Modern Physics Lecture Material PDF

MODERN PHYSICS LECTURE MATERIAL MODELS OF THE ATOM Modern atomic theory provides a reasonably satisfactory explanation of the properties of matter, the mechanisms of chemical change and the interaction of matter and energy. Such a theory emerged from the synthesis of the work of several scientists...

MODERN PHYSICS LECTURE MATERIAL MODELS OF THE ATOM Modern atomic theory provides a reasonably satisfactory explanation of the properties of matter, the mechanisms of chemical change and the interaction of matter and energy. Such a theory emerged from the synthesis of the work of several scientists only a few of which will be discussed here. John Dalton (1766-1844) is generally credited as the father of atomic theory but the early Greek philosophers were the originators of the concept of atoms. They taught that matter is composed of atoms and is therefore finitely divisible. John Dalton considered the atom as constituting the simplest component of matter. He viewed the atom as indestructible, tiny hard spheres. However, the discovery of radioactivity by Henry Becquerel in 1891 showed that atoms are complex rather than indivisible. Radioactivity showed that atoms are not indivisible or indestructible but can disintegrate forming atoms of different elements. The discovery of cathode rays in electric discharge tubes (1895) by William Crookes revealed that negatively charge electrons were components of the atom. By 1900, it was already established that matter consist of atoms but nothing was known about the structure of the atom. It was known that the atoms contain electrons but that on the whole the atom was electrically neutral. This neutrality means that there must exist within the atom enough positively charge components. This led Sir J.J. Thomson to propose an atomic model which visualized the atom as a homogeneous sphere of positive charge inside of which are embedded negatively charged electrons. Figure 1: Thomson’s Model of the atom 1  𝑒 𝑒 J.J Thomson also determined the ratio of the charge to mass, , of the electrons and found to be 𝑚 𝑚 identical for all cathode-ray particles, irrespective of the kind of gas in the tube or the metal the electrodes are made of. Specific charge (e/m) of Electrons by J J. Thomson’s Method Working Principle If a beam of cathode rays (fast moving electrons) is subjected to pass through an electric as well as magnetic field then the electrons experience a force due to each field. If the arrangement of the two fields is made in such a way that the force due to them nullify each other and hence overall deflection of the beam becomes zero. The diagram, shown above, represents the modified version of the apparatus used by J. J. Thomson. The apparatus essentially consists of a discharge tube G at an extremely low pressure. It has a cathode, made up of a filament coated with materials like barium oxide (BaO) and a low-tension voltage is applied across it in order to produce the thermal electrons. A small slit is placed in front of the cathode gun in order to produce a collimated beam. D and E are two electrodes, through which a high electric field is applied perpendicular to the cathode beam (in the plain of the paper). Further, a magnetic field is applied perpendicular to the electric field (perpendicular to the plain of the paper) as well as the direction of the beam. S is the screen coated with zinc sulphide, which acts like a fluorescent material. When electron beam strikes the screen, the scintillation is produced. 2 Working and Theory Under given conditions, cathode ray meet the ‘M’ point on the screen in the absence of any electric or magnetic field. Now, we apply the electric field, as a result of which, the beam suffers a deflection. After this, we make the deflected beam to experience a force due to magnetic field, which is in such a direction that the beam gets readjusted to the point ‘M’. This means the electrons of the rays experience an equal and opposite force due to the electric and magnetic fields acting simultaneously. Let m and e be the mass and charge of electron and v be the velocity of the electron when it comes out of the slit. The force on the electron due to the electric field can be given by F = eE The force on the electron due to the magnetic field is given by F = Bev Since, the two forces balance each other, we can write eE = Bev Or velocity , 𝐸 𝑣= 𝐵 ……………………………………………………………………………………….(i) The electrons at the cathode possess potential energy of eV joules higher than at the anode. As the electrons proceed towards the anode, the potential energy of the electrons gets converted into the kinetic energy. Thus, we can write 1 2 𝑚𝑣 2 = 𝑒𝑉 ………………………………………………………………………………(ii) which gives, ………………………………………………………………………....(iii) 3  𝑒 Since, we can easily determine the values of E, B and V the value of specific ratio, of the electron can be 𝑚 𝑒 determined by using the equation (iii). Thomson found the value of 𝑚 of the electron equal to 1.77 × 𝑒 1011 𝐶𝑘𝑔−1. The accurate value of 𝑚 for the electron is found to be 1.7589 × 1011 𝐶𝑘𝑔−1. The field produced by the coils is given by the equation: μo 8NI B= (5√5r) Where N is the number of turns on the coil, 𝐼 is the current in the coils and r the radius of the coils. Example Electrons, accelerated from rest through a potential difference of 3000 V, enter a region of uniform magnetic field, the direction of the field being at right angle to the motion of the electrons. If the flux density is 0.010 T, Calculate the radius of the electron’s orbit. Solution 𝑚𝑣 2 Magnetic force = Bev = 𝑟 Where r is the radius of the orbit. 1 But also 2 𝑚𝑣 2 = 𝑒𝑉 Therefore, 2𝑉𝑚 𝑟 = √ 𝐵2 𝑒 = 1.85 mm Rutherford Model – the Nuclear Idea and the Planetary Model Ernest Rutherford (1911) and his co-workers performed experiments whose results contradicted J.J Thomson’s model. In these experiments a beam of positively charged alpha (𝛼) particles was directed at a thin sheet of a metal foil. It was found that most of the alpha-particles passed through the foil without deflection as if the foil were mostly empty space. Only a few of them were diverted from their paths. Some of these few actually rebounded backward. 4  Figure 2: Scattering of 𝛼 – particle by a metal foil This scattering of alpha-particles by the metal foil was explained as a repulsion from a heavy positively charged nucleus present at the center of the atom of the metal foil. This follows because an abrupt change in path (as noted for a few 𝛼 – particles) of a relatively heavy and positively charged 𝛼 – particle can result only from its hitting or from its close approach to another particle (the nucleus) with a highly concentrated, positive charge. All these contradicted Thomson’s atomic model which supposed that the distribution of the charges was diffuse. Hence, Rutherford proposed his model of the atom. Rutherford proposed a planetary model of the atom which suggested that the atom consists of a positively charged heavy core called the nucleus where most of the mass of the atom was concentrated. Around this nucleus negatively charged electrons circle in orbits much as planets move around the sun. Each nucleus much be surrounded by a number of electrons necessary to produce an electrically neutral atom. This model was a major step toward how we view the atom today. It however, had two problems. According to Newtonian Physics, such an atom as Rutherford’s would collapse by spiraling into the nucleus, since there is an attractive force between the oppositely charged nucleus and electrons. Further experiments indicated that charged particles moving in a field of opposite charge lose energy by emitting radiation. But atoms in their normal state neither collapsed nor emitted radiation. Thus the two main difficulties of Rutherford model are these: 1. It predicts that light of a continuous range of frequencies will be emitted, whereas experiment shows line spectra instead of continuous spectra. 2. It predicts that atoms are instable – electrons quickly spiral into the nucleus – but we know that atoms in general are stable, since the matter around us is stable. Clearly Rutherford’s model was not sufficient to explain experimental observations. Some sort of modification was needed and this was provided by Neils Bohr. 5  Figure 3: Rutherford’s planetary model The Bohr model – Idea of Quantized Energy State This model is an elucidation of Rutherford’s model, which he did by putting forward some postulates. Bohr Postulates 1. Only those orbits are allowed or permissible in which the angular momentum of the electron about the orbit is conserved. ℎ The angular momentum is an integral multiple of 2𝜋 , that is 𝑛ℎ = 𝑚𝑣𝑛 𝑟𝑛 2𝜋 Where n is an integer (quantum number). The quantity 𝑚𝑣𝑛 𝑟𝑛 is the angular momentum of the electron 𝑛ℎ in its nth orbit. That is, angular momentum 𝑣𝑛 𝑟𝑛 = 2𝜋 , where n = 1, 2, 3, The speed of the electron is v, its mass is m, and h is Planck’s constant, 6.63 × 10−34 𝐽. 𝑆.When an electron is in one of these stationary orbits, it does not radiate energy even though it is continuously accelerated. 6  2. When an electron jumps from a higher level or orbit to a lower energy level, it emits radiation in quanta or definite amount. Also, an electron absorbs radiation or energy when it jumps from a lower level to a higher level. The excitation energy for two levels, n = 1 and n = 2 is given by ℎ𝑓 = 𝐸2 − 𝐸1 SUCCESS OF BOHR’S MODEL 1. For the explanation of the structure of the hydrogen atom 2. For the explanation of the existence of line spectra of the hydrogen atom 3. Bohr’s theory provides evidence for the existence of energy levels in the atom. LIMITATIONS 1. It could not interpret the details of spectra of complex atoms because he used a simple atom of hydrogen for his model. 2. There is no theoretical basis for selecting the allowed states or orbits 3. The radius of an orbit could not be checked experimentally. 4. The motion of the nucleus was not considered. 5. The model could not explain the existence of fine structure of the spectral lines. 6. It could not explain the Zeeman effect and stark effect Shell or orbit Electron Nucleus Figure 4: Bohr model of hydrogen atom 7 Figure 5: An electron emits a photon when it move to a lower energy level [Line spectrum from an atom] Figure 6: An electron absorbs energy when it transfers to a higher energy level (excitation) [Line spectrum from an atom] 8 ELECTRON CLOUD MODEL This model visualizes the atom as consisting of a tiny nucleus of radius of order of 10−5 𝑚. The electron is visualized as being in rapid motion within a relatively large region around the nucleus, but spending most of its time in certain high-probability regions. Thus the electron is not considered as a ball revolving around the nucleus but as a particle or wave with a specified energy having only a certain probability or being in a given region in the space outside the nucleus. The electron is visualized as spread out around the nucleus in a sort of electron cloud. The cloud of negative charge is considered being dense in region of high electron probability and more diffuse in region of low probability. LIMITATIONS 1. It is abstract 2. It involves difficult mathematics (wave mechanics) Figure 7: Electron-cloud Model 9 ENERGY QUANTIZATION Some of the important postulates of the Bohr’s model of the atom was that the electrons moved around the nucleus in specified orbits. In these orbits they can move without radiating energy. They can acquire or loss energy only in discrete units called quanta. Thus Bohr suggests that electrons in the atom exist in discrete (quantize) energy states. Therefore, we noted that there exist energy levels in atom, which follows the arranged accordingly as shown below. 0 Increasing negative energy values Figure 8: Energy level for Hydrogen 10 The energy level in an atom is governed by the principal quantum number, n. The energy increases upwards as the levels come closer to each other. Photon or radiation is emitted whenever electron makes a transition. The energy of the photon is given as; 𝑚ℎ𝑓 𝐸= 𝜆 ℎ𝑐 Where m = number of quanta. For one quanta, 𝐸 = ℎ𝑓 = 𝜆 Where h = Planck’s constant, f = frequency, c = velocity of light and 𝜆 = wavelength. ATOMIC ENERGY LEVELS Electrons in atoms are believed to be arranged around their nuclei in positions known as energy levels or electron orbits or electron shells. The electron is held in its orbit by the force of attraction between it and the nucleus. Each electron in a given atom possesses a specified amount of energy that is related to the distance between the electron and the nucleus. Since the electrons in an atom are restricted to individual orbits they are said to occupy energy levels. Electrons in the orbit closest to the nucleus have the highest energy, that is, it takes more energy to remove such electrons from the first energy level than to remove electrons from any of the other higher levels. Such electrons in the closest orbit to the nucleus are said to be in the ground state or lowest energy level. The energy of an electron is given by the relation: 1 𝐸 = − 𝑛2 𝑅 Where n is the electron’s quantum number and R is a constant. The minus sign signifies that one must do work on the electron to remove it from the atom. From the energy level diagram above for hydrogen, the energy state or levels are numbered n = 1, 2, 3, … stating from the ground state (𝐸0 ) as n = 1. When the electron is infinitely far from the atom (or is completely remove from the atom) the atom is said to be ionized. The energy of this state with n = ∞ is defined as zero. When an atom is heated or bombarded with an energetic particle, its energy is increased and the atom is said to be excited. For an atom to be excited, its energy must be increased by a definite amount. During excitation, one or more of the atomic electrons may move to a higher energy level. The electron possesses this higher energy as long as it remains in the excited state. Because excited states are unstable, the electron soon jumps back to its original orbit or energy level and emits a photon or quantum of light with a characteristic frequency, f, (or wavelength, 𝜆) in the process, according to the relation ℎ𝑐 𝐸𝑛 − 𝐸0 = ℎ𝑓𝑛 = 𝜆𝑛 Where 𝐸𝑛 is the energy in the excited state and 𝐸0 is the energy in the ground state (n = 1). Not all jumps are to the ground state or ground level. Electrons can also jump from one allowed higher energy level to a lower allowed energy level. The energy of each radiated photon is therefore proportional to the length of its corresponding arrow as shown below 11 EXAMPLE An electron jumps from one energy level to another in an atom radiating 4.5 × 10−19 joules. If Planck’s constant is 6.6 × 10−34 𝐽𝑠, what is the wavelength of the radiation? Take velocity of light = 3 × 108 𝑚𝑠 −1. SOLUTION ∆𝐸 = 𝐸𝑛 − 𝐸𝑚 = ℎ𝑓 ∆𝐸 = 4.5 × 10−19 = 6.6 × 10−34 × 𝑓 𝑐 But f= 𝜆 3×108 ∆𝐸 = 4.5 × 10−19 = 6.6 × 10−34 × 𝜆 6.6×10−34 ×3×108 𝜆= 4.5×10−19 = 4.4 × 10−7 m 12 EXAMPLE The figure below gives the energy level for mercury atom: 0eV E4 - 1.6eV E3 - - 3.7eV E2 - 5.5eV E1 - 10.4eV If a bombarding electron has energy 6.7 eV, i. to what level will the mercury atom be excited ii. What will be that wavelength of the light emitted if the mercury atom drops from this excited state to the second level? [Take h = 6.626 x 10 -34J; c = 3 x10 8 m/s,]. [Note: 1eV = 1.6 x 10 -19] Solution i. Since the atom is normally at ground state the energy received is En -E1 = 6.7eV = En – (- 10.4) En = 6.7eV – 10.4 eV En = - 3.7eV This is the energy of the 3rd level. For a drop from E3 to E2 E3 – E2 = -3.7 – (-5.5) = 1.8 eV ii. hc 1.8eV  hv   hc  1.8eV 6.626x10 34 x3x108 = 1.8 x1.6 x10 19 = 6.9 x 10 – 7 m 13 EXAMPLE An atom excited to an energy level 𝐸2 = −2.42 × 10−19 𝐽 falls to the ground state 𝐸0 = −21.8 × 10−19 𝐽. Calculate the frequency and the wavelength of the emitted photon. [Take h = 6.6 × 10−34 𝐽𝑠] SOLUTION ∆𝐸 = 𝐸2 − 𝐸0 = ℎ𝑓 −2.42 × 10−19 − (−21.8 × 10−19 ) = ℎ𝑓 19.38 × 10−19 = 6.6 × 10−34 × 𝑓 19.38×10−19 𝑓= 6.6×10−34 𝑓 = 2.94 × 1015 𝐻𝑧 𝑐 3 × 108 𝜆= = 𝑓 2.94×1015 = 1.02 × 10−7 𝑚 FRANK – HERTZ EXPERIMENT Experiment performed by Frank and Hertz gave one of the most direct proofs of the existence of energy levels or electron shells within an atom. In these experiments it was possible to measure the energy necessary to raise an electron from the ground state in an atom to an outer orbit or higher state (excitation) or to remove it from the atom entirely (ionization). They bombarded atoms in form of mercury or sodium vapour with electrons emitted from a hot filament, and accelerated to higher energy (several electron volts). The electrons were accelerated towards a grid plate. The distance between the filament and the grid plate was arranged to be much greater than the mean distance between atomic collisions (the mean free path). This was to ensure that the electrons would make collisions before reaching the grid. The potentiometer was used to vary the p.d. (V) between the filament and the grid. This is the electron accelerating p.d. The anode close to the grid was made negative in potential relative to the grid by applying a small p.d. between the anode and the grid. The electrons accelerated by the high p.d. acquire high kinetic energy before reaching the grid. They are however subjected to a retarding field before passing through the grid to the anode. The number per second reaching the anode was measured by an electrometer. As p.d. was slowly increasing from zero, the speed and number of electrons reaching the grid increased and the anode plate current increased until the p.d. reached a value 𝑉1. As the p.d. was further increased, the current dropped to a minimum, rises again to another p.d. 𝑉2. This pattern of drop and rise in the current was repeated as the p.d. was further increased. 14 Figure 9: Frank-Hertz Experiment THE HYDROGEN ATOM For simplicity, we will now consider only the hydrogen atom with its single electron. We take an electron that is at rest outside the atom to have zero energy, and so the levels within the atom have negative energy values. Each level is given a number (quantum number for that level) with the higher numbers representing the states of greater energy. When an electron in the atom falls from one of the upper energy levels to one lower down, energy is emitted in the form of radiation. The frequency, f, of the emitted radiation is given by Planck’s formula E = hf Table 1: The values of the energy levels can be calculated from Bohr’s formula as shown in the table below n Energy (eV) Energy (J) 1 −13.60 −2.18 × 10−18 2 −3.39 −5.42 × 10−19 3 −1.54 −2.42 × 10−19 4 −0.85 −1.36 × 10−19 5 −0.54 −8.71 × 10−19 The hydrogen atom has a diameter of about 0.1 nm; it consists of a proton as the nucleus (with a radius of about 10−15 𝑚) and a single electron. The hydrogen atom is considered as one electron circulating around a single proton. For the electron’s de Broglie wave to resonate or fit in an orbit of radius r, the following bust be true 𝑛ℎ 2𝜋 = 𝑚𝑣𝑛 𝑟𝑛 15 Where n is an integer. The quantity 𝑚𝑣𝑛 𝑟𝑛 is the angular momentum of the electron in its nth orbit. The speed of the electron is v, its mass is m, and h is Planck’s constant. The centripetal force that holds the electron in orbit is supplied by coulomb attraction between the nucleus and the electron. Hence, 𝑘0 𝑒 2 𝑚𝑣𝑛2 𝐹= 𝑟2 = 𝑚𝑎 = 𝑟𝑛 𝑚𝑣𝑛2 𝑘0 𝑒 2 𝑟𝑛 = 𝑟2 We can solve and get radii of stable orbit as 𝑟𝑛 = (0.053𝑛𝑚) 𝑛2 The energy of the atom when it is in the nth state (that is, with its electron in the nth orbit configuration) is 13.6 𝐸𝑛 = − 𝑛2 𝑒𝑉 For a nucleus with charge Ze orbited by a single electron, the corresponding relation are 𝑛2 13.6 𝑟𝑛 = (0.053𝑛𝑚) ( ) and 𝐸𝑛 = − 𝑒𝑉 𝑍 𝑛2 Where Z is called the atomic number of the nucleus. EXAMPLE What wavelength does a hydrogen atom emit as its excited electron falls from the n = 5 state to the n = 2 state? Give your answer to three significant figures. [1240 nm corresponds to 1.00 eV] SOLUTION 13.6 From the Bohr model we know that the energy levels of the hydrogen atom are given by 𝐸𝑛 = − 𝑛2 𝑒𝑉, therefore 13.6 𝐸5 = − 𝑒𝑉 = −0.54𝑒𝑉 52 13.6 𝐸2 = − 22 𝑒𝑉 = −3.40𝑒𝑉 The energy difference between these states is 3.40 – 0.54 = 2.86 eV. Because 1240 nm correspond to 1.00 eV in inverse proportion, we have, for the wavelength of the emitted photon, 1.00 𝑒𝑉 𝜆 = (2.86 𝑒𝑉) (1240 𝑛𝑚) = 434 𝑛𝑚 EXAMPLE Calculate the energy released and the wavelength of emitted radiation when an electron falls from level n = 3 (- 1.51 eV) to n = 2 (-3.4 eV) 16 SOLUTION Energy difference (E) = - 1.51 – (-3.4) = 1.89 eV = 3.0 × 10−19 𝐽 ℎ𝑐 6.6 × 10−34 × 3 × 108 Therefore, wavelength emitted , 𝜆 = 𝐸 = 3.0 × 10−19 = 6.6 × 10−7 = 660 𝑛𝑚 BOHR’S ATOMIC RADIUS Hydrogen atom consists of one electron and one proton. 𝑒 (−𝑒) −𝑒 2 Electrostatic force of attraction = = 4𝜋𝜀𝑜 𝑟2 4𝜋𝜀𝑜 𝑟2 𝑚𝑉 2 Centripetal force acting on the electron = 𝑟 Electrostatic force = Centripetal force 𝑚𝑉 2 𝑒2 = 𝑟 4𝜋𝜀𝑜 𝑟2 1 Note that K.E = 2 𝑚𝑉 2 The expression for the radius of the orbit of an atom is given by: 𝜀𝑜 𝑛 2 ℎ 2 𝑟= …………………………………………..(1) 𝜋𝑚𝑍𝑒 2 Where 𝜀𝑜 is a constant = 8.85 x 10 -12 C2/Nm, n is the orbit level, m is the mass, e is the charge of the electron and Z is the atomic number of the element. THE TOTAL ENERGY OF ORBITING ELECTRON The total energy, E of orbiting electron is given by −𝑍𝑒 2 𝐸= ………………………………………………(2) 8𝜋𝜀𝑜 𝑟2 From (1) and (2) The total energy in the nth orbit is 𝑚𝑒 4 𝑍2 𝐸=− 2 8𝜋𝜀2𝑜 𝑛2 ℎ Thus, when an electron falls from one energy level E2 to another E1 radiating of frequency f will be emitted. Bohr showed that the energy difference (E2 – E1 = hf) was given by the equation 17  𝑚𝑒 4 𝑍2 1 1 ℎ𝑐 𝐸 = 𝐸2 − 𝐸1 = 2( 2 − 𝑛2 ) = ℎ𝑓 = 8𝜋𝜀2𝑜 ℎ 𝑛2 1 𝜆 The frequency of radiation is given by 𝑚𝑒 4 𝑍2 1 1 𝑓= 2 3 (𝑛2 − 𝑛2 ) 8𝜋𝜀𝑜 ℎ 2 1 1 𝑓 𝑚𝑒 4 𝑍2 1 1 𝜆 = 𝑐 = 2 3 (𝑛2 − 𝑛2 ) 8𝜋𝜀𝑜 ℎ 2 1 1 where 𝜆 is called the wave number Example – 34 Calculate the radius of the first Bohr’s orbit of a hydrogen atom (h = 6.62 X 10 JS, 𝜀𝑜 = 8.85 × 10−12 𝐶 2 /𝑁𝑚, m = 9.1 X 10 – 31Kg, e = 1.6 X 10 – 19C). Solution From 𝜀𝑜 𝑛 2 ℎ 2 𝑟= 𝜋𝑚𝑍𝑒 2 – 34 For the Bohr’s first orbit, n = 1 and for a hydrogen atom, Z = 1, h = 6.62 X 10 , 𝜀𝑜 = 8.85 × 10−12 𝐶 2 /𝑁𝑚, m = 9.1 X 10 – 31Kg, e = 1.6 X 10 – 19C 12 × (6.62 × 10−34 )2 × 8.85 × 10−12 𝑟= = 0.53 × 10−10 𝑚 or 𝑟 = 0.53𝑂 𝐴 (1𝑂 𝐴 = 10−10 𝑚) 3.142 × 9.1 × 10−31 × (1.6 × 10−19 )2 ATOMIC SPECTRA; COLOUR AND LINE FREQUENCY TYPES OF SPECTRUM There are two main types of spectrum (a) Emission spectrum: where light is given out by a source (b) Absorption spectrum: where light from a source is absorbed when it passes through another material, usually a gas or a liquid. Emission spectra These show different characteristic that depend on the nature of the source. (i) A continuous spectrum contains all the wavelengths in a certain region of the spectrum. A continuous spectrum in the visible region may be emitted by a hot solid at a temperature above some 800 K. The Sun emits a more or less continuous spectrum, as does a piece of iron heated in a flame. (ii) A line spectrum is a spectrum that consists of a number of well defined lines each having a particular frequency or wavelength or colour. 18 Absorption spectra If light from a white light source is passed through a gas then the continuous spectrum is crossed by a series of dark lines that correspond exactly to the lines observed in the emission spectrum, which involve transitions to the ground state. This is because in a gas almost all the atoms are in their ground state. Absorption occurs by the electrons in the atom absorbing the energy from the incoming radiation and then reradiating it in all directions. Thus the energy that was originally travelling in one direction is spread out, and compared with the rest of the spectrum these wavelength appear dark. SPECTRAL LINES The spectral lines emitted by excited isolated hydrogen atom occur in series. Typical is the series that appears at visible wavelengths, the Balmer series shown below. Other series exist; one, in the ultraviolet, is called Lyman series; there are others in the infrared, the one closest to the visible portion of the spectrum being the Paschen series. There wavelengths are given by the following formulas: 1 1 1 Lyman: 𝜆 = 𝑅 (12 − 𝑛2 ) 𝑛 = 2, 3, … 1 1 1 Balmer: 𝜆 = 𝑅 (22 − 𝑛2 ) 𝑛 = 3, 4, … 1 1 1 Paschen: 𝜆 = 𝑅 (32 − 𝑛2 ) 𝑛 = 4, 5, … Where R = 1.0974 × 107 𝑚−1 is called the Rydberg constant. 19 Figure 10: The Balmer series—the spectral lines in the visible region of hydrogen's emission spectrum— corresponds to electrons relaxing from n=3-6 energy levels to the n=2 energy level. 20 Figure 11: Emission of radiation from atoms EXAMPLE When a hydrogen atom is bombarded, the atom may be raised into a higher energy state. As the excited electron falls back to the lower energy levels, light is emitted. What are the three longest wavelength spectral lines emitted by the hydrogen atom as it returns to the n = 1 state from higher energy state? Give your answers to three significant figures. SOLUTION Using Figure 11. 𝑛 = 2 → 𝑛 = 1: ∆𝐸2,1 = −3.4 − (−13.6) = 10.2 𝑒𝑉 𝑛 = 3 → 𝑛 = 1: ∆𝐸3,1 = −1.5 − (−13.6) = 12.1 𝑒𝑉 𝑛 = 4 → 𝑛 = 1: ∆𝐸4,1 = −0.85 − (−13.6) = 12.8 𝑒𝑉 To fine the corresponding wavelengths 21  ℎ𝑐 ∆𝐸 = ℎ𝑓 = 𝜆 For n = 2 to n = 1 transition, ℎ𝑐 (6.63 × 10−34 𝐽.𝑠)(2.998 × 108 𝑚/𝑠) 𝜆= Δ𝐸2,1 = (10.2 𝑒𝑉)(1.60 × 10−19 𝐽/𝑒𝑉) = 1.22 𝑛𝑚 The other lines are found in the same way to be 102 nm and 96.9 nm. These are the first three lines of the Lyman series. EXAMPLE The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from the n = ∞ state to the n = 2 state. What is the wavelength of this line (to three significant figures)? SOLUTION From Figure 11, ∆𝐸 = 3.40 − 0 = 3.40 𝑒𝑉. ℎ𝑐 The corresponding wavelength, 𝜆 = Δ𝐸 = 365 𝑛𝑚 EXAMPLE Unexcited hydrogen atoms are bombarded with electrons that have been accelerated through 12.0 V. What wavelengths will the atoms emit? SOLUTION When an atom in the ground state is given 12.0 eV of energy, the most these electrons can supply, the atom can be excited no higher than 12.0 eV above the ground state. Only one state exists in this energy region, the n = 2 state. Hence, the only transition possible is 𝑛 = 2 → 𝑛 = 1: ∆𝐸2,1 = 13.6 − 3.4 = 10.2 𝑒𝑉 The only emitted wavelength will be 1.00 𝑒𝑉 𝜆 = (10.2 𝑒𝑉) (1240 𝑛𝑚) = 122 𝑛𝑚 Which is the longest-wavelength line in the Lyman series. EQUATIONS OF THE RUTHERFORDS MODEL We shall be looking at two equations: 1. The distance of closest approach, b or the impact parameter in the Rutherford’s experiment is related to the scattering angle 𝜃 by the formula: 𝜃 4𝜋𝜀𝑜 𝑏𝑇 𝐶𝑜𝑡 = 2 𝑍𝑒 2 22 Where T = KE of the ∝ −particle e = electronic charge Z = atomic number of the target material (gold) The distance of closest approach is the closest or nearest distance the ∝ −particle comes nearer to the nucleus. 2. The number of ∝ −particle scattered through an angle 𝜃 in one minute is given by the equation 7.9×10−24 N(θ) = θ T2 Sin4 2 Example Calculate the impact parameter for a beam of alpha particle whose scattering angle is 460 and kinetic energy 10 eV, if gold of atomic number 79 is used. Solution 𝜃 𝑍𝑒 2 𝐶𝑜𝑡 2 𝑏= 4𝜋𝜀𝑜 𝑇 79 ×(1.6 ×10−19 ×𝐶𝑜𝑡 230 ×9×109 ) = 10×1.6×10−19 = 2.68 × 10−8 𝑚 EXPLANATION OF CERTAIN TERMS USED IN ENERGY QUANTIZATION 1. Ground State: This is the least or smallest energy state of the atom. It is the state with energy of the level for n = 1, where n = principal quantum number. Atom are most stable in the ground state. Electrons have the tendency to remain in the ground state. 2. Excited State: This is the state other than the ground state, that is a state for n = 2 and above. Electron has more energy in the excited state and hence, the atom is unstable. Energy is liberated when an electron goes from a higher level to a lower level. 3. Excitation Energy: This is the energy required or absorbed to raise the state of the atom. It is the difference in energy between two given levels. The energy is quantized. Excitation energy ∆𝐸 is ℎ𝑐 represented by the expression: ∆𝐸 = 𝐸𝑖 − 𝐸𝑓 = ℎ𝑓 = 𝜆 , 𝑖 and 𝑓 are initial and final level respectively e.g. for levels 1 and 2, ∆𝐸 = 𝐸2 − 𝐸1. If 𝐸1 = −13.6 𝑒𝑉 and 𝐸2 = −3.43 𝑒𝑉 then ∆𝐸 = −3.43 + 13.6 = 10.2 𝑒𝑉. 4. Ionization Energy: This is the minimum energy required to remove an electron from a given atom to infinity. For the hydrogen atom, the ionization energy occurs for n = 1 and 𝐸1 = −13.6 𝑒𝑉 The outermost electrons in an atom are easily removed. 5. Band Spectra: This is obtained when the atom makes a transition from one level to another. It is unresolved lines joining together. 23 QUANTUM MECHANICS OF THE HYDROGEN ATOM; QUANTUM NUMBER: Quantum mechanics is a much more sophisticated and successful theory than Bohr’s. Yet in a few details, they agree. Quantum mechanics predicts the same basic energy levels for hydrogen atom, as does the Bohr Theory. That is, 13.6 𝑍2 𝑒𝑉 𝐸𝑛 = − 𝑛2 , 𝑛 = 1,2,3, … , ----------------------------- (*) The lowest energy level (n = 1) for hydrogen (Z = 1) is 𝐸1 = −13.6 𝑒𝑉 Since 𝑛2 appears in the denominator from equation (*), the energies of the larger orbits in hydrogen (Z = 1) are given by 13.6 𝑒𝑉 𝐸𝑛 = − , 𝑛 = 1,2,3, … , 𝑛2 where n is an integer There are four quantum numbers used to completely describe the structure of all atoms. These are: a. Principal Quantum Number, n: This is used to determine the energy and energy levels of the electrons with values 𝑛 = 1, 2, 3, … ∞ b. Orbital Quantum Number, 𝒍: is related to the magnitude of the angular momentum of the electron; 𝑙 can take an integer values from 0 to (n - 1) that is, 0, 1, 2, … , 𝑛 − 1. For the ground state, 𝑛 − 1, 𝑙 can only be zero. For 𝑛 = 3, 𝑙 can be 0.1 or 2. The actual magnitude of the angular momentum 𝐿 is related to the quantum number 𝑙 by ℎ 𝐿 = √𝑙(𝑙 + 1) ℏ , Where ℏ = 2𝜋 c. Magnetic Quantum Number, 𝒎 𝒍 : This specifies the number of orbitals present in each sub- energy level. It also defines the possible orientations of the angular momentum in space. It can take on integer values ranging from – 𝑙 to +𝑙. For example, if 𝑙 = 2, then 𝑚𝑙 can be −2, −1, 0, +1, or +2. Since angular momentum is a vector, it is not surprising that both its magnitude and its direction would be quantized. For 𝑙 = 2, the five different directions allowed can be represented by the diagram below. This limitation on the direction of 𝐿⃗ is often called space quantization. In quantum mechanics, the direction of the angular momentum is usually specified by giving its component along the 𝑧 axis (this choice is arbitrary). Then 𝐿𝑧 is related to 𝑚𝑙 by the equation 𝐿𝑧 = 𝑚𝑙 ℏ The value of 𝐿𝑥 𝑎𝑛𝑑 𝐿𝑦 are not definite, however. The name for 𝑚𝑙 derives not from theory (which relate it to 𝐿𝑧 ), but from experiment. It was found that when a gas-discharge tube was placed in a magnetic field, the spectral lines were split into several very closely spaced lines. This splitting, known as the Zeeman Effect, implies that the energy levels must be split, and thus that the energy of a state depends not only on 𝑛 but also on 𝑚𝑙 when a magnetic field is applied – hence the name magnetic quantum number. 24  Figure 12: Quantization of angular momentum direction for 𝑙 = 2 d. Spin Quantum Number, 𝒎𝒔 : This takes care of the spin of the electron about its axis as it 1 1 moves round the nucleus. The two possible values of 𝑚𝑠 (+ 2 𝑎𝑛𝑑 − 2) are often said to be spin up and spin down, referring to the two possible directions of the spin angular momentum Table 2: Quantum Numbers for an Electron Name Symbol Possible values Principal 𝑛 1, 2, 3, … , ∞ Orbital 𝑙 For a given 𝑛: 𝑙 can be 0, 1, 2, … , 𝑛 − 1. Magnetic 𝑚𝑙 For a given 𝑛 and 𝑙: 𝑚𝑙 can 𝑏𝑒 𝑙, 𝑙 − 1, … , 0, … , −1 Spin 𝑚𝑠 1 1 For each set of 𝑛, 𝑙, and 𝑚𝑙 : 𝑚𝑠 can 𝑏𝑒 + 2 𝑜𝑟 − 2. 25  Figure 3: Zeeman Effect EXAMPLE How many different states are possible for an electron whose principal quantum number is 𝑛 = 3? SOLUTION For n = 3, 𝑙 can have the values 𝑙 = 2, 1, 0. 𝑓𝑜𝑟 𝑙 = 2, 𝑚𝑙 can be 2, 1, 0, −1, −2, which is five different 1 1 possibilities. For each of these, 𝑚𝑠 can be either up or down (+ 2 𝑎𝑛𝑑 − 2); so for 𝑙 = 2, there are 2 × 1 1 5 = 10 states. For 𝑙 = 1, 𝑚𝑙 can be 1, 0, -1, and since 𝑚𝑠 = + 2 𝑎𝑛𝑑 − 2 for each of these, we have 6 more possible states. Finally, for 𝑙 = 0, 𝑚𝑙 can only be 0, and these are only 2 states corresponding to 1 1 𝑚𝑠 = + 2 𝑎𝑛𝑑 − 2. The total number of states is 10 + 6 + 2 = 18, as detailed in the table below. 26  𝑛 𝑙 𝑚𝑙 𝑚𝑠 𝑛 𝑙 𝑚𝑙 𝑚𝑠 3 2 2 1 3 1 1 1 2 2 3 2 2 1 3 1 1 1 − − 2 2 3 2 1 1 3 1 0 1 2 2 3 2 1 1 3 1 0 1 − − 2 2 3 2 0 1 3 1 −1 1 2 2 3 2 0 1 3 1 −1 1 − − 2 2 3 2 −1 1 3 0 0 1 2 2 3 2 −1 1 3 0 0 1 − − 2 2 3 2 −2 1 2 3 2 −2 1 − 2 EXAMPLE 𝐸 and 𝐿 for n = 3. Determine (a) the energy and (b) the orbital angular momentum for an electron in each hydrogen atom states of the example above. SOLUTION (a) The energy of a state depends only on n, except for the very small corrections mentioned above, which we will ignore. Energy is calculated as in the Bohr theory, 13.6 𝑒𝑉 13.6 𝑒𝑉 𝐸𝑛 = − 𝑛2 = − 32 = −1.51 𝑒𝑉 (b) Using 𝐿 = √𝑙(𝑙 + 1) ℏ For 𝑙 = 0 𝐿 = √𝑙(𝑙 + 1) ℏ = 0 For 𝑙 = 1 𝐿 = √𝑙(1 + 1) ℏ = √2ℏ = 1.49 × 10−34 J. s (c) For 𝑙 = 2 𝐿 = √2(2 + 1) ℏ = √6ℏ Note: Atomic angular momentum are generally given as a multiple of ℏ (√2ℏ or √6ℏ in this case), rather than in SI unit. 27 MAGNETIC MOMENT OF AN ORBITING ELECTRON From Zeeman Effect and magnetic dipole moment 𝜇 = 𝐼𝐴 -------------------------------------------------------------------- (1) That a plane current loop with vector 𝐴 carrying current 𝐼 has a magnetic moment 𝜇 given in equation (1). When a magnetic dipole moment 𝜇 is placed in a magnetic field 𝐵 ⃗ , the field exerts a torque 𝜏 = 𝜇 × 𝐵 ⃗ on the dipole. The potential energy U associated with this interaction is given by ⃗ = −𝜇𝐵𝑐𝑜𝑠 𝜙 U = −𝜇. 𝐵 (potential energy for a magnetic dipole) ⃗ --------------------------------------------------------------(2) So U = 𝜇. 𝐵 Using equation (1) and (2) and the Bohr model to look at the interaction of hydrogen atom with a magnetic field. The orbiting electron (speed, v) is equivalent to a current loop with radius r and area 𝜋𝑟 2. The average current 𝐼 is the average charge per unit time equal to the charge magnitude e divided by the time T for one revolution, given by 2𝜋𝑟 𝑇= 𝑉 𝑒𝑉 Thus 𝐼 = 2𝜋𝑟 and from eqn (1), the magnitude 𝜇 of the magnetic moment is 𝑒𝑉 𝑒𝑉𝑟 𝜇 = 𝐼𝐴 = 2𝜋𝑟 𝜋𝑟 2 = 2 ------------------------------------------------------------------- (3) We can also express this in terms of the magnitude L of the orbital angular momentum From 𝐿 = 𝑚𝑣𝑟 for a particle in a circular orbit, so 𝑒 𝜇 = 2𝑚 𝐿 ------------------------------------------------------------------------ (4) The ratio of the magnitude of 𝜇 to the magnitude of 𝐿⃗ is 𝜇 𝑒 𝐿 = 2𝑚 and is called the gyromagnetic ration In the Bohr model, 𝑛ℎ 𝐿 = 2𝜋 = 𝑛ℏ, where n = 1, 2, … for n = 1 state (a ground state) Eqn (4) becomes 𝑒 𝜇 = 2𝑚 ℏ , This quantity is a natural unit for magnetic moment; it is called one Bohr magneton denoted by 𝜇𝐵 𝑒ℏ 𝜇𝐵 = 2𝑚 ------------------------------------ (5) (definition of the Bohr magneton) Evaluating (5) and (4) gives 28  𝑒𝑉 𝜇𝐵 = 5.788 × 10−5 𝑇 = 9.274 × 10−24 𝐽/𝑇 or A.m 2 ⃗ is directed along the + Z – axis. Then from eqn (2), the interaction energy U Suppose the magnetic field 𝐵 of the atom’s magnetic moment with the field is 𝑈 = −𝜇𝑧 𝐵------------------------------------------------------------------- (6) Where 𝜇𝑧 is the Z- component of the vector 𝜇 Now we use eqn (4) to find 𝜇𝑧 , recalling that e is the magnitude of the electron charge and that the actual charge is – e. Because the electron charge is negative, the orbital angular momentum and magnetic moment vectors are opposite. We find 𝑒 𝜇𝑧 = − 2𝑚 𝐿𝑧 ------------------------------------------------- (7) For the Schrodinger wave functions, 𝐿𝑧 = 𝑚𝐿 ℏ, with 𝑚𝐿 = ±1, ±2, … ± 𝑙 so 𝑒 𝑒ℏ 𝜇𝑧 = − 2𝑚 𝐿𝑧 = −𝑚𝑙 2𝑚 --------------------------------- (8) Finally, we can express the interaction energy (6), as 𝑒ℏ 𝑈 = −𝜇𝑧 𝐵 = −𝑚𝐿 2𝑚 B ------------------------------------------ (9) 𝑚𝑙 = 0, ±1, ±2, … ± 𝑙 (Orbital magnetic interaction energy) 𝑒ℏ In terms of Bohr magneton 𝜇𝐵 = 2𝑚 𝑈 = 𝑚𝐿 𝜇𝐵 𝐵 (Orbital magnetic interaction energy) ------------------------ (10) Example An atom in a state with 𝑙 = 1 emits a photon with wavelength 600.00 nm as it decays to a state with 𝑙 = 0. If the atom is placed in a magnetic field with magnitude B = 2.00 T, determine the shifts in the energy levels and in the wavelength resulting from the interaction of the magnetic field and the atom’s orbital magnetic moment. Solution The energy of a 600 nm photon is ℎ𝑐 4.14 ×10−15 𝑒𝑉 ×3×108 𝑚/𝑠 𝐸= 𝜆 = 600×10−9 𝑚 = 2.07𝑒𝑉 If no net magnetic field, the difference in energy between 𝑙 = 0 and 𝑙 = 1 levels is 2.07 eV. With a 2.00 T field present equation (10) shows that there is no shift of the 𝑙 = 0 state (which has 𝑚𝑙 = 0) 29 For the 𝑙 = 1 state, the splitting of the levels is given by 𝑒𝑉 𝑈 = 𝑚𝑙 𝜇𝐵 𝐵 = 𝑚𝑙 (5.788 𝑋 10−5 )(2.00 𝑇) 𝑇 = 𝑚𝑙 (1.16 𝑋 10−4 𝑒𝑉) = 𝑚𝑙 (1.85 𝑋 10−23 𝑒𝑉) When 𝑙 = 1, the possible values of 𝑚𝑙 are – 1, 0, and + 1 and the three corresponding levels are separated by equal intervals of 1.16 𝑋 10−4 𝑒𝑉. This is a small fraction of the 2.07 eV photon energy 1.16×10−4 𝑒𝑉 2.07 𝑒𝑉 = 5.60 × 10−5 The corresponding wavelength shift are approximately (5.60 × 10−5 )(600 𝑛𝑚) = 0.034 𝑛𝑚. The original 600.000 nm line is split into a triplet with wavelength 599.966, 600,000, and 600, 034 nm THE STERN-GERLACH EXPERIMENT When a beam of neutral atom is passed through a non-uniform magnetic field, some anomalies are observed. Atoms were deflected according to the orientation of their magnetic moments with respect to the field. The Stern- Gerlach experiment demonstrate the quantization of angular momentum in a very direct way. If there were only orbital angular momentum, the deflections would split the beam into an odd number (2𝑙 + 1) of different components. However, some atomic beams were split into an even number of component. If we use a different symbol 𝑗 for an angular momentum quantum number, setting 1 3 5 2𝑗 + 1 equal to an even number given 𝑗 = 2 , , 2 2 ,... , suggesting a half integer angular momentum. 30