnetic radiation emitted by transmitter. Which radiation consists of different kinds of waves part of the electromagnetic spectrum of much smaller wavelengths, smaller units does it belong to? are used. Fig.2.7 shows various types of Solution electro-magnetic radiations which differ from The wavelength, λ, is equal to c/ν , where one another in wavelengths and frequencies. c is the speed of electromagnetic In vaccum all types of electromagnetic radiation in vacuum and ν is the radiations, regardless of wavelength, travel ν (a) (b) Fig. 2.7 (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible region is only a small part of the entire spectrum. 2015-16 36 CHEMISTRY frequency. Substituting the given values, we have 1 1 ν= = λ 5800×10–10 m c λ= ν =1.724×106 m –1 3.00 × 10 8 m s –1 =1.724×104 cm –1 = 1368 kHz (b) Calculation of the frequency (ν ) 3.00 × 10 m s 8 –1 c 3 ×108 m s–1 = ν= = = 5.172 ×1014 s– 1 1368 × 10 3 s –1 –10 λ 5800 ×10 m = 219.3 m This is a characteristic radiowave wavelength. 2.3.2 Particle Nature of Electromagnetic Radiation: Planck’s Quantum Problem 2.4 Theory The wavelength range of the visible Some of the experimental phenomenon such spectrum extends from violet (400 nm) as diffraction* and interference** can be to red (750 nm). Express these explained by the wave nature of the wavelengths in frequencies (Hz). electromagnetic radiation. However, following (1nm = 10–9 m) are some of the observations which could not Solution be explained with the help of even the Using equation 2.5, frequency of violet electromagentic theory of 19th century light physics (known as classical physics): c 3.00 × 10 8 m s –1 (i) the nature of emission of radiation from ν = = hot bodies (black -body radiation) λ 400 × 10– 9 m (ii) ejection of electrons from metal surface = 7.50 × 1014 Hz when radiation strikes it (photoelectric Frequency of red light effect) c 3.00 × 108 ms–1 (iii) variation of heat capacity of solids as a ν= = = 4.00 × 1014 Hz λ 750 × 10–9m function of temperature The range of visible spectrum is from (iv) line spectra of atoms with special 4.0 × 1014 to 7.5 × 1014 Hz in terms of reference to hydrogen. frequency units. It is noteworthy that the first concrete Problem 2.5 explanation for the phenomenon of the black Calculate (a) wavenumber and (b) body radiation was given by Max Planck in frequency of yellow radiation having 1900. This phenomenon is given below: wavelength 5800 Å. When solids are heated they emit radiation over a wide range of wavelengths. Solution For example, when an iron rod is heated in a (a) Calculation of wavenumber (ν ) furnace, it first turns to dull red and then progressively becomes more and more red as λ =5800Å =5800 × 10–8 cm the temperature increases. As this is heated = 5800 × 10–10 m further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. In terms of * Diffraction is the bending of wave around an obstacle. ** Interference is the combination of two waves of the same or differ ent frequencies to give a wave whose distribution at each point in space is the algebraic or vector sum of disturbances at that point resulting from each interfering wave. 2015-16 STRUCTURE OF ATOM 37 frequency, it means that the frequency of to its frequency ( ν ) and is expressed by emitted radiation goes from a lower frequency equation (2.6). to a higher frequency as the temperature E = hν (2.6) increases. The red colour lies in the lower frequency region while blue colour belongs to The proportionality constant, ‘h’ is known the higher frequency region of the as Planck’s constant and has the value electromagnetic spectrum. The ideal body, 6.626×10–34 J s. which emits and absorbs radiations of all With this theory, Planck was able to frequencies, is called a black body and the explain the distribution of intensity in the radiation emitted by such a body is called radiation from black body as a function of black body radiation. The exact frequency frequency or wavelength at different distribution of the emitted radiation (i.e., temperatures. intensity versus frequency curve of the Photoelectric Effect radiation) from a black body depends only on In 1887, H. Hertz performed a very interesting its temperature. At a given temperature, experiment in which electrons (or electric intensity of radiation emitted increases with decrease of wavelength, reaches a maximum current) were ejected when certain metals (for example potassium, rubidium, caesium etc.) value at a given wavelength and then starts were exposed to a beam of light as shown in decreasing with further decrease of Fig.2.9. The phenomenon is called wavelength, as shown in Fig. 2.8. Fig.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. Fig. 2.8 Wavelength-intensity relationship Max Planck (1858 – 1947) The above experimental results cannot be Max Planck, a German physicist, explained satisfactorily on the basis of the received his Ph.D in theoretical wave theory of light. Planck suggested that physics from the University of atoms and molecules could emit (or absorb) Munich in 1879. In 1888, he was energy only in discrete quantities and not in appointed Director of the Institute a continuous manner, a belief popular at that of Theoretical Physics at the time. Planck gave the name quantum to the University of Berlin. Planck was awarded the Nobel smallest quantity of energy that can be Prize in Physics in 1918 for his quantum theory. emitted or absorbed in the form of Planck also made significant contributions in electromagnetic radiation. The energy (E ) of thermodynamics and other areas of physics. a quantum of radiation is proportional 2015-16 38 CHEMISTRY Photoelectric effect. The results observed in (intensity) may shine on a piece of potassium this experiment were: metal for hours but no photoelectrons are (i) The electrons are ejected from the metal ejected. But, as soon as even a very weak surface as soon as the beam of light yellow light (ν = 5.1–5.2 × 1014 Hz) shines on strikes the surface, i.e., there is no time the potassium metal, the photoelectric effect lag between the striking of light beam is observed. The threshold frequency (ν0) for and the ejection of electrons from the potassium metal is 5.0×1014 Hz. metal surface. Einstein (1905) was able to explain the (ii) The number of electrons ejected is photoelectric effect using Planck’s quantum proportional to the intensity or theory of electromagnetic radiation as a brightness of light. starting point, (iii) For each metal, there is a characteristic Shining a beam of light on to a metal minimum frequency,ν0 (also known as surface can, therefore, be viewed as shooting threshold frequency) below which a beam of particles, the photons. When a photoelectric effect is not observed. At a photon of sufficient energy strikes an electron frequency ν > ν 0, the ejected electrons in the atom of the metal, it transfers its energy come out with certain kinetic energy. instantaneously to the electron during the The kinetic energies of these electrons collision and the electron is ejected without increase with the increase of frequency any time lag or delay. Greater the energy of the light used. possessed by the photon, greater will be transfer of energy to the electron and greater All the above results could not be the kinetic energy of the ejected electron. In explained on the basis of laws of classical other words, kinetic energy of the ejected physics. According to latter, the energy electron is proportional to the frequency of content of the beam of light depends upon the electromagnetic radiation. Since the the brightness of the light. In other words, striking photon has energy equal to h ν and number of electrons ejected and kinetic the minimum energy required to eject the energy associated with them should depend electron is h ν0 (also called work function, W0 ; on the brightness of light. It has been Table 2.2), then the difference in energy observed that though the number of electrons (h ν – h ν0 ) is transferred as the kinetic energy ejected does depend upon the brightness of of the photoelectron. Following the light, the kinetic energy of the ejected conservation of energy principle, the kinetic electrons does not. For example, red light [ν energy of the ejected electron is given by the = (4.3 to 4.6) × 1014 Hz] of any brightness equation 2.7. Albert Einstein, a Ger m a n 1 bor n American physicist, is hν = hν0 + m e v2 (2.7) 2 regar ded by many as one of the two great physicists the where m e is the mass of the electron and v is world has known (the other the velocity associated with the ejected is Isaac Newton). His thr ee electron. Lastly, a more intense beam of light resear ch papers (on special consists of larger number of photons, relativity, Br ownian motion consequently the number of electrons ejected and the photoelectric ef fect) Albert Einstein is also larger as compared to that in an (1879 - 1955) which he published in 1905, experiment in which a beam of weaker while he was employed as a technical intensity of light is employed. assistant in a Swiss patent of fice in Ber ne have profoundly influenced the development Dual Behaviour of Electromagnetic of physics. He r eceived the Nobel Prize in Radiation Physics in 192 1 for his explanation of the The particle nature of light posed a photoelectric effe ct. dilemma for scientists. On the one hand, it 2015-16  STRUCTURE OF ATOM 39 Table 2.2 Values of Work Function (W0 ) for a Few Metals Metal Li Na K Mg Cu Ag W0 /eV 2.42 2.3 2.25 3.7 4.8 4.3 could explain the black body radiation and the number of photons emitted per second photoelectric effect satisfactorily but on the by the bulb. other hand, it was not consistent with the Solution known wave behaviour of light which could account for the phenomena of interference Power of the bulb = 100 watt –1 and diffraction. The only way to resolve the = 100 J s dilemma was to accept the idea that light Energy of one photon E = hν = hc/λ possesses both particle and wave-like properties, i.e., light has dual behaviour. 6.626 × 10−34 J s × 3 × 108 m s−1 Depending on the experiment, we find that = 400 × 10−9 m light behaves either as a wave or as a stream of particles. Whenever radiation interacts with = 4.969 ×10− 19 J matter, it displays particle like properties in Number of photons emitted contrast to the wavelike properties (interference and diffraction), which it 100 J s−1 −19 = 2.012 ×1020 s−1 exhibits when it propagates. This concept was 4.969 ×10 J totally alien to the way the scientists thought Problem 2.8 about matter and radiation and it took them a long time to become convinced of its validity. When electromagnetic radiation of It turns out, as you shall see later, that some wavelength 300 nm falls on the surface microscopic particles like electrons also of sodium, electrons are emitted with a exhibit this wave-particle duality. kinetic energy of 1.68 ×105 J mol–1. What is the minimum energy needed to remove Problem 2.6 an electron from sodium? What is the maximum wavelength that will cause a Calculate energy of one mole of photons photoelectron to be emitted ? of radiation whose frequency is 5 ×1014 Hz. Solution Solution The energy (E) of a 300 nm photon is given by Energy (E) of one photon is given by the expression hν = hc / λ E = hν 6.626 × 10 −34 J s × 3.0 × 108m s–1 = h = 6.626 ×10–34 J s 300 × 10−9 m ν = 5×10 14 s–1 (given) = 6.626 × 10-19 J E = (6.626 ×10–34 J s) × (5 ×1014 s–1) The energy of one mole of photons –19 23 –1 = 3.313 ×10–19 J = 6.626 ×10 J × 6.022 ×10 mol Energy of one mole of photons = 3.99 × 105 J mol–1 –19 23 –1 = (3.313 ×10 J) × (6.022 × 10 mol ) The minimum energy needed to remove = 199.51 kJ mol–1 one mole of electrons from sodium Problem 2.7 = (3.99 –1.68) 105 J mol –1 5 –1 = 2.31 × 10 J mol A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate The minimum energy for one electron C:\Chemistry XI\Unit-2\Unit-2(2)-Lay-3(reprint).pmd 27.7.6, 16.10.6 (Reprint) 2015-16 40 CHEMISTRY longest wavelength is deviated the least while 2.31 ×105 J mol –1 the violet light, which has shortest wavelength = 6.022 × 1023 electrons mol –1 is deviated the most. The spectrum of white = 3.84 ×10− 19 J light, that we can see, ranges from violet at 7.50 × 1014 Hz to red at 4×1014 Hz. Such a This corresponds to the wavelength spectrum is called continuous spectrum. hc Continuous because violet merges into blue, ∴λ = blue into green and so on. A similar spectrum E is produced when a rainbow forms in the sky. 6.626 × 10 −34 J s × 3.0 × 108m s− 1 Remember that visible light is just a small = 3.84 ×10− 19 J portion of the electromagnetic radiation (Fig.2.7). When electromagnetic radiation = 517 nm interacts with matter, atoms and molecules (This corresponds to green light) may absorb energy and reach to a higher Problem 2.9 energy state. With higher energy, these are in The threshold frequency ν 0 for a metal an unstable state. For returning to their is 7.0 ×1014 s–1. Calculate the kinetic normal (more stable, lower energy states) energy of an electron emitted when energy state, the atoms and molecules emit radiation of frequency ν =1.0 ×10 15 s –1 radiations in various regions of the hits the metal. electromagnetic spectrum. Solution Emission and Absorption Spectra According to Einstein’s equation The spectrum of radiation emitted by a 2 substance that has absorbed energy is called Kinetic energy = ½ mev =h(ν – ν0 ) –34 15 –1 an emission spectrum. Atoms, molecules or = (6.626 ×10 J s) (1.0 × 10 s – 7.0 ions that have absorbed radiation are said to ×1014 s–1) be “excited”. To produce an emission –34 14 –1 = (6.626 ×10 J s) (10.0 ×10 s – 7.0 spectrum, energy is supplied to a sample by ×1014 s–1) heating it or irradiating it and the wavelength = (6.626 ×10–34 J s) × (3.0 ×1014 s–1) (or frequency) of the radiation emitted, as the –19 sample gives up the absorbed energy, is = 1.988 ×10 J recorded. 2.3.3 Evidence for the quantized* An absorption spectrum is like the Electronic Energy Levels: Atomic photographic negative of an emission spectra spectrum. A continuum of radiation is passed through a sample which absorbs radiation of The speed of light depends upon the nature certain wavelengths. The missing wavelength of the medium through which it passes. As a which corresponds to the radiation absorbed result, the beam of light is deviated or by the matter, leave dark spaces in the bright refracted from its original path as it passes continuous spectrum. from one medium to another. It is observed that when a ray of white light is passed The study of emission or absorption through a prism, the wave with shorter spectra is referred to as spectroscopy. The wavelength bends more than the one with a spectrum of the visible light, as discussed longer wavelength. Since ordinary white light above, was continuous as all wavelengths (red consists of waves with all the wavelengths in to violet) of the visible light are represented the visible range, a ray of white light is spread in the spectra. The emission spectra of atoms out into a series of coloured bands called in the gas phase, on the other hand, do not spectrum. The light of red colour which has show a continuous spread of wavelength from * The restriction of any pr operty to discrete values is called quantization. 2015-16 STRUCTURE OF ATOM 41 red to violet, rather they emit light only at minerals were analysed by spectroscopic specific wavelengths with dark spaces methods. The element helium (He) was between them. Such spectra are called line discovered in the sun by spectroscopic spectra or atomic spectra because the method. emitted radiation is identified by the Line Spectrum of Hydrogen appearance of bright lines in the spectra When an electric discharge is passed through (Fig, 2.10) gaseous hydrogen, the H 2 molecules Line emission spectra are of great dissociate and the energetically excited interest in the study of electronic structure. hydrogen atoms produced emit Each element has a unique line emission electromagnetic radiation of discrete spectrum. The characteristic lines in atomic frequencies. The hydrogen spectrum consists spectra can be used in chemical analysis to of several series of lines named after their identify unknown atoms in the same way as discoverers. Balmer showed in 1885 on the finger prints are used to identify people. The basis of experimental observations that if exact matching of lines of the emission spectral lines are expressed in terms of spectrum of the atoms of a known element wavenumber (ν ), then the visible lines of the with the lines from an unknown sample hydrogen spectrum obey the following quickly establishes the identity of the latter, formula : German chemist, Robert Bunsen (1811-1899) was one of the first investigators to use line  1 1  –1 spectra to identify elements. ν = 109,677  2 − 2  cm (2.8) 2 n  Elements like rubidium (Rb), caesium (Cs) thallium (Tl), indium (In), gallium (Ga) and where n is an integer equal to or greater than scandium (Sc) were discovered when their 3 (i.e., n = 3,4,5,....) (a) (b) 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. 2015-16 42 CHEMISTRY The series of lines described by this formula i) The electron in the hydrogen atom can are called the Balmer series. The Balmer move around the nucleus in a circular series of lines are the only lines in the hydrogen path of fixed radius and energy. These spectrum which appear in the visible region of paths are called orbits, stationary states the electromagnetic spectrum. The Swedish or allowed energy states. These orbits are spectroscopist, Johannes Rydberg, noted that arranged concentrically around the all series of lines in the hydrogen spectrum nucleus. could be described by the following expression ii) The energy of an electron in the orbit does : not change with time. However, the  1 1  −1 ν = 109,677  2 − 2  cm (2.9) Table 2.3 The Spectral Lines for Atomic n  1 n 2  Hydrogen where n 1=1,2........ n2 = n 1 + 1, n1 + 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 shows the L yman, Balmer and Paschen series of transitions for hydrogen atom. Of all the elements, hydrogen atom has the simplest line spectrum. Line spectrum becomes more and more complex for heavier atom. There are however certain features which are common to all line spectra, i.e., (i) line spectrum of element is unique and (ii) there is regularity in the line spectrum of each element. The questions which arise are : What are the reasons for these similarities? Is it something to do with the electronic structure of atoms? These are the questions need to be answered. We shall find later that the answers to these questions provide the key in understanding electronic structure of these elements. 2.4 BOHR’S MODEL FOR HYDROGEN ATOM Neils Bohr (1913) was the first to explain quantitatively the general features of hydrogen atom structure and its spectrum. Though the theory is not the modern quantum mechanics, it can still be used to rationalize many points in the atomic Fig. 2.11 T ransitions of the electron in the structure and spectra. Bohr’s model for hydrogen atom (The diagram shows hydrogen atom is based on the following the Lyman, Balmer and Paschen series postulates: of transitions) 2015-16 STRUCTURE OF ATOM 43 electron will move from a lower stationary commonly known as Bohr’s frequency state to a higher stationary state when rule. required amount of energy is absorbed iv) The angular momentum of an electron by the electron or energy is emitted when in a given stationary state can be electron moves from higher stationary expressed as in equation (2.11) state to lower stationary state (equation h 2.16). The energy change does not take m e v r =n. n = 1,2,3..... (2.11) place in a continuous manner. 2π Thus an electron can move only in those Angular Momentum orbits for which its angular momentum is Just as linear momentum is the product integral multiple of h/2π that is why only of mass (m) and linear velocity (v), angular certain fixed orbits are allowed. momentum is the product of moment of The details regarding the derivation of inertia (I ) and angular velocity (ω). For an energies of the stationary states used by Bohr, electron of mass me, moving in a circular are quite complicated and will be discussed path of radius r around the nucleus, in higher classes. However, according to angular momentum = I × ω Bohr’s theory for hydrogen atom: Since I = mer 2 , and ω = v/r where v is the a) The stationary states for electron are linear velocity, numbered n = 1,2,3.......... These integral ∴angular momentum = mer2 × v/r = mevr numbers (Section 2.6.2) are known as Principal quantum numbers. iii) The frequency of radiation absorbed or b) The radii of the stationary states are emitted when transition occurs between expressed as : two stationary states that differ in energy rn = n 2 a0 (2.12) by ∆E, is given by : where a 0 = 52,9 pm. Thus the radius of the first stationary state, called the Bohr ∆E E2 − E1 orbit, is 52.9 pm. Normally the electron ν = = (2.10) h h in the hydrogen atom is found in this Where E1 and E2 are the energies of the orbit (that is n=1). As n increases the lower and higher allowed energy states value of r will increase. In other words r espectively. This expression is the electron will be present away from the nucleus. c) The most important property associated Niels Bohr (1885–1962) with the electron, is the energy of its stationary state. It is given by the Niels B o hr, a Danish expression. physicist received his Ph.D. f rom the University of  1  En = − R H  2  n = 1,2,3.... (2.13) Copenhagen in 1911. He n  then spent a year with J.J. where RH is called Rydberg constant and its Thomson and Er nest Rutherfor d in England. value is 2.18×10–18 J. The energy of the lowest In 1913, he retur ned to Copenhagen wher e state, also called as the ground state, is he remained for the rest of his life. In 1920 1 he was named Director of the Institute of E1 = –2.18×10 –18 ( 2 ) = –2.18×10 –18 J. The 1 theor etical Physics. After first World War, Bohr worked energetically for peaceful uses energy of the stationary state for n = 2, will of atomic energy. He received the first Atom s 1 be : E 2 = –2.18×10–18J ( 2 )= –0.545×10 –18 J. for Peace award in 1957. Bohr was awar ded 2 the Nobel Prize in Physics in 1922. Fig. 2.11 depicts the energies of different 2015-16 44 CHEMISTRY stationary states or energy levels of hydrogen where Z is the atomic number and has values atom. This representation is called an energy 2, 3 for the helium and lithium atoms level diagram. respectively. From the above equations, it is evident that the value of energy becomes more What does the negative electronic negative and that of radius becomes smaller energy (E n) for hydrogen atom mean? with increase of Z. This means that electron The energy of the electron in a hydrogen will be tightly bound to the nucleus. atom has a negative sign for all possible orbits (eq. 2.13). What does this negative e) It is also possible to calculate the sign convey? This negative sign means that velocities of electrons moving in these the energy of the electron in the atom is orbits. Although the precise equation is lower than the energy of a free electron at not given here, qualitatively the rest. A free electron at rest is an electron magnitude of velocity of electro n that is infinitely far away from the nucleus increases with increase of positive charge and is assigned the energy value of zero. on the nucleus and decreases with Mathematically, this corresponds to increase of principal quantum number. setting n equal to infinity in the equation 2.4.1 Explanation of Line Spectrum of (2.13) so that E∞=0. As the electron gets Hydrogen closer to the nucleus (as n decreases), En becomes larger in absolute value and more Line spectrum observed in case of hydrogen and more negative. The most negative atom, as mentioned in section 2.3.3, can be energy value is given by n=1 which explained quantitatively using Bohr’s model. corresponds to the most stable orbit. We According to assumption 2, radiation (energy) call this the ground state. is absorbed if the electron moves from the orbit of smaller Principal quantum number When the electron is free from the influence to the orbit of higher Principal quantum of nucleus, the energy is taken as zero. The number, whereas the radiation (energy) is electron in this situation is associated with the emitted if the electron moves from higher orbit stationary state of Principal Quantum number to lower orbit. The energy gap between the = n = ∞ and is called as ionized hydrogen atom. two orbits is given by equation (2.16) When the electron is attracted by the nucleus ∆E = Ef – Ei (2.16) and is present in orbit n, the energy is emitted and its energy is lowered. That is the reason Combining equations (2.13) and (2.16) for the presence of negative sign in equation  R   R  (2.13) and depicts its stability relative to the ∆ E =  − H2  −  − H2  (where n and n reference state of zero energy and n = ∞.  n f   ni  i f d) Bohr’s theory can also be applied to the stand for initial orbit and final orbits) ions containing only one electron, similar 1 1   1 1  to that present in hydrogen atom. For ∆E = R H  2 − 2  = 2.18 ×10 −18 J  2 − 2  n  i n f  n  i n f  example, He+ Li2+ , Be3+ and so on. The energies of the stationary states (2,17) associated with these kinds of ions (also The frequency (ν ) associated with the known as hydrogen like species) are given absorption and emission of the photon can by the expression. be evaluated by using equation (2.18)  Z2  E n = − 2.18 ×10 −18  2  J (2.14) ∆ E RH  1 1  n  ν = =  −  and radii by the expression h h  n 2i n f2  52.9 (n 2 ) 2.18 × 10 − 18 J  1 1  rn = pm (2.15) =  −  (2.18) Z 6.626 × 10 −34 J s  n i2 n f2  2015-16 STRUCTURE OF ATOM 45  1 1  It is an emission energy = 3.29 ×1015  2 − 2  Hz (2.19) The frequency of the photon (taking  ni n f  energy in terms of magnitude) is given and in terms of wavenumbers (ν ) by ∆E ν = RH  1 − 1  ν = ν= c hc  n 2 n 2  i f (2.20) h 4.58×10– 19 J 3.29 ×10 s  1 15 −1 1  = = − 2 6.626×10–34 J s 8 −s  2 3 × 10 m s  n i n f  = 6.91×1014 Hz  1 1  c 3.0 × 108 m s −1 = 1.09677 × 107  2 − 2  m − 1 (2.21) λ= = = 434 nm  ni nf  ν 6.91× 1014 Hz In case of absorption spectrum, nf > ni and Problem 2.11 the term in the parenthesis is positive and Calculate the energy associated with the energy is absorbed. On the other hand in case first orbit of He+. What is the radius of of emission spectrum n i > nf , ∆ E is negative this orbit? and energy is released. The expression (2.17) is similar to that Solution used by Rydberg (2.9) derived empirically (2.18 × 10 −18 J)Z 2 using the experimental data available at that En = − atom–1 n2 time. Further, each spectral line, whether in absorption or emission spectrum, can be For He +, n = 1, Z = 2 associated to the particular transition in (2.18 ×10−18 J)(22 ) hydrogen atom. In case of large number of E1 = − 2 = −8.72 × 10−18 J 1 hydrogen atoms, different possible transitions can be observed and thus leading to large The radius of the orbit is given by number of spectral lines. The brightness or equation (2.15) intensity of spectral lines depends upon the (0.0529 nm)n 2 number of photons of same wavelength or rn = Z frequency absorbed or emitted. Since n = 1, and Z = 2 Problem 2.10 (0.0529 nm)12 rn = = 0.02645 nm What are the frequency and wavelength 2 of a photon emitted during a transition from n = 5 state to the n = 2 state in the 2.4.2 Limitations of Bohr’s Model hydrogen atom? Bohr’s model of the hydrogen atom was no Solution doubt an improvement over Rutherford’s Since ni = 5 and nf = 2, this transition nuclear model, as it could account for the gives rise to a spectral line in the visible stability and line spectra of hydrogen atom region of the Balmer series. Fr o m and hydrogen like ions (for example, He+, Li2+ , equation (2.17) Be3+ , and so on). However, Bohr’s model was too simple to account for the following points. 1 1 ∆E = 2.18 × 10 −18 J  2 − 2  i) It fails to account for the finer details 5 2  (doublet, that is two closely spaced lines) −19 = − 4.58 ×10 J of the hydrogen atom spectrum observed 2015-16 46 CHEMISTRY by using sophisticated spectroscopic Louis de Broglie (1892 – 1987) techniques. This model is also unable to explain the spectrum of atoms other than Louis de Broglie, a French hydrogen, for example, helium atom which physicist, studied history as an possesses only two electrons. Further, undergraduate in the early Bohr’s theory was also unable to explain 1910’s. His interest turned to the splitting of spectral lines in the science as a result of his presence of magnetic field (Zeeman effect) assignment to radio or an electric field (Stark effect). communications in World War I. ii) It could not explain the ability of atoms to He received his Dr. Sc. from the University of form molecules by chemical bonds. Paris in 1924. He was professor of theoretical physics at the University of Paris from 1932 untill In other words, taking into account the his retirement in 1962. He was awarded the points mentioned above, one needs a better Nobel Prize in Physics in 1929. theory which can explain the salient features of the structure of complex atoms. which is based on the wavelike behaviour of 2.5 TOWARDS QUANTUM MECHANICAL electrons just as an ordinary microscope MODEL OF THE ATOM utilises the wave nature of light. An electron In view of the shortcoming of the Bohr’s model, microscope is a powerful tool in modern attempts were made to develop a more scientific research because it achieves a suitable and general model for atoms. Two magnification of about 15 million times. important developments which contributed It needs to be noted that according to de significantly in the formulation of such a Broglie, every object in motion has a wave model were : character. The wavelengths associated with 1. Dual behaviour of matter, ordinary objects are so short (because of their large masses) that their wave properties 2. Heisenberg uncertainty principle. cannot be detected. The wavelengths 2.5.1 Dual Behaviour of Matter associated with electrons and other subatomic particles (with very small mass) can however The French physicist, de Broglie in 1924 be detected experimentally. Results obtained proposed that matter, like radiation, should from the following problems prove these also exhibit dual behaviour i.e., both particle points qualitatively. and wavelike properties. This means that just as the photon has momentum as well as Problem 2.12 wavelength, electrons should also have What will be the wavelength of a ball of momentum as well as wavelength, de Broglie, mass 0.1 kg moving with a velocity of 10 from this analogy, gave the following relation m s–1 ? between wavelength (λ) and momentum (p) of a material particle. Solution According to de Brogile equation (2.22) h h λ= = (2.22) mv p h (6.626 × 10−34 Js) λ= = where m is the mass of the particle, v its mv (0.1 kg)(10 m s−1 ) velocity and p its momentum. de Broglie’s = 6.626×10–34 m (J = kg m2 s–2) prediction was confirmed experimentally Problem 2.13 when it was found that an electron beam undergoes diffraction, a phenomenon The mass of an electron is 9.1×10–31 kg. characteristic of waves. This fact has been put If its K.E. is 3.0×10–25 J, calculate its to use in making an electron microscope, wavelength. 2015-16 STRUCTURE OF ATOM 47 Solution the other hand, if the velocity of the electron is known precisely (∆(vx ) is small), then the Since K. E. = ½ mv 2 position of the electron will be uncertain 1/2 1/ 2  2 × 3.0 × 10 −25 kg m 2 s −2  (∆x will be large). Thus, if we carry out some 2K.E.  v =   =  physical measurements on the electron’s  m   9.1 × 10 −31 kg  position or velocity, the outcome will always = 812 m s–1 depict a fuzzy or blur picture. The uncertainty principle can be best h 6.626 × 10−34 Js understood with the help of an example. λ= = m v (9.1× 10−31kg)(812 m s−1 ) Suppose you are asked to measure the thickness of a sheet of paper with an = 8967 × 10–10 m = 896.7 nm unmarked metrestick. Obviously, the results Problem 2.14 obtained would be extremely inaccurate and Calculate the mass of a photon with meaningless, In order to obtain any accuracy, wavelength 3.6 Å. you should use an instrument graduated in Solution units smaller than the thickness of a sheet of the paper. Analogously, in order to determine λ = 3.6 Å = 3.6 × 10− 10 m the position of an electron, we must use a Velocity of photon = velocity of light meterstick calibrated in units of smaller than the dimensions of electron (keep in mind that h 6.626 × 10−34 Js an electron is considered as a point charge m= = λν (3.6 × 10–10 m)(3 × 108 m s−1) and is therefore, dimensionless). To observe an electron, we can illuminate it with “light” = 6.135 × 10–29 kg or electromagnetic radiation. The “light” used must have a wavelength smaller than the 2.5.2 Heisenberg’s Uncertainty Principle dimensions of an electron. The high Werner Heisenberg a German physicist in  h 1927, stated uncertainty principle which is momentum photons of such light  p =   λ the consequence of dual behaviour of matter would change the energy of electrons by and radiation. It states