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“Atomic Structure” Illustrative problem 1 Radio city broadcasts on a frequency of 5,090 KHz.What is the wavelength of electromagnetic radiation emitted by the transmitter? c λ= ν c is 3 × 10 8 m / s 3 × 108 λ= = 5090 × 103 0.589 × 102 = 58.9 m Illustrative Problem 2 The ratio of the energy of a photon of 2000 Å wavelength radiation to that of 6000 Å (a) ¼ (b) ½ radiation is (b) 4 (d) 3 Solution: hc E = hν = λ hc hc = E1 = E2 λ1 λ2 0 6000 A E1 λ 2 3 ∴ == = 0 E2 λ1 2000 A Hence, answer is (d). Illustrative Problem 3 The energy of the electron in the second and third Bohr orbits of the hydrogen atom is -5.42 X 10-12 and –2.41 X 10-12 respectively. Calculate the wavelength of the emitted radiation, when the electron drops from third to second orbit. Solution ∆E = E2 − E1 =−5.42 × 10−12 − (−2.41 × 10−12 )ergs = −3.01 × 10−12 ergs. According to Planck’s quantum theory E = hυ E =h c λ = h 6.6 × 10−27 ergs Solution c ∴λ = h E = 6.62 × 10−27 × 3 × 108 3.01 × 10−12 = λ 6.6 × 10−5 cm = 6.6 × 103 A0 1 A0 = 10−8 cm Class Exercise - 1 Which of the following fundamental particles are present in the nucleus of an atom? (a) Alpha particles and protons (b) Protons and neutrons (c) Protons and electrons (d) Electrons, protons and neutrons Solution The nucleus of an atom is positively charged and almost the entire mass of the atom is concentrated in it. Hence, it contains protons and neutrons. Hence, answer is (b). Class Exercise - 2 The mass of the proton is (a) 1.672 × 10–24 g (b) 1.672 × 10–25 g (c) 1.672 × 1025 g (d) 1.672 × 1026 g Solution The mass of the proton is 1.672 × 10–24 g Hence, answer is (a). Class Exercise - 3 Which of the following is not true in case of an electron? (a) It is a fundamental particle (b) It has wave nature (c) Its motion is affected by magnetic field (d) It emits energy while moving in orbits Solution An electron does not emit energy while moving in orbit. This is so because if it would have done that it would have eventually fallen into the nucleus and the atom would have collapsed. Hence, answer is (d). Class Exercise - 4 Positive charge of an atom is (a) concentrated in the nucleus (b) revolves around the nucleus (c) scattered all over the atom (d) None of these Solution Positive charge of an atom is present entirely in the nucleus. Hence, answer is (a). Class Exercise - 5 Why only very few a-particles are deflected back on hitting a thin gold foil? Solution Due to the presence of a very small centre in which the entire mass is concentrated. Class Exercise - 6 Explain why cathode rays are produced only when the pressure in the discharge tube is very low. Solution This is happened because at higher pressure no electric current flows through the tube as gases are poor conductor of electricity. Class Exercise - 7 If a neutron is introduced into the nucleus of an atom, it would result in the change of (a) number of electrons (b) atomic number (c) atomic weight (d) chemical nature of the atom Solution Neutrons contribute in a major way to the weight of the nucleus, thus addition of neutron would result in increase in the atomic weight. Hence, answer is (c). Class Exercise - 8 The concept of stationary orbits lies in the fact that (a) Electrons are stationary (b) No change in energy takes place in stationary orbit (c) Electrons gain kinetic energy (d) Energy goes on increasing Solution When an electron revolves in a stationary orbit, no energy change takes place. Energy is emitted or absorbed only when the electron jumps from one stationary orbit to another. Hence, answer is (b). Class Exercise - 9 What is the energy possessed by 1 mole of photons of radiations of frequency 10 × 1014 Hz? Solution E = hv E = 6.6 × 10–34 × 10 × 1014 E = 66 × 10–20 = 6.6 × 10–19 joules ∴ energy of 1 mole of photons = 6.6 × 10–19 × 6.023 × 1023 = 39.7518 × 104 = 397.518 kJ/mol The Atomic Mass Scale and Mass Weights atomic Mass = (0.691)(62.9 amu) 63 Cu isotope 16 The Atomic Weight Scale and Atomic Weights atomic weight = (0.691)(62.9 amu) + (0.309)(64.9 amu) 63 Cu isotope 17 65 Cu isotope The Atomic Weight Scale and Atomic Weights atomic weight = (0.691)(62.9 amu) + (0.309)(64.9 amu) 63 Cu isotope atomic weight = 63.5 amu for copper 18 65 Cu isotope Atomic Number Atomic number (Z) of an element is the number of protons in the nucleus of each atom of that element. Element # of protons Atomic # (Z) Carbon 6 6 Phosphorus 15 15 Gold 79 79 Question 2 The table on the next slide shows the five isotopes of germanium found in nature, the abundance of each isotope, and the atomic mass of each isotope. Calculate the atomic mass of germanium. Answer 72.591 amu The wavelength of the peak in the intensity graph is given by Wien’s law (T must be in kelvin): Wien’s Displacement Law EXAMPLE: Finding peak wavelengths QUESTIONS: EXAMPLE :Finding peak wavelengths EXAMPLE : Finding peak wavelengths EXAMPLE :Finding peak wavelengths The Electron Volt • Consider an electron accelerating (in a vacuum) from rest across a parallel plate capacitor with a 1.0 V potential difference. • The electron’s kinetic energy when it reaches the positive plate is 1.60 x 10−19 J. • Let us define a new unit of energy, called the electron volt, as 1 eV = 1.60 x 10−19 J. Energy of an electron QUESTION: EXAMPLE 38.5 Energy of an electron EXAMPLE 38.5 Energy of an electron EXAMPLE : Energy of an electron Example 1 Write out the electron configuration of the elements (i) carbon (Z = 6) (ii) magnesium (Z = 12) (iii) argon (Z = 18) Solution (i) 1s2 2s2 2p2 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p6 Example 2 Determine which of the following are in the excited state (i) 1s2 2s2 2p5 3s1 (ii) 1s2 2s2 2p6 3s2 3p6 4s1 (iii) 1s2 2s2 2p3 (iv) 1s2 2s2 2p6 3s1 3p5 Solution (i) and (iv) are in excited states Mass Number Mass number is the number of protons and neutrons in the nucleus of an isotope: Mass # = p+ + n0 p+ n0 e- Mass # 8 10 8 18 Arsenic - 75 33 42 33 75 Phosphorus - 31 15 16 15 31 Nuclide Oxygen - 18 Information in the Periodic Table The number at the bottom of each box is the average atomic mass of that element. This number is the weighted average mass of all the naturally occurring isotopes of that element. Question 1 How does the atomic number of an element differ from the element’s mass number? Answer The atomic number of an element is the number of protons in the nucleus. The mass number is the sum of the number of protons and neutrons. The Atomic Weight Scale and Atomic Weights The atomic weight of an element is the weighted average of the masses of its stable isotopes Example 5-2: Naturally occurring Cu consists of 2 isotopes. It is 69.1% 63Cu with a mass of 62.9 amu, and 30.9% 65Cu, which has a mass of 64.9 amu. Calculate the atomic weight of Cu to one decimal place. Calculating Atomic Mass Calculating Atomic Mass Copper exists as a mixture of two isotopes. The lighter isotope (Cu-63), with 29 protons and 34 neutrons, makes up 69.17% of copper atoms. The heavier isotope (Cu-65), with 29 protons and 36 neutrons, constitutes the remaining 30.83% of copper atoms. Calculating Atomic Mass The atomic mass of Cu-63 is 62.930 amu, and the atomic mass of Cu-65 is 64.928 amu. Use the data above to compute the atomic mass of copper. Calculating Atomic Mass First, calculate the contribution of each isotope to the average atomic mass, being sure to convert each percent to a fractional abundance. Calculating Atomic Mass The average atomic mass of the element is the sum of the mass contributions of each isotope. Can you tell me: how many atoms are there in 43.2 grams of iron? Using unit analysis: The unit of the answer is atoms. The unit of the information given is grams. But there is no conversion factor for grams ---> atoms. Reason: the mass of an atom is different for every element. Solution: use the mole concept. 6.02x10 23 atoms Fe 1 mol Fe = 4.66x 10 23 atoms Fe x 43.2g Fe x 1 mol Fe 55.85 g Fe Electron Orbits Consider the planetary model for electrons which move in a circle around the positive nucleus. The figure below is for the hydrogen atom. r FC - e- Coulomb’s law: + Nucleus mv 2 e2 = r 4πε 0 r 2 FC = Centripetal FC: e 2 4πε 0 r 2 Radius of Hydrogen atom mv FC = 2 r r= 2 e 2 4πε 0 mv 2 Example 1: Use the Balmer equation to find the wavelength of the first line (n = 3) in the Balmer series. How can you find the energy? 1 1 R 2 + 2 ; n = 3 R = 1.097 x 107 m-1 = λ 2 n 1 1 1 1 = R 2 + 2 = R(0.361); λ= 0.361R λ 2 3 1 λ = 656 nm λ= 7 -1 0.361(1.097 x 10 m ) 1 The frequency and the energy are found from: c = fλ and E = hf Example 2: Find the radius of the Hydrogen atom in its most stable state (n = 1). n ε 0h rn = π me 2 2 2 r= m = 9.1 x 10-31 kg 2 (1) (8.85 x 10 e = 1.6 x 10-19 C -12 Nm 2 C2 -31 π (9.1 x 10 r = 5.31 x 10-11 m )(6.63 x 10−34 J ⋅ s) 2 kg)(1.6 x 10-19 C) 2 r = 53.1 pm Total Energy of an Atom The total energy at level n is the sum of the kinetic and potential energies at that level. E= K +U; But we recall that: 2 vn = e 2ε 0 nh 2 1 K= mv ; 2 n ε 0h rn = π me 2 Total energy of Hydrogen atom for level n. 2 2 e2 U= 4πε 0 r Substitution for v and r gives expression for total energy. me 4 En = − 2 2 2 8ε 0 n h Energy for a Particular State It will be useful to simplify the energy formula for a particular state by substitution of constants. m = 9.1 x 10-31 kg e = 1.6 x 10-19 C εo = 8.85 x 10--12 C2/Nm2 h = 6.63 x 10-34 J s me 4 (9.1 x 10-31kg)(1.6 x 10-19 C) 4 En = − 2 2 2 = − -12 C2 2 2 8ε 0 n h 8(8.85 x 10 Nm2 ) n (6.63 x 10-34 Js) 2 2.17 x 10-18 J En = − 2 n Or −13.6 eV En = n2 Balmer Revisited Total energy of Hydrogen atom for level n. me 4 En = − 2 2 2 8ε 0 n h Negative because outside energy to raise n level. When an electron moves from an initial state ni to a final state nf, energy involved is: 1 = λ me 4 2 3 2 8ε 0 h cn f 1 1 me 4 2 − 2 ; If R= 2 3 2 8ε 0 h cn f n f ni hc = E= E0 − E f ; λ Balmer’s Equation: 1 −me 4 me 4 = 2 2 2 + 2 2 2 λ hc 8ε 0 h n0 8ε 0 h n f 1 1 1 = 1.097 x 107 m -1 R 2 − 2 ; R = n n λ 0 f 1 Energy Levels We can now visualize the hydrogen atom with an electron at many possible energy levels. Emission Absorption The energy of the atom increases on absorption (nf > ni) and de-creases on emission (nf < ni). Energy of nth level: −13.6 eV E= n2 The change in energy of the atom can be given in terms of initial ni and final nf levels: 1 1 −13.6 eV 2 − 2 E= n n 0 f Spectral Series for an Atom The Lyman series is for transitions to n = 1 level. The Balmer series is for transitions to n = 2 level. The Pashen series is for transitions to n = 3 level. The Brackett series is for transitions to n = 4 level. n =1 n =2 n =3 n =4 n =5 n =6 1 1 −13.6 eV 2 − 2 E= n f n0 Example 3: What is the energy of an emitted photon if an electron drops from the n = 3 level to the n = 1 level for the hydrogen atom? 1 1 −13.6 eV 2 − 2 E= n n 0 f Change in energy of the atom. 1 1 E= −13.6 eV 2 − 2 = −12.1 eV 1 3 ∆E = -12.1 eV The energy of the atom decreases by 12.1 eV as a photon of that energy is emitted. You should show that 13.6 eV is required to move an electron from n = 1 to n = ∞. Summary (Cont.) Spectrum for nf = 2 (Balmer) 434 nm n=3 n=4 Emission spectrum 653 nm n 6 n=5 486 nm 410 nm The general equation for a change from one level to another: Balmer’s Equation: 1 1 = 1.097 x 107 m -1 R 2 − 2 ; R = n n λ 0 f 1 Questions 1. a. Sketch a cathode-ray tube and label it with the following parts: partially evacuated glass vessel, electrically charged plates, cathode, anode, fluorescent screen, high voltage source. b. Explain how it works. Questions 2. Do you think the hypothesis inferring from the observations are reasonable? Why or why not? Explain in each hypothesis Questions 3. How can the observations from the cathode ray tube experiment lead to the new atomic model “the plum pudding model” Plum Pudding Model • in 1904 J.J. Thomson proposed the plum pudding model of atom. • According to this model the electrons were randomly stuck into a ball of positively charge matter. positively charge matter electrons © 2004 by Pearson Education Inc. • A more modern name for this model might be chocolate-chip muffin model. Discovering the Electron 57 The Discovery of the Electron Thomson found q/m for cathode ray particles to be 1.76 x 1011 C/kg This is close to 2000 x q/m for a hydrogen ion Thomson’s conclusion: cathode-ray particles (today called electrons) are approximately 1/2000 the mass of a hydrogen ion (today called protons) The Discovery of the Electron Since Thomson didn’t know the charge or mass of these particles couldn’t use qV = 1 2 mv2 to determine their speed He balanced Fe versus Fm was undeflected so that particle path The Discovery of the Electron Then, Fe = Fm q E = qv B E v= B Using only a magnetic field, Fm = Fc v2 qv B = m r q v = m Br The Discovery of the Electron Examples: Practice Problem 1, = v E 6.0 × 10 4 N C 4 = = 2.4 × 10 2.50 T B m s Practice Problem 1, 1.0 × 106 m s q v = = = 1.0 × 108 C kg m B r 1.0 T × 0.0100 m √ The potential energy of electrons is converted to kinetic energy Since change in energy is the Electrons are times attracted positively charged voltage the to charge plate. then They½mv²=qV accelerate towards it and small percentage escape plate through small Therefore v=the √(2qV/m) hole, creating electron beam. 120 Volts Plate is heated and electrons boil off. Velocity= 0 Potential Energy= ½ mv^2 6.3 Volts We now know that v= √(2qV/m), so we can now easily find the velocity of our beam of electrons. q(charge) of an electron= -1.6•10^-19 V(volts)=120 m(mass) of an electron=9.11•10^-31 kg Therefore: v=√(2)(-1.6•10^-19)(120)/(9.11•10^-31) v=√4.215•10^13 v=649•10^6 m/s In order to predict the angle at which the electrons are deflected, we must first measure the force that the magnetic field inserts upon the beam To do this, we use the equation: F=qvB Magnetic field The force is always Perpendicular to the magnetic field And the velocity of the electrons Electrons Like Solar Wind, the electrons in the CRT beam are deflected when entering a magnetic field, therefore the electron beam “bends.” In order to find the force of the magnetic field, we must first calculate its strenghth. Since F=qvB and, according to Newton’s second law, F=m•v²/r, we can deduce that qvB=m•v²/r Or B=mv/qr mass= 9.11•10^-31 kg velocity= 6.492•10^6 m/s Charge= 1.6•10^-19 C And if the measured distance of the electron beam from the magnets Is .075 meters Then B= (9.11•10^-31)(6.492•10^6)/(1.6•10^-19)(.075) B=2.772•10^-6 tesla Now that we know the strength of the magnetic field at the electron beam, we can Calculate the force which the field exerts upon the electrons. F=qvB F=(6.49•10^6) (1.6•10^-19)(2.772•10^-6 F=2.879•10^-18 N Examples 1. In a Cathode ray tube an electron travels through an electrical field of 5.80 x 103 N/C and a magnetic field of 6.60 x 10-3 T undeflected. a. what is the velocity of the electron? b. If the electric field is turned off the magnetic field causes the electron to travel in a circular path, what is the radius of the circular path of the electron? 2. An electron travels through a CRT and a magnetic field of 4.20 x 10-3 T, the electron travels a circular path with a radius of 4.90 x 10-3 m. a. What is the velocity of the electron? b. What was the potential difference in the CRT? 3. An unknown particle travels in a CRT tube through a magnetic field of 4.60 x 10-2 T at a velocity of 5.20 x 105 m/s, the magnetic field causes the particle to travel in a circular path with a radius of 0.235 m. What is the charge to mass ratio of this particle? Deflection of Electrons in a Uniform Electric Field (1) Consider an electron beam directed between two oppositely charged parallel plates as shown below. With a constant potential difference between the two deflecting plates, the trace is curved towards the positive plate. + d - Deflection of Electrons in a Uniform Electric Field (2) The force acting on each electron in the field is given by eVP F = eE = d where E = electric field strength, Vp = p.d. between plates, d = plate spacing. Deflection of Electrons in a Uniform Electric Field (3) The vertical displacement y is given by 1 2 1 eV p 2 y = at = ( )t 2 2 md 1 eV p x 2 = ( ) 2 2 md v This is the equation for a parabola. Deflection of electrons in a uniform magnetic field Deflection of Electrons in a Uniform Magnetic Field The force F acting on an electron in a uniform magnetic field is given by F = Bev Since the magnetic force F is at right angles to the velocity direction, the electron moves round a circular path. Deflection of Electrons in a Uniform Magnetic Field The centripetal acceleration of the electrons is Bev a= m v 2 Bev Hence a = = r m mv r= eB which gives Determination of Specific Charge Using a Fine Beam Tube Ber Since v = m (For an electron moving in a uniform magnetic field) and the kinetic energy of the electron provided by the electron gun is 1 2 mv = eV 2 Where V is the anode voltage. Determination of Specific Charge Using a Fine Beam Tube (3) 1 Ber 2 So ) = eV m( 2 m Rearrange the equation gives e 2V = 2 2 m B r The value of the specific charge of an electron is now known accurately to be (1.758803 ± 0.000003) × 1011 C/kg Thomson’s e/m Experiment (1) Thomson’s apparatus for measuring the ratio e/m + × × v × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × - Thomson’s e/m Experiment (2) A beam of electron is produced by an electron gun with an accelerating voltage V. The electron beam is arranged to travel through an electric field and a magnetic field which are perpendicular to each other. The apparatus is set-up so that an electron from the gun is undeflected. Thomson’s e/m experiment (3) As the electron from the gun is undeflected, this gives FE = FB i.e. eE = Bev On the other hand, eE E ⇒v= B v Bev 1 mv 2 = eV 2 Combining the equations, we get 2 e E = m 2VB 2 A F B +Q +V d To move the charge from A to B a force equal to F would be needed. The work done moving Q from A to B is W. W = Fd But E = F/Q or F = EQ So, W = EQd V Potential difference is defined as work done per unit charge: V = W/Q d position So: V = W/Q = EQd /Q V = Ed E = V/d (Uniform fields) Acceleration: gravitational and electrical An accelerating charge A falling ball starts at high gravitational potential energy gravitational field g starts at high electrical potential energy charge q mass m height h electric field + gains kinetic energy potential difference V gains kinetic energy ends at low gravitational potential energy + – ends at low electrical potential energy + electric potential gravitational potential ball falls down gravitational potential hill ΔEp=mgΔh gain of kinetic energy = loss of potential energy = mgh charge ‘falls down’ electrical potential hill ΔW = QΔV gain of kinetic energy = loss of potential energy = qV An electric field accelerates a charge as a gravitational field accelerates a mass E Can you come up with an expression for the charge on the droplet? Identify the forces on a charge in an electric field Draw the forces acting on the drop F= Eq mg q= E Can you come up with an expression for the charge on the droplet? F= mg Identify the forces on a charge in an electric field Draw the forces acting on the drop electric force upward = weight downward EQ = mg Q = mg/E = 1.8 x 10-15 x 9.81 / 1.9 x 104 = 9.4 x 10-19 C 1) ΔEp=m ΔW = Q gΔh ΔV Quantization of Charge Example: Practice Problem 1, page 763 (Pearson Physics) Fe = Fg qE =mg q × 5.0 × 105 N C = 2.4 × 10 −14 kg × 9.81m s 2 q 4.7 × 10 −19 C = 4.7 × 10 −19 C − e = 3 1.60 × 10 −19 C e− Example A beam of electrons travels an undeflected path in a cathode ray tube. E is 7.0 x 103 N/C. B is 3.5 x 10-2 T. What is the speed of the electrons as they travel through the tube? What we know: E = 7.0 x 103 N/C B=3.5 x 10-2 T Equation: v = E/B Substitute: v = (7.0 x 103N/A s) / (3.5 x 10-2 N/A m) Solve! v = 2.0 x 105 m/s Example An electron of mass 9.11 x 10-31 kg moves with a speed of 2.0 x 105 m/s across a magnetic field. The magnetic induction is 8.0 x 10-4 T. What is the radius of the circular path followed by the electrons while in the field? What we know: M = 9.11 x 10-31 kg B=8.0 x 10-4 T v=2.0 x 105 m/s Equation: Bqv = mv2/r so r = (mv) / (Bq) Substitute: R = (9.11x10-31kg)(2.0x105 m/s) (8.0x10-4N/Am)(1.6x10-19As) Solve! r = 1.4 x 10-3 m Modifications of the Bohr Theory – Elliptical Orbits Sommerfeld extended the results to include elliptical orbits Retained the principle quantum number, n Added the orbital quantum number, ℓ Determines the energy of the allowed states ℓ ranges from 0 to n-1 in integer steps All states with the same principle quantum number are said to form a shell The states with given values of n and ℓ are said to form a subshell Example A negatively charged polystyrene sphere of mass 3.3 x 10-15 kg is held at rest between 2 parallel plates separated by 5.0 mm when the potentialdifference between them is 170 V. How many excess electrons are on the sphere? Solution: Using qV/d = mg gives q = 9.5 x 10-19 C Taking e to be 1.6 x 10-19, q corresponds to 5.9 excess electrons Because we cannot have 0.9 of an electron, the answer must be 6 electrons, the discrepancy must be due to experimental inaccuray or rounding errors Example Electrons are accelerated to a speed of 8.4 x 106 m/s. they then pass into a region of uniform magnetic flux density 0.50 mT. The path of the electrons in the field is a circle with a radius 9.6 cm. Calculate: a) the specific charge of the electron b) the mass of the electron, assuming the charge on the electron is 1.6 x 10-19 C Solution a) e/m = v/Br = 1.8 x 1011 C/kg b) using e/m, m = 9.1 x 10-31 kg Example It is required to select charged ions which have a speed of 4.2 x 106 m s-1. the electric field strength in the velocity selector is 3.2 x 104 V m-1. Calculate the magnetic field strength required. Solution v = E/B = 7.6 x 10-3 T Example A charged particle has mass 6.7 x 10-27 kg and charge +3.2 x 10-19 C. It is travelling at speed 2.5 x 108 m s-1 when it enters a region of space where there is a uniform magnetic field of flux density 1.6 T at right angles to its direction of motion. Calculate a) the force on the particle due to the magnetic field b) the radius of its orbit in the field Solution a) F = Bqv sin θ = 1.3 x 10-10 N b) centripetal force is provided by the electromotive force mv2/r = Bqv r = 3.2 m