Pra-U STPM Physics Penggal 3 2018 CB039248c

Physics Term 3 STPM Chapter 24 Quantum Physics 245 24 (ii) Hence deduce in terms of λ, a and α, the condition for strong diffraction in this direction. (iii) Describe qualitatively what happens to the direction of the strongly diffracted beam if the wavelength λ of the incident electron beam is shorter. (c) Diffraction experiments using low energy electron beams provide information about the arrangement of atoms very near the surface of a crystal. Suggest a reason why the spacing of atoms near the surface may be expected to differ from that well inside the crystal. (d) Diffraction experiments using X-rays of wavelength similar to that of the electron beam give information about the arrangement of atoms in the body of the crystal, rather than at the surface. Suggest an explanation for this. Solution: (a) (i) Arrangements of atoms • Atoms in a crystal are arranged orderly in atomic planes. • The separation between atoms is of the same order of magnitude as the wavelength of X-rays or electron beams. (ii) Characteristics of X-rays and electron beams. • X-rays are waves. • Under certain circumstances, an electron beam exhibits wave properties such as diffraction. • Wavelengths of X-rays and electron beams are of the same order of magnitude as the separation between atoms in a crystal. (b) (i) a A B R P Q α α α Path difference between PA and QB = PR = a sin α (ii) For strong diffraction, path difference = nλ a sin α = nλ n = 1, 2, 3,... (iii) When the wavelength λ of the electron beam is shorter, from the equation a sin α = nλ When λ decreases, a decreases for each value of n. More orders of strong diffraction would be obtained. (c) Atoms on the crystal surface are more energetic than those atoms well inside the crystal. Hence they vibrate with greater amplitude. The mean separation of atom at the surface is greater than those inside the crystal. (d) X-rays are not charged. Hence the interaction between X-rays and atoms in the crystal is negligible. Whereas electrons are negatively charged, and interact with ions in the crystal. Incident beam Row of atoms Crystal surface a a P Q A B α α

Physics Term 3 STPM Chapter 24 Quantum Physics 246 24 Quick Check 2 1. Which statement about de Broglie wavelength is correct? A de Broglie wavelength of a photon is inversely proportional to its frequency. B de Broglie wavelength of a particle is proportional to its momentum. C de Broglie wavelength of an electron is greater than that a proton with the same speed. D de Broglie wavelength of a man running the 100 m race is greater than that of alpha particle. 2. What is the velocity of a particle of mass m and of de Broglie wavelength λ? A h mλ C mλc2 h B mλ h D ( 2hc mλ ) 1 2 3. The intensity of a beam of monochromatic light is doubled. Which of the following, if any, represents the change in the momentum of a photon of the light? A Doubled B Halved C Four times D Remains unchanged 4. In 1923, de Broglie suggested that an electron of momentum p processes properties correspond to a wave of wavelength λ . Which of the following graphs shows the relation between λ and p? A C 0 p λ 0 p λ B D 0 p λ 0 p λ 5. The wave nature of electrons is suggested by experiments on A photoelectric effect B electron diffraction by crystalline material C production of X-rays D line spectra of atoms 6. Experiments on the diffraction of electrons by G.P. Thomson were challenged because of the possibility that the electrons detected by photographic plates could be X-rays. He proved otherwise by placing bar magnets on each side of the electron beam as shown in the above fi gure. The effect of the magnetic fi eld is to defl ect the ring in the direction A P C R B Q D S 7. The de Broglie wavelength of a particle of mass m and kinetic energy E is A h (2 mE) C h 2mE B 2mE h D h mE 8. If de Broglie wavelengths of each of the following particles are the same, which particle has the greatest speed? A proton C 2 1 H nucleus B α-particle D electron 9. An electron and a proton of mass me and mp respectively are accelerated from rest through the same potential difference. Which of the following is equal to the ratio of the de Broglie wavelength of the accelerated electron to that of the accelerated proton? A me mp C ( me mp ) 2 B ( me mp ) 1 2 D ( mp me ) 1 2 S N Photographic plate Ring Crystal Electron S P Q R

Physics Term 3 STPM Chapter 24 Quantum Physics 247 24 10. If k is Boltzmann constant and h is Planck constant. Which of the following gives the de Broglie wavelength for neutron of mass m at a temperature T? A h mkT C h 5mkT B h 3mkT D h 6mkT 11. An electron microscope has higher resolving power than an optical microscope because A electrons are smaller than visible quanta B the electron wavelength is much shorter than that of visible light C the electron beam can be focussed much better than light D electrons are not easily dispersed compared to light 12. Find (a) the energy, (b) the momentum, of photons of light of wavelength 500 nm. 13. Under certain conditions, electrons show wave properties. Find the wavelength of an electron of kinetic energy 1 × 10–18 J. Comment on the wavelength you have calculated in relation to the size of an atom. 14. Electron diffraction experiments show that the wavelength associated with a certain electron beam is 0.15 nm. Find the momentum of an electron in the beam. Through what potential difference should the electrons be accelerated from rest to acquire this momentum. 15. (a) What is meant by wave-particle duality? (b) Derive an expression for de Broglie wavelength of an electron in terms of its kinetic energy E, the electron mass me , and the Planck constant h. (c) Electrons are accelerated from rest by a potential difference of 1 000 V in a vacuum. After passing through a very thin crystal, many are observed to emerge with a defl ection of 5°. (i) Calculate the wavelength associated with the electrons. (ii) If the accelerating potential is now doubled, calculate the ratio of the new electron speed to the old. (iii) What is the new angle of defl ection? 16. (a) Under certain circumstances particles behave as waves. With the aid of a labelled diagram, outline an experiment to show the wave nature of a beam of electrons. (b) A beam of electrons and a beam of X-rays produce the same diffraction pattern using the same object. Deduce an expression for the potential difference V required to accelerate electrons in the electron beam from rest in terms of λ, the wavelength of the X-ray, e charge of electron, m mass of electron and h, Planck constant. (c) Write an approximate value for the separation of neighbouring atoms in a typical crystalline solid. If the electron wavelength of a electron beam is of the same order of magnitude as the atomic separation, estimate the momentum of the electron in the beam. 24.3 Atomic Structure Learning Outcomes Students should be able to: • state Bohr’s postulates for a hydrogen atom • derive an expression for the radii of the orbits in Bohr’s model • derive the formula En = – Z2e4m 8ε2 0 h 2 n 2 for Bohr’s model • explain the production of emission line spectra with reference to the transitions between energy levels • explain the concepts of excitation energy and ionisation energy 1. At the beginning of the twentieth century scientists were still uncertain of the structure of the atom. 2016/P3/Q12

Physics Term 3 STPM Chapter 24 Quantum Physics 248 24 2. In 1911, Rutherford uses α-particles to bombard gold foil. From the results of this α-scattering experiment, Rutherford suggested the idea of a nucleus is an atom. 3. Rutherford suggested that an atom consists of a massive central nucleus which is positively charged and around the nucleus are electrons so that the atom is neutral. 4. Based on Rutherford’s model of the atom, a model of the hydrogen atom would consists of a nucleus that contains a proton and an electron going around the proton. (Figure 24.11) Figure 24.11 Classical model of the hydrogen atom Electron Nucleus + – 5. The centripetal force that enables the electron to go around the nucleus is the electrostatic force between the positively charged nucleus and the negatively charged electron. 6. This model has its weakness. According to Maxwell’s theory of electromagnetic radiation, an electron in circular motion accelerates and emits electromagnetic waves. Energy dissipates from the atom. This causes the electron to spiral into the nucleus. Bohr’s Postulates 2006/P1/Q43, 2012/P1/Q44, 2013/P3/Q20 1. In 1913 Niels Bohr presented his postulates to explain for the emission of line spectrum from hydrogen atoms. 2. Bohr’s first postulate states that in a hydrogen atom the electron is able to orbit around the nucleus in certain allowed discrete orbits known as stable energy levels such that the angular momentum of the electron is given by angular momentum, L = (mv)r = nh 2π where h = Planck constant n = integer (quantum number) This means that the angular momentum of the electron is quantised. 3. Normally the electron is in the lowest energy level, known as the ground state, as it is the most stable state. 4. The electron in the ground state may acquire sufficient energy to be in one of the higher energy states, known as the excited states. 5. Figure 24.12 shows the various energy levels in the hydrogen atom. At any instant, the electron may be in one of energy levels. Figure 24.12 E6 = – 0.38 eV E5 = – 0.54 eV E4 = – 0.85 eV E3 = – 1.5 eV E2 = – 3.4 eV E1 = – 13.6 eV E• = 0 Free electron Ground state Excited states Figure 24.10 Rutherford’s model of the atom + Electron Nucleus

Physics Term 3 STPM Chapter 24 Quantum Physics 249 24 6. Bohr’s second postulate When the electron drops from a higher energy level E2 to a lower energy level E1 , the difference in energy (E2 – E1 ) is emitted as a photon of electromagnetic radiation of frequency f is given by E2 – E1 = hf Figure 24.13 + Electron Nucleus e.m. Radiation Radii of Orbits 1. Figure 24.14 shows the electron in the hydrogen atom revolving round the nucleus of charge +e in an orbit of radius r with a speed v. 2. The centripetal force on the electron is the electrostatic force between the electron of charge –e and the proton in the nuclues of charge +e. Hence mv2 r = e × e 4πε0 r 2 m = mass of electron v2 = e2 4πε0 rm ............................. ① From Bohr’s first postulate angular momentum, mvr = nh 2π m2 v 2 r2 = n2 h2 4π2 m2 e2 4πε0 rm r2 = n2 h2 4π2 Radius of orbit, r = h2 ε0 πme2 n2 3. Using r = ε0 h2 πme2 n2 when n = 1, r1 = (8.85 × 10–12) (6.63 × 10–34)2 π × (9.11 × 10–31) (1.60 × 10–19)2 = 5.31 × 10–11 m when n = 2, r2 = (5.31 × 10–11) (22 ) = 2.12 × 10–10 m when n = 3, r3 = (5.31 × 10–11) (32 ) = 4.78 × 10–10 m Info Physics Niels Bohr was a Danish physicist who made contributions to understanding of atomic structure and quantum mechanics. He received the Nobel Prize in Physics in 1922. Figure 24.14 + Electron Proton +e –e – r v F

Physics Term 3 STPM Chapter 24 Quantum Physics 250 24 Energy Levels in Hydrogen Atom 2010/P1/Q46, 2011/P2/Q7, 2009/P2/Q3, 2017/P3/Q17 1. The electron revolving around the nucleus in an orbit of radius r has kinetic energy and electric potential energy. 2. From mv2 r = e2 4πεo r2 Kinetic energy, 1 2 mv2 = e2 8πεor 3. Electric potential energy = (–e) × (+e) 4πεo r = –e2 4πεo r 4. Total energy, En = kinetic energy + electrical potential energy = e2 8πεo r – e2 4πεo r = – e2 8πεo r r = ε0 h2 πme2 n2 = – e2 8πεo × π me2 εo h2 n2 En = – me4 8εo 2 h2 n2 n = 1, 2, 3, … This shows that the energy En is quantized. 5. When n = 1, the ground state E1 = – me4 8εo 2 h2 = – (9.11 × 10–31) (1.60 × 10–19)4 8 (8.85 × 10–12)2 (6.63 × 10–34)2 J = – 13.6 eV In general, En = – 13.6 n2 eV when n = 2, E2 = – 13.6 22 eV = –3.4 eV n = 3, E3 = – 13.6 32 eV = –1.5 eV n = 4, E4 = – 13.6 42 eV = –0.85 eV n = 5, E5 = – 13.6 52 eV = –0.54 eV n = 6, E6 = – 13.6 62 eV = –0.38 eV n = ∞ E∞ = 0 (electron is free)

Physics Term 3 STPM Chapter 24 Quantum Physics 251 24 6. Figure 24.15 shows diagrammatically the various energy levels in the hydrogen atom. Figure 24.15 Energy levels in hydrogen atom E6 = – 0.38 eV E5 = – 0.54 eV E4 = – 0.85 eV E3 = – 1.5 eV E2 = – 3.4 eV E1 = – 13.6 eV Free electron Ground state Excited states 1st excited state n = • n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 7. For an atom of proton number Z, there are Z protons in the nucleus. Using identical method as shown above, it can be shown that the energy En is given by En = – Z2 e4 m 8εo 2 h2 n2 8. The atom is said to be in the ground state if the electron is at the lowest energy level, E1 . The atom is most stable. 9. Excited state. The atom is in the excited state when the electron is at any one of the energy levels higher than the ground state, such as E2 , E3 , … 10. The energy required by the electron to move up from the ground state, n = 1 to the first excited state, n = 2 is known as the excitation energy of the atom. If the excitation energy = e × V then V is known as the excitation potential. 11. The ionisation energy of the atom is the energy required by an electron in the ground state to escape completely from the attraction of the nucleus. If the ionisation energy = e × V then V is known as the ionisation potential. 12. The energy of hydrogen atom in the ground state is E1 = –13.6 eV. For the electron in the ground state to escape completely from the attraction of the nucleus, its final energy E∞ = 0. Hence the ionization energy of the hydrogen atom is 13.6 eV, and its ionization potential 13.6 V. 13. There are a number of ways that an atom may be excited or ionized. For an atom to be excited, the electron in the ground state must gain energy that is exactly equal to the energy difference between the initial energy level of the electron and the final energy level. The various ways that the electron can gain energy are (a) Bombardment by an electron (i) Figure 24.16 shows a hydrogen discharge tube which contains hydrogen gas at low pressure. When a high voltage is applied across the tube, ionisation Figure 24.16 Excitation and ionization a hydrogen discharge tube + – + – High voltage Anode Discharge tube Hydrogen gas at low pressure Cathode

Physics Term 3 STPM Chapter 24 Quantum Physics 252 24 initially occurs near the cathode and anode. The electrons and positive ions produced are accelerated to the anode and cathode respectively. When a fast electron collides with the electron in the ground state, energy is transferred to the electron in the ground state. This causes the electron to transit to a higher energy level. The atom is said to be excited. If the electron is ejected from the atom, then the atom is ionised. (ii) The excited electron does not stay long in the excited state but falls back to a lower energy level. In the process a photon equals to the energy difference between the initial higher energy level and the lower energy level is emitted. Similarly the electron from an ionised atom does not stay free for long but fall back to the ground state directly or in steps. In the process photons of different wavelengths are emitted. (b) Collisions between molecules in a fl ame Collisions between gas molecules which have higher thermal energy in a fl ame can cause excitation and ionization. For this reason, the space around a Bunsen fl ame, or candle fl ame is full of charge. (c) Absorption of a photon One way the electron can be excited or ionized is to absorb a photon. The energy of the photon hf must be equal to the energy difference between the initial lower energy level and the fi nal higher energy level. Since the excited electron falls back to the lower energy level by emitting a photon equal to the energy difference between the two energy levels, an atom would absorb photons which have the same frequencies as the frequencies of the photons it emits. Example 9 The energies of the six lowest levels of the hydrogen atom are given below: Level n 1 2 3 Energy/J –2.2 × 10–18 –5.3 × 10–19 –2.4 × 10–19 Level n 4 5 6 Energy/J –1.3 × 10–19 –8.0 × 10–20 –6.0 × 10–20 When an electron drops from level n = 6 to n = 1, the wavelength of the photon emitted is 9.1 × 10–8 m. What is the wavelength of the photon emitted by the transition from n = 4 to n = 3? Solution: For the transition n = 6 to n = 1 E6 – E1 = hf = hc λ1 ..............................① For the transition n = 4 to n = 3 E4 – E3 = hc λ2 ..............................② Figure 24.17 1153 E1 = – 13.6 eV n = 1 Ground state E1 E2 E3 Incident photon Excitation Emission of photons

Physics Term 3 STPM Chapter 24 Quantum Physics 253 24 —① ② : λ2 λ1 = E6 – E1 E4 – E3 λ2 = (–6.0 × 10–20) – (–2.2. × 10–18) (–1.3 × 10–19) – (–2.4 × 10–19) × (9.1 ×10–8) = 1.8 × 10–6 m Example 10 Transitions of electron between three energy levels in a particular atom give rise to three spectral lines of wavelengths in increasing magnitudes λ1 , λ2 and λ3 . (a) Sketch a diagram to show the transitions that produces these lines. (b) Deduce an equation relating λ1 , λ2 , and λ3 . Solution: (a) E3 λ3 λ2 λ1 E2 E1 (b) Referring to the diagram, E3 – E2 = hc λ3 ..............................① E2 – E1 = hc λ2 ..............................② ① + ② : E3 – E1 = hc + 1 λ3 1 λ2 E3 – E1 = hc λ1 hc λ1 = hc + 1 λ3 1 λ2 Hence, 1 λ1 = 1 λ3 + 1 λ2 Example 11 The fi rst excitation energy of the hydrogen atom is 10.2 eV. Explain what is meant by the statement. What is the speed of the slowest electron that could cause excitation of a hydrogen atom? Solution: E3 E2 E1 1st excited state Ground state

Physics Term 3 STPM Chapter 24 Quantum Physics 254 24 The fi rst excitation energy of the hydrogen atom is 10.2 eV means that the energy required by the electron in the ground state, E1 to move up to the fi rst excited state E2 is 10.2 eV. For the slowest electron to cause excitation of a hydrogen atom, its kinetic energy must be completely transferred to the electron in the ground state, and suffi cient for the latter to reach the fi rst excited state. Hence 1 2 mv2 = 10.2 eV v = 2 × 10.2 × (1.6 × 10–19) 9.11 × 10 √ –31 = 1.89 × 106 m s–1 Example 12 A hydrogen atom in the ground state is assumed to have zero energy, and its energies in the fi rst, second and third excited states are 1.635 × 10–18 J, 1.936 × 10–18 J, and 2.024 × 10–18 J respectively. (a) What is the wavelength of incident photon that would excite the atom from the fi rst excited state to the second excited state? (b) In the hydrogen atom spectrum, there is a blue line of wavelength 5.17 × 10–7 m. Transition between which energy levels would produce this wavelength? Solution: (a) Ground state : E1 = 0 First excited state E2 = 1.635 × 10–18 J Second excited state, E3 = 1.936 × 10–18 J Third excited state, E4 = 2.024 × 10–18 J If λ = wavelength produced, then hc λ = E3 – E2 λ = hc E3 – E2 = (6.63 × 10–34) × (3.0 × 108 ) (1.936 – 1.635) × 10–18 = 6.61 × 10–7 m Quick Check 3 1. An electron in a hydrogen atom makes a transition from an energy level of energy E1 to another of energy E2 and a photon is emitted. The wavelength of the emitted photon is A h (E1 – E2 ) C h c(E1 – E2 ) B hc (E1 – E2 ) D E1 – E2 h 2. The energies of four levels of the hydrogen atom are level P : –13.60 eV level Q : –3.40 eV level R : –1.50 eV level S : –0.85 eV The spectral line of wavelength 488 nm is produced when an electron makes the transition from level (b) When λ = 5.17 × 10–7 m E = hc λ = (6.63 × 10–34) × (3. × 108 ) J 5.17 × 10–19 J ≈ (2.024 – 1.635) 10–19 J = E4 – E2 Transition is from E4 to E2

Physics Term 3 STPM Chapter 24 Quantum Physics 255 24 A Q to P C R to Q B R to P D S to Q 3. The minimum energy to ionize an atom is the energy required to A add one electron to the atom B remove one outermost electron from the atom C remove one innermost electron from the atom D remove all the electrons from the atom 4. Fluorescent paint, which appears to glow in daylight, because A ultraviolet light is absorbed by the paint and some of the absorbed energy is re-emitted as visible light. B the paint is heated by the daylight and gives out visible light. C light falls on the paint from many directions but is diffracted only in the direction of the eye. D the long wavelength infrared radiation in sunlight causes some colours to become brighter. 5. 0.0 eV – 1.6 eV – 3.7 eV – 5.5 eV –10.4 eV The above figure shows a number of energy levels of an atom. If an electron of kinetic energy 7.0 eV collides elastically with an atom in the ground state, what is the kinetic energy of the electron after collision? A 0.3 eV C 3.3 eV B 1.5 eV D 3.4 eV 6. The energy levels in a hydrogen atom are given by E = – 13.6 n2 eV where n = 1, 2, 3, … the energy required to excite an electron from the ground state to the first excited state is A 3.4 eV C 10.2 eV B 4.5 eV D 13.6 eV 7. An electron of mass m and kinetic energy E collides with an atom X in the ground state. When the excited atom returns to the ground state, a photon of wavelength λ is emitted. If h is Planck constant, c speed of light, which of the following gives the velocity of the electron after collision? A 1 m E – hc λ B 2 m E – hc λ C 2 m hc λ – E D 1 m hc λ – E Emission Line Spectrum 2008/P1/Q44, 2009/P1/Q45 1. Evidence of the existence of discrete electron energy levels in atoms is provided by line spectrum. 2. When an electron in the excited state E2 drops to a lower energy level E1 , the energy difference (E2 – E1 ) is emitted as a photon of frequency f given by hf = (E2 – E1 ) 3. Since the energy levels are quantized, photons of discrete frequencies (or wavelengths) are emitted and forms a line spectrum.

Physics Term 3 STPM Chapter 24 Quantum Physics 256 24 Figure 24.18 Eα = 0 E1 = – 13.6 eV E2 = – 3.4 eV E3 = – 1.5 eV E4 = – 0.85 eV E5 = – 0.54 eV E6 = – 0.38 eV Lyman series ( Ultraviolet ) Balmer series (Visible light and ultraviolet radiation) Paschen series (Infrared) 4. Figure 24.18 shows the transitions of the electron in the hydrogen atom that produce the line spectrum of hydrogen. 5. All the possible transitions are grouped into series. Each series consists of transitions ending on a particular level. 6. The Lyman series consists of transitions ending on the ground state, n = 1. The photons emitted are ultraviolet radiations. 7. The Balmer series, ending on the first excited state, n = 2. The Balmer series consists of four spectral lines which are visible as shown in Figure 24.19. The other lines of shorter wavelengths are ultraviolet radiations. Figure 24.19 Visible spectral lines in Balmer series Violet Red 4 3 2 1 8. The Paschen series ends at n = 3, and are all infrared radiations. Absorption Spectrum 2010/P1/Q44 Figure 24.20 White light Decreased intensity for certain frequency Reemission in all directions Gas Discharged tube Info Physics Absorption spectrum recorded by Hubble’s Space Telescope is used to determine the distance and chemical composition of gas clouds in galaxies and in intergalactic space.

Physics Term 3 STPM Chapter 24 Quantum Physics 257 24 1. Figure 24.20 shows a beam of white light passing through a discharged tube containing a gas, such as hydrogen. When observed through an optical spectroscope fi tted with a diffraction grating, dark lines are seen against the continue spectrum of white light. This constitutes the absorption spectrum of the gas in the discharge tube. 2. The dark lines in the absorption spectrum correspond to the emission line spectrum of the gas. 3. When white light goes through the gas, atoms in the gas absorb photons of defi nite wavelengths which would raise the electron in each atom to a higher excited state. When the excited atoms fall back to the lower energy state, the energy absorbed is reemitted as photons in all directions. Hence, the intensity of light that emerges from the gas in the direction of the incident beam shows decrease intensity for certain wavelengths. Hence the dark lines is in the absorption spectrum. 4. By studying the absorption spectrum of light from a star, such as the sun, physicists are able to determine the types of gas on the surface of the star. Example 13 E5 E4 E3 E2 E1 1 2 3 4 5 The fi gure shows fi ve energy levels of an atom. Five transitions of the electron between the energy levels are shown, each produces a photon of defi nite frequency. Draw a sketch to show the relative positions of the spectral lines due to these transitions. Denote the low and high frequency ends of the spectrum. Solution: High frequency Low frequency 1 2 3 4 5 Explanation: Using the equation of energy difference between two energy levels = hf, the energy difference (E5 – E1 ) is biggest, hence the photon emitted has the highest frequency i.e. line 1. The difference between (E5 – E1 ) and (E4 – E1 ) is small, hence line 2 is close to line 1. Whereas the difference between (E3 – E1 ) and (E5 – E2 ) is large. Hence, the bigger separation between line 3 and line 4.

Physics Term 3 STPM Chapter 24 Quantum Physics 258 24 Example 14 (a) (i) What is a photon? (ii) Show that the energy E of a photon and its wavelength λ are related by Eλ = 1.99 × 10–16 J nm. 91.2 95.0 97.3 102.6 121.6 λ/nm (b) The fi gure represents part of the emission spectrum of atomic hydrogen. It contains a series of lines, the wavelengths of some of which are marked. There are no lines in the series with wavelengths less than 91.2 nm. (i) In which region of the electromagnetic spectrum are these lines? (ii) Using the relation between E and λ given above, fi nd the photon energy for all the wavelengths marked. (iii) Use the answers to (ii) to map a partial energy level diagram for hydrogen. Show and label clearly, the electron transitions responsible for the emission lines labelled in the fi gure given above. (iv) Another line in the hydrogen spectrum occurs at a wavelength of 434.1 nm. Identify and label on your diagram the transition responsible for this line. (c) Emission spectra are produced using a discharge tube containing the gas. Explain the physical processes occuring in the gas which lead to the excitation of the gas atoms and the emission of light. Solution: (a) (i) The photon is a discrete quantity of energy of an electromagnetic radiation. (ii) Energy of a photon, E = hc λ Eλ = (6.63 × 10–34) (3.0 × 108 ) = 1.99 × 10–25 J m = (1.99 × 10–25) (109 nm) J = 1.99 × 10–16 J nm (b) (i) The range of wavelengths shown : 91.2 nm to 121.6 nm is shorter than the wavelength of violet light, 400 nm. Hence they are in the ultraviolet region of the electromagnetic spectrum. (ii) Using Eλ = 1.99 × 10–16 J nm E = 1.99 × 10–16 (λ/nm) J λ/nm Energy E = (1.99 × 10–16)/λ 91.2 2.18 × 10–18 J 95.0 2.09 × 10–18 J 97.3 2.05 × 10–18 J 102.6 1.94 × 10–18 J 121.6 1.64 × 10–18 J

Physics Term 3 STPM Chapter 24 Quantum Physics 259 24 (iii) E• 0 E5 E4 E3 E2 E1 95.0 nm 91.2 nm – 0.09 fi 10–18 J – 0.13 fi 10–18 J – 0.24 fi 10–18 J – 0.54 fi 10–18 J – 2.18 fi 10–18 J 97.3 nm 102.6 nm 121.6 nm 434.1 nm Explanation λ = 91.2 nm is caused by the transition E∞ = 0 to E1 = –2.18 × 10–18 J. Start by drawing the energy levels E∞ and E1 The energy difference of the photon of wavelength 91.2 nm and 95.0 nm is (2.18 – 2.09) × 10–18 J = 0.09 × 10–18 J Hence E5 = –0.09 × 10–18 J Similarly E6 = –(2.18 – 2.05) × 10–18 J = –0.13 × 10–18 J and so on. (iv) For λ = 434.1 nm E = 1.99 × 10–16 434.1 = 4.58 × 10–19 J E5 – E2 = [–0.09 – (–0.54)] × 10–19 J = 4.5 × 10–19 J Hence the transition responsible for λ = 434.1 nm is from E5 to E2 as shown in the fi gure for (b) (iii). (d) When a high potential difference is applied across the discharge tube, ionisation of the gas atoms initially occurs near the electrodes. The electrons and positive ions produced are accelerated to the anode and cathode respectively. Collisions between fast electrons and atoms in the ground state excite or ionise other atoms. When the excited atoms fall back to lower energy states, photons of defi nite frequencies are emitted. Example 15 When two hydrogen atoms of suffi cient energy collide, one or both atoms may be raised to the fi rst excited state. (a) Explain why such a collision is not elastic? (b) How may the excitation be detected? (c) If the fi rst excitation potential for hydrogen is 10 V, estimate the temperature of hydrogen gas when excitation by collision can occur.

Physics Term 3 STPM Chapter 24 Quantum Physics 260 24 Solution: (a) The collision is not elastic because the total kinetic energy of the two hydrogen atoms after collision is less than their total kinetic energy before collision. Part of the kinetic energy is used to raise the hydrogen atom to the fi rst excited state. (b) Excitation can be detected because electron in the excited atom does not stay long in the higher energy level. It falls back to the lower energy level, and in the process a photon is emitted. The emission of a photon shows that the atom had be excited. (c) Using Maxwell equipartition of energy, kinetic energy of hydrogen atom at temperature T is 3 2 kT. If two hydrogen atoms collide, and causes one atom to be excited, then kinetic energy of 2 atoms = excitation energy of 1 atom 2 × 3 2 kT = eV T = 10 × (1.6 × 10–19) 3 × (1.4 × 10–23) = 3.80 × 104 K Alternatively, if both atoms are excited, then 2 × 3 2 kT = 2 (eV) T = 3 2 eV k = 7.60 × 104 K Quick Check 4 1. Which of the following experimental phenomena provides evidence for discrete electron energy levels in atoms? A the spectrum of a tungsten fi lament lamp B the spectrum of a sodium discharge lamp C the photoelectric effect D the emission of β-particle by radioactive atoms 2. A parallel beam of white light passes through a metal vapour, dark lines appear in the spectrum of the emergent light because A white light does not contain certain wavelengths B certain wavelengths are absorbed by the vapour and not radiated C certain wavelength are absorbed and re-radiated uniformly in all directions D destructive interference occurs in certain directions 3. The fi gure, drawn to scale, represents the energy levels for an electron in a certain atom E2 E4 E3 E1 The transistion from E3 to E1 produces a green line. What transition could give rise to a red line? A E4 to E3 B E3 to E2 C E4 to E2 D E2 to E1

Physics Term 3 STPM Chapter 24 Quantum Physics 261 24 4. The figure shows some energy levels E1 to E6 of a hydrogen atom. E4 E5 E6 E3 E2 E1 –0.38 eV –0.54 eV –0.85 eV –1.5 eV –3.4 eV –13.6 eV Which of the following transitions produces a photon of wavelength in the ultraviolet region of the electromagnetic spectrum? A E2 – E1 C E4 – E3 B E3 – E2 D E6 – E5 5. White light from a tungsten filament lamp is passed through sodium vapour lamp and viewed through a diffraction grating. Which of the following describes correctly the first order spectrum seen? A Dark lines on a white background. B Dark lines on a coloured background. C Coloured lines on a white background. D Coloured lines on a black background. 6. E2 E4 E3 E1 The figure shows four energy levels of an atom. Five transitions are shown, each produces a photon of definite frequency. Which of the spectra below best corresponds to the transitions shown? A B C D 7. The energy levels of an hydrogen atom is given by the expression En = – me4 8e2 0h2 n2 , n =1,2,3,... where m and e are the mass and charge of the electron and h is Planck's constant. Which is the correct expression for the de Broglie wavelength of an electron in the nth energy level? A e0 h2 n 2me2 C e0 h2 n2 me4 B 2e0 h2 n me2 D 2e0 h2 n2 me4 8. The figure shows four energy levels A, B, C, and D within an atom and an electron transition from level A to level C which results in the emission of a photon of light. – 0.496 fi 10–19 J – 3.136 fi 10–19 J – 14.72 fi 10-19 J A B C D 0 (a) Copy the figure and show on it all the other possible transitions between these four levels which result in photon emissions. (b) Calculate the wavelength of the light emitted as a result of the electron transition from A to C. (c) Which other transition may result in visible light being emitted? (d) When a transition takes place from level A, B, or C to level D, in which part of the electromagnetic spectrum will the radiation occur? 9. One of the lines in the line spectrum of cadmium has wavelength 467.8 nm. What is the energy difference, in eV, between the two energy levels which give rise to this line? 10. The ground state of a helium atom is –54.4 eV, and the energy forthe energy level E1 is – 13.6 eV. (a) Calculate the wavelength of the photon emitted when a helium atom undergoes transition of the energy level E1 to the ground state. (b) Name the type of electromagnetic radiation emitted.

Physics Term 3 STPM Chapter 24 Quantum Physics 262 24 11. An element whose ionisation energy is 9.50 eV has a line spectrum with lines at 2.00 eV, 6.50 eV and 8.50 eV. (a) Draw a diagram to show the energy levels and show the transitions that give rise to the three lines above. Label the ground state. (b) Use your diagram to deduce the wavelengths that may be absorbed by atoms of the element in the ground state. 12. The figure shows four energy levels in a hydrogen atom. The transition shown emits light of wavelength 486 nm. Increasing energy (a) Copy the figure, and show on it, (i) another transition which results in the emission of light of longer wavelength, (ii) a transition which results in the emission of infrared radiation, (iii) a transition which results from absorption. Label the transitions clearly (b) Calculate the energy change when an electron under the transition that results in the emission of wavelength 486 nm. 13. (a) The figure shows the line spectrum of hydrogen. Increasing wavelength (i) Explain how a line of the spectrum is produced. (ii) Why the lines in the spectrum have discrete wavelengths? (iii) Some of the lines in the spectrum are of higher intensity than others. Explain why. (b) The figure shows a fluorescent tube which is coated on the inside with a powder. The tube is filled with mercury vapour at low pressure. A coil of high inductance and a starter switch are connected in series with the tube and a.c. supply. The starter switch is normally close and opens when heated. A high voltage is required to ionise the mercury vapour. Coil of high inductance Starter switch Mercury vapour at low pressure a.c.supply Explain (i) how a high voltage is produced across the tube when the switch S is closed. (ii) how ultraviolet radiation is produced by the mercury vapour. (iii) how the ultraviolet radiation causes the emission of light from the powder that coats the tube. 14. (a) The energy levels in a hydrogen atom is given by En = – 13.6 n2 where n is an integer known as the quantum number. Explain (i) what is meant by an energy level, and the significant of the negative sign. (ii) evidence for the existence of discrete energy levels in hydrogen atom. (b) (i) Define ionisation energy of the hydrogen atom. (ii) What is the energy required to ionise a hydrogen atom in the excited state n = 4? (c) An electron with kinetic energy 2.10 × 10–18 J collides with a hydrogen atom in the ground state. After the collision the hydrogen is in the second excited state. (i) Find the kinetic energy of the electron after collision with the hydrogen atom. (ii) The excited hydrogen atom then falls back to the ground state. Calculate the possible wavelengths of the photons emitted.

Physics Term 3 STPM Chapter 24 Quantum Physics 263 24 24.4 X-Rays Students should be able to: • interpret X-ray spectra obtained from X-ray tubes • explain the characteristic line spectrum and continuous spectrum including λmin in X-rays • derive and use the equation λmin = hc eV • describe X-ray diffraction by two parallel adjacent atomic planes • derive and use Bragg’s law 2d sin θ = mλ Learning Outcomes Production of X-rays 2007/P1/Q41 Figure 24.21 a.c. Transformer Focussing cylinder Cathode X- rays Vacuum Anode Cooling fins Tungsten target 1. X-rays are electromagnetic radiations of short wavelengths, 10–9 m to 10–11 m. 2. X-rays are produced when fast electrons are retarded during collisions with a heavy metal. 3. Figure 24.21 shows an X-ray tube. When the cathode is heated by the current in it, electrons are emitted due to thermionic emission. 4. A focussing cylinder collimates the electrons into a narrow beam. 5. The electrons are accelerated by the high potential difference, more than 10 kV, between the cathode and copper anode. 6. The fast electrons collide with a tungsten target embedded in the copper anode. Tungsten is used because it is one of the most effi cient metals for the production of X-rays. 7. Nevertheless less than 1% of the total energy of the colliding electrons is converted into X-rays. The rest of the energy is converted into heat. 8. Hence the target metal must have a high melting point. Heat is readily conducted away to the cooling fi ns by the copper anode. In high power X-ray tube, a radiator system is used to cool the anode. 9. If V = accelerating potential difference between the cathode and the anode, kinetic energy gained by electron, 1 2 mv2 = eV m = mass of electron e = electronic charge v = fi nal velocity of electron 2015/P3/Q12

Physics Term 3 STPM Chapter 24 Quantum Physics 264 24 10. If the kinetic energy of the electron is completely converted into a photon of X-ray, then eV = hfmax = hc λmin where f max = maximum frequency of X-ray photon λmin = minimum wavelength of X-ray photon h = Planck constant c = speed of light in a vacuum. 11. The minimum wavelength λmin = hc eV X-rays of shorter minimum wavelength can be obtained by increasing the potential difference V between the cathode and anode. 12. X-rays of shorter wavelengths have greater penetrating power, and are known as hard X-rays. 13. The intensity of the X-rays increases if more X-ray photons are peoduced per second. The intensity increases if the number of electrons per second colliding with the target is increased. This can be achieved by increasing the current through the cathode. Example 16 Calculate the shortest wavelength of X-rays produced by the collision of electrons with the screen of a television tube if the accelerating potential is 2.0 kV. Solution: Using the equation , λmin = hc eV = (6.63 × 10–34) (3.0 × 108 ) (1.60 × 10–19)(2.0 × 103 ) = 6.22 × 10–10 m Example 17 The potential difference between the cathode and the target of an X-ray tube is 50 kV and the anode current is 20 mA. Only 1% of the total power supply to the X-ray tube is emitted as X-rays. (a) What is the maximum frequency of the X-rays emitted? (b) At what rate heat must be removed from the target so that its temperature remains constant? Solution: (a) Using eV = hfmax f max = (1.60 × 10–19)(50 × 103 ) 6.63 × 10–34 = 1.21 × 1019 Hz (b) In order that the temperature of the anode remains constant, rate at which heat must be removed = rate at which heat is generated = 99% (input power) = 99 100 × (20 × 10–3)(50 × 103 ) = 990 W

Physics Term 3 STPM Chapter 24 Quantum Physics 265 24 Properties of X-rays 1. X-rays are not defl ected by an electric fi eld or a magnet fi eld. Hence X-rays are not charged particles. 2. Photographic plates are darkened by X-rays. 3. X-rays caused fl uorescent materials to glow. X-ray photons excite atoms in the fl uorescent material. When the excited atoms drop back to the lower energy level, light is emitted. 4. Absorption of X-ray photons by atoms in matter causes ionisation. The free electrons produced by ionisation disperse energy throughout the matter through collisions. 5. X-rays have high penetrating power. Elements with high atomic numbers are able to absorb more X-rays because the atoms are closely packed. 6. X-rays are electromagnetic waves of high frequency, hence short wavelength. X-rays are diffracted by a crystal as shown in Figure 24.22. Figure 24.22 (b) X-ray diffraction pattern Photographic plate S1 S2 X- ray tube Lead plates Photographic plate Slit Crystal (a) X-ray spectrometer 7. Uses of X-rays. (a) Soft X-rays of low penetrating power are used for X-ray photography. X-rays penetrate easily soft tissues such as the fl esh, whereas the bones which are of higher density absorb more X-rays. Hence the image of the bones on the photographic plate is less exposed compared to that of the soft tissues. (b) Hard X-rays are used in radio therapy for destroying cancerous cells. It is found that cancerous cells are more easily damaged by X-rays than healthy ones. (c) X-rays are used to study the structure of crystals as in Bragg’s X-ray diffraction. Quick Check 5 1. When the potential difference across an X-ray tube is V, X-rays of minimum wavelength λmin are emitted. Which graph represents the relation between λmin and V? A B C D

Physics Term 3 STPM Chapter 24 Quantum Physics 266 24 2. In an X-ray tube, electrons of each of charge q are accelerated through a potential difference V. If h is Planck constant and c is the speed of light in vacuum, the minimum wavelength of X-ray emitted is A qV hc B hc qV C hq cV D hcV q 3. The value of the minimum wavelength of X-rays produced by an X-ray tube may be reduced by A increasing the current through the cathode B cooling the anode C using a target element of higher atomic number D increasing the potential difference across the X-ray tube 4. An X-ray tube operates at 20 kV with an anode current of 8 mA. The heat capacity of the anode is 40 J K–1. What is the initial rate of increase of temperature of the anode? A 0.25 K s–1 C 16 K s–1 B 4.0 K s–1 D 160 K s–1 5. The penetrating power of an X-ray increases if its A frequency is higher B intensity is higher C amplitude is higher D wavelength is higher 6. An X-ray tube operates at 159 kV. (a) What is the maximum energy for an X-ray photon? (b) Calculate the maximum frequency of the X-ray photon. X-ray Spectrum 2008/P1/Q45, 2010/P1/Q47, 2009/P2/Q7, 2012/P1/Q45, 2014/P3/Q12, 20, 2016/P3/Q13 1. When an X-ray tube operates at a constant potential difference V between the cathode and anode, X-rays emitted do not consist of only one wavelength, but consists of various wavelengths forming a spectrum. The shortest wavelength λmin is given by λmin = hc eV ............................. ① but consist of various wavelengths, forming a spectrum. Equation ① gives the value of the shortest wavelength. Figure 24.23 50 kV 20 kV 5 kV Intensity 1.40 O 0.6 1.54 2.4 Wavelength/ fi 10–10 m 2. Figure 24.23 shows the graph of intensity against wavelength for X-rays emitted from an X-ray tube with copper as target. 3. The X-ray spectrum consists of (a) a continuous background spectrum, (b) a number of characteristic lines with high intensity, the characteristic X-ray line spectrum.

Physics Term 3 STPM Chapter 24 Quantum Physics 267 24 4. The continuous background spectrum has a cut-off or minimum wavelength λmin. For an accelerating potential difference of 20 kV, the value of λmin = hc eV = (6.63 × 10–34)(3 × 108 ) (1.60 × 10–19) (20 × 103 ) = 0.6 × 10–10 m 5. A photon of X-ray of the minimum wavelength λmin is produced when the energy of an electron accelerated through the X-ray tube is completely converted into a photon of X-ray. This happens when the electron colliding with the target is decelerated and stopped in a single collision. 6. The majority of the electrons colliding with the target are stopped after a few collisions with atoms of the target. For each collision a fraction of the energy of the electron is converted into a photon of X-ray. Since the fraction of the energy of the colliding electron which is converted into X-ray photon differs for different electrons, X-rays of different wavelengths are produced. These X-rays formed the continuous background spectrum. 7. The characteristic X-ray line spectrum consists of a number series, each having lines of high intensity. 8. The K-series consists of two lines Kα and Kβ lines and the L-series consists of the Lα and Lβ lines. The wavelengths of these lines are characteristics of the target element. For copper as target, the wavelengths of the Kα-line and Kβ -line are 1.54 × 10–10 m and 1.40 × 10–10 m respectively. (a) (b) K Lβ β Kα Lα K M N L Nucleus Kβ Lβ Kα Lα K M N L Transition of electron Figure 24.24 Production of characteristic X-rays 9. A photon of the Kα-line is produced when an electron from the cathode collides with an atom of the target and excites the electron in the K-shell to a higher energy level. When an electron from the L-shell drops into the K-shell to fill the vacancy, a Kα-photon is emitted (Figure 24.24 (a)) since the energy difference between the L-shell and the K-shell of copper atom is discrete, the Kαphoton has a definite wavelength, 1.54 × 10–10 m. There is a large number of copper atoms that experience this type of transition. Hence the high intensity of the Kα-line. 10. A photon of the Kβ -line is produced when an electron from the M-shell drops into the K-shell to fill the vacancy (Figure 24.24 (a)) since the energy difference between the M-shell and K-shell is bigger, the Kβ -photon has shorter wavelength, 1.40 × 10–10 m for copper. The intensity of the Kβ -line is lower than that of the Kα-line because the number of target atoms undergoing this transition is smaller. The probablility of an electron from the M-shell falling to the K-shell is less than that of an electron from the L-shell to the K-shell. The L-shell being closer to the K-shell. 11. Figure 24.24 shows the transitions responsible for the emission of the L-series which consists of the Lα-line and Lβ -line.

Physics Term 3 STPM Chapter 24 Quantum Physics 268 24 12. In 1913, Moseley measured the wavelengths of the characteristic X-rays for various elements. The results of his experiments showed that elements can be arranged in a defi nite sequence on the basis of the wavelengths of their characteristic X-rays. The sequence follows the positions of the elements in the periodic table. Figure 24.25 shows a plot of √ f against atomic number Z, where f is the frequency of the Kα-line. 13. Before Moseley experiments, the elements in the periodic table were arranged in order of atomic masses. Except for a number of discrepancies, the elements fall into the various groups. Members of each group show identical chemical properties. Moseley suggested that the elements should be arranged according to their atomic numbers Z, the number of protons in the nucleus and not according to their atomic masses. 14. An example of such discrepancies is when Cl, K and Ar are arranged in order of their atomic masses, the sequence is chlorine (35.5)-potassium (39.1)-argon (39.9) Potassium would be in the group of noble gases, and argon in the group of alkali metals, which is wrong. If arranged according to the frequencies of the characteristic Kα-lines, the sequence is chlorine- argon - potassium which is correct. Example 18 Figure (a) shows the important parts of a hot cathode X-ray tube, and Figure (b) shows the X-ray spectrum from such a tube. (a) State the important properties for materials used as the cathode, and anode of such a tube. (b) Describe how the emission of electrons from the cathode produces an X-ray spectrum consisting of a continuous spectrum P, and the line spectrum Q. (c) Find the potential difference between the target and cathode necessary to produce the X-ray spectrum shown in Figure (b). (d) Explain why cooling water is necessary for the target. (e) State and explain the safety precautions need to be undertaken by radiographers manning X-ray instruments. Solution: (a) Properties of cathode • Low value for the work function, so that electrons are easily emitted. • Low specifi c heat capacity, so that it heats up fast. (a) (b) λ 0 Q P Intensity 6.0fi10–11 m C A Cathode Target Cooling water Figure 24.25 0 Atomic number Z f

Physics Term 3 STPM Chapter 24 Quantum Physics 269 24 Properties of target • High melting point because a lot of heat is produced when fast electrons collide with the target. • Large atomic number, so that effi ciency of producing X-ray increases. • Good conductor of heat, so that heat is conducted away quickly. (b) The continuous spectrum P is produced when fast electrons are decelerated on hitting the target. Different electrons are stopped after different number of collisions with the target atoms, and different fractions of the electron energy are converted into X-ray photons. Hence X-rays of different wavelengths are emitted. A photon of line spectrum, say the Kα-line, is produced when • an electron in the K-shell of the target atom is excited to a higher energy level, • an electron from the L-shell transits to the K-shell, • the energy difference (EL – EK) is emitted as a photon of Kα-line, • the number of target atoms that make this transition is large. Hence the high intensity. (c) Using eV = hc λmin . Potential difference, V = (6.63 × 10–34) (3.0 × 108 ) (1.60 × 10–19)(6 × 10–11) = 2.07 × 104 V (d) Less than 1% of the total energy of electrons striking the target is converted into X-rays. Hence 99% of the power supplied to the X-ray tube is dissipated as heat. The large amount of heat generated is conducted away from the target by the cooling water to prevent the target from melting. (e) Safety precautions • The X-ray tube must be immediately switch off after used. • Lead shields are used to focus the X-rays so that unnecessary exposure is avoided. Quick Check 6 1. The potential difference between the cathode and the target of an X-ray tube is 10 kV. What is the minimum wavelength of the X-rays produced? A 4.0 × 10–11 m C 1.2 × 10–10 m B 8.0 × 10–11 m D 1.1 × 10–9 m 2. A photon of characteristic X-ray is produced when A an electron from the cathode is decelerated on collision with the target. B an electron in the target atom drops from a higher energy level to a lower energy level. C kinetic energy of an accelerated electron is converted into an X-ray photon. D an electron in a target atom is excited to a higher energy level. 3. Which of the following represents the spectrum of X-rays from a X-ray tube? A C Intensity Wavelength 0 Intensity Wavelength 0 B D Intensity Wavelength 0 Intensity Wavelength 0

Physics Term 3 STPM Chapter 24 Quantum Physics 270 24 4. The graph below shows how the intensity I of X-rays from an X-ray tube varies with the wavelength. 0 I λ λmin λmin represents the wavelength A of the characteristic Kα-line of the target B that corresponds to all the energy supplied to an electron in the accelerating electric field been converted to an X-ray photon C of the incident photon that wound ionize an atom of the target D below which X-rays from the X-ray tube would not be able to emerge from the tube 5. Which of the following graphs best represents the spectra produced by an X-ray tube for two different accelerating potentials of V1 and V2 (V1 > V2 )? A Intensity 0 λ V1 V2 B 0 λ Intensity V1 V2 C 0 λ Intensity V1 V2 D 0 λ Intensity V1 V2 6. The figure shows the spectrum produced by an X-ray tube with a constant accelerating potential. Wavelength, λ 0 Intensity, I P Q R S If the accelerating potential is increased A the maximum Q remains at the same wavelength B the peaks R and S shift to smaller values of wavelength C the peaks R and S shift to larger values of wavelength D the cut-off wavelength at P becomes smaller 7. The figure shows the X-ray spectrum produced by an X-ray tube. Wavelength Characteristic lines 0 Intensity λ1 λ2 The values of the wavelengths λ1 and λ2 depends on A the potential difference between the cathode and target of the X-ray tube B the temperature of the cathode C the current due to the beam of electrons D the type of metal of the target 8. Which statement explains the production of a wavelength in the continuous X-ray spectrum? A The ionization of an electron in the target atom. B It is the de Broglie wavelength of an electron moving from the cathode. C The energy from the transition of an electron from a higher energy level in the target atom. D The transformation of the kinetic energy of an electron from the cathode during deceleration.

Physics Term 3 STPM Chapter 24 Quantum Physics 271 24 X-ray Diffraction – Bragg’s Law 2012/P1/Q46, 2012/P2/Q13, 2013/P3/Q13, 2017/P3/Q12 1. Figure 24.26 shows a beam of monochromatic X-ray incident on a crystal. Atoms in a crystal are arranged orderly in atomic planes. Each atom in the crystal diffracts the incident X-ray. 2. The figure shows X-rays being diffracted by atoms in the top plane. Similarly X-rays are also being diffracted by atoms in the other atomic planes. 3. Figure 24.27 shows the X-rays being diffracted by atoms in different atomic planes S1 , S2 , S3 , and S4 which are separated uniformly by a distance d. 4. In certain directions, X-rays diffracted by atoms in successive atomic planes produces constructive interference and strong diffracted beams are detected. 5. The condition for constructive interference is: path difference between X-ray diffracted by atoms in two successive atomic planes = mλ m = 1, 2, 3, ... and λ = wavelength of X-ray. From the notations shown in Figure 24.27, path difference = AB + BC = 2d sin θ Hence 2d sin θ = mλ m = 1, 2, 3, ... is known as the order of diffraction d = separation between atomic planes θ = glancing angle This equation is known as Bragg equation. Ionisation chamber C1 C2 Figure 24.28 Exam Tips Bragg equation is for diffraction of X-rays, not reflection. In reflection of light, the reflected light is detected for all angles of incident. The angle of reflection = angle of incident. In diffraction of X-rays, strong X-rays are only detected when the X-rays are incident along certain directions only. 6. Only X-rays incident at glancing angle θ that satisfies Bragg equation is strongly “reflected”. Figure 24.27 S1 S2 S3 S4 d P A B C d θ θ Figure 24.26 Atom Incident radiation Diffracted radiation Atomic planes

Physics Term 3 STPM Chapter 24 Quantum Physics 272 24 7. Bragg used the X-ray spectrometer shown in Figure 24.28 to measure the separation between atomic planes in a crystal. 8. X-rays from an X-ray tube are collimated into a fi ne beam using thick lead plates with narrow slits C1 and C2 . The crystal is placed on a turntable fi tted with a circular scale to measure the angle θ. Diffracted X-rays are detected by an ionisation chamber. Example 19 (a) X-rays can be diffracted by a crystalline solid. What does this fact suggest about (i) X-rays, (ii) the crystalline solid? (b) The fi gure shows atoms in a crystal arranged in atomic planes separated by a distance of 0.30 nm. Explain why no strong diffracted X-rays can be detected from this crystal if the wavelength is greater than a certain maximum value λmax and fi nd the value of λmax. Solution: (a) (i) If X-rays are diffracted, it shows that X-rays are waves since diffraction is a wave property. (ii) The atoms in the crystalline solid are arranged in parallel planes with the atomic plane separation of the same order of magnitud as the wavelength of X-rays. (b) Using Bragg equation 2d sin θ = mλ when the value of m is smallest, i.e. m = 1, the wavelength λ given is maximum. 1 × λmax = 2d sin 90° λmax = 2 × 0.30 nm = 0.60 nm Example 20 X-rays of wavelength 100 pm are incident on a crystal. The largest angle from the normal to the crystal for Bragg diffraction is 60°. What is the wavelength of the incident X-rays if the largest angle from the normal for Bragg diffraction is 25°. Solution: When the angle of incidence, 60° is the largest, the Bragg angle (90° – 60°) = 30° is smallest. When m = 1, λ’ = 100 pm, θ = 30°. Hence 2d sin 30° = 1 × (100 pm) ............................① When m = 1, θ = (90° – 25°) = 65°. 2d sin 65° = 1 × λ’ ............................② — ② ① : sin 65° sin 30° = λ’ (100 pm) λ’ = 181 pm Crystal 30° 30° 60°

Physics Term 3 STPM Chapter 24 Quantum Physics 273 24 Example 21 The density of NaCl is 2.17 g cm–3 and the mass of a mole is 58.5 g. In the solid form, each molecule of NaCl consists of two ions, one Na+ and another Cl– arranged alternately in a cubic arrangement. (a) Calculate the number of ions in a crystal of NaCl in the form of a cube of lengths 1.0 cm. (b) What is the separation between two neighbouring ions? (c) Find the smallest Bragg angle for X-rays of wavelength 1.54 × 10–10 m. (d) What is the maximum order of diffraction that can be obtained from these planes of ions? Solution: (a) In 1 mole or 58.5 g of NaCl, there are 6.02 × 1023 molecules and number of ions = 2 × (6.02 × 1023). A NaCl crystal of cube 1.0 cm3 is of mass 2.17 g. Number of ions in cube = 2.17 58.5 × (2 × 6.02 × 1023) = 4.47 × 1022 (b) Number of ions along an edge of the cube of length 1.0 cm = (4.47 × 1022) = 3.55 × 107 ions Hence separation between ions, d = (1.0 × 10–2) (3.55 × 107 ) = 2.82 × 10–10 m (c) Using 2d sin θ = nλ when m = 1, θ is smallest sin θ = λ 2d = (1.54 × 10–10) 2(2.82 × 10–10) θ = 15°50′ (d) Using 2d sin θ = mλ m  2d sin 90° λ  2 × (2.82 × 10–10) × 1.0 (1.54 × 10–10)  3.7 Maximum order of diffraction = 3. Quick Check 7 1. X-rays of wavelength λ are incident on a crystal. The fi gures below show the direction of incident beam and the fi rst order diffracted beam. The atoms in the crystal are represented by dots. Which one of the fi gures shows correctly the angle θ and the distance d as in Bragg equation? A θ d B θ d C θ d D θ d

Physics Term 3 STPM Chapter 24 Quantum Physics 274 24 2. X-rays of wavelength 1.5 × 10–10 m incident on a crystal produces third order diffraction when the glancing angle is 60°. What is the distance between atomic planes of the crystal? A 2.6 × 10–10 m B 5.2 × 10–10 m C 9.0 × 10–10 m D 2.9 × 10–9 m 3. X-rays are diffracted by crystals but are not diffracted by a diffraction grating because A ions in the crystal lattice are wellarranged B lines of the diffraction grating are not able to reflect X-rays C the wavelength of X-rays are of the same order of magnitude as the separation between atomic planes in the crystal D X-rays have high penetrating power, and the diffraction grating is thin 4. The distance between a set of atomic planes in a crystal is 0.300 nm. The crystal is used in an X-ray spectrometer. How many order of diffraction can be detected from these planes if the incident X-rays is of wavelength 0.154 nm? 5. The structure of a crystal of which the separation between atomic planes is 0.30 nm is being studied using X-rays. The first order diffraction is obtained for Bragg angle 30°. (a) Calculate the wavelength of the X-rays. (b) What is the minimum accelerating potential across the X-ray tube required to produce X-ray photons of this wavelength? 6. (a) Explain how X-rays are strongly diffracted in certain directions by a crystalline solid. (b) With the aid of labelled diagram identify the symbols m, d, and θ in Bragg equation mλ = 2d sin θ (c) In copper, the atoms are arranged in a ‘face-centred lattice’ as shown in Figure (a). (b) a B P D C A (a) A a R B S T G U H Q P D C E F There is an atom at each of the corner of the cube (A, B, C, D, E, F, G, H) and at the centre of each face (P, Q, R, S, T, U). Figure (b) shows the atoms on a face of the cube. The atoms are assumed to be hard spheres in contact along the diagonals APC, and BPD. All atoms in the cube shown in Figure (a) are shared with neighbouring cubes. X-ray diffraction measurements show that the cube side is equal to 0.36 nm. (i) Find the diameter of a copper atom. (ii) What is the number of effective number of whole atoms within the cube? (iii) Determine the fraction of the volume of the cube occupied by copper atoms. (iv) The density of copper is 9.0 × 103 kg m–3 and the mass of a mole of copper is 0.064 kg. Find the number of copper atoms in a mole.

Physics Term 3 STPM Chapter 24 Quantum Physics 275 24 24.5 Nanoscience Students should be able to: • explain the basic concept of nanoscience • state the applications of nanoscience in electronics devices. Learning Outcomes 1. One nanometer is 10–9 m. The diameter of an atom is about 0.1 nm. 2. Nanoscience is the sciences of developing materials at the atomic and molecular level. 3. Nanotechnology involves the manipulation of materials on an atomic or molecular scale. Nanotechnology is making significant contributions to the fields of electronics, biotechnology, organic chemistry and molecular biology. 4. Materials reduced to nanoscale can show different properties compared to properties they have on a macroscale. 5. An example is the carbon nanotube which was accidentally discovered by a Japanese researcher at NEC in 1990. A carbon nanotube resembles a cylinder one to two nanometers in diameter by any number of millimeters in length. 6. With a tensile strength of 10 times greater than steel at about one quarter the weight, nanotubes are considered the strongest material for their weight known to mankind. 7. Currently it is used to strengthen plastics and carbon fi bers, as well as reinforcing structural materials in buildings, cars and airplane. 8. Applications of carbon nanotechnology in electronics: • Nanotubes may replace silicon in electronic circuits. Nanotube transistors and non-volatile memory chips (NRAM) had been created. Nanotubes are used in the construction of sensors and display screens with better colour, contrast and defi nition. • Combining gold nanoparticles with organic molecules to create a transistor known as a NOMFET (Nanoparticle Organic Memory Field-Effect Transistor). • Nanoelectronics holds some answers for how the capabilities of electronics devices can be increased while reducing their weight and power and power consumption. • Intel had developed a 32 nm computer chip that has high performance and is energy effi cient. Quick Check 8 Figure 24.29 Important Formulae 1. Einstein’s equation: maximum K.E., Emax = hf – W 2. Maximum K.E., Emax = eVs where Vs = stopping potential 3. Work function, W = hf0 where f 0 = threshold frequency 4. Crompton’s effect – particle property of wave: Momentum of photon, p = mc = h λ 1. What is nanoscience? 2. Give two uses of nanoscience in electronics. 2017/P3/Q13

Physics Term 3 STPM Chapter 24 Quantum Physics 276 24 5. de Broglie’s wavelength of a particle of mass m moving with a velocity v: λ = h mv 6. Bohr’s postulates: angular momentum, L = mvr = —– nh 2π Frequency, f of EM radiation emitted or absorbed is given by E2 – E1 = hf 7. Energy levels in hydrogen atom, En = – ——13.6 n2 eV n = 1, 2, 3, … 8. Maximum frequency, f max and shortest wavelength, λmin of X-rays from an X-ray tube with accelerating potential V are given by hfmax = hc λmin = eV 9. Bragg’s law: 2d sin θm = mλ STPM PRACTICE 24 1. The work function of a metal is 3.60 eV. Photons of energy 4.20 eV is incident on the metal surface. Which statement about the kinetic energy of the photoelectrons emitted is correct? A The minimum kinetic energy of the photoelectrons is 0.60 eV B The maximum kinetic energy of the photoelectrons is 0.60 eV C The mean kinetic energy of the photoelectrons is 0.60 eV D The kinetic energy of all the photoelectrons is 0.60 eV 2. The threshold frequency of the cathode of a photocell is f 0 . Light of frequency 3f 0 is incident on the metal surface. What is the stopping potential? A hf0 e C 3hf0 e B 2hf0 e D 4hf0 e 3. The variation of energy E of a photon with its wavelength λ is shown by the graph A C E 0 λ E 0 λ B D E 0 λ E 0 λ 4. Which statement is not true about a photon? A It has a rest mass. B It is a discrete amount of energy. C It has a discrete amount of momentum. D It exerts a force when refl ected from a surface. 5. The work functions of four metals P, Q, R and S are given in the table below. Metal P Q R S Work function / eV 1.82 2.28 2.41 3.02 Which metal (s) will show the photoemission light of wavelength 540 nm? A P C P, Q and R B P and Q D P, Q, R and S 6. Electron, proton, hydrogen atom and alphaparticle have the same kinetic energy. The particle that has the shortest de Broglie’s wavelength is A Electron B Proton C Hydrogen atom D Alpha-particle

Physics Term 3 STPM Chapter 24 Quantum Physics 277 24 7. Which of the following shows that electrons exhibit wave property? A Electrons have mass B Electrons travel in straight line C Electrons are diffracted by a crystal D Electrons are deflected by magnetic field 8. Four energy levels of an atom are shown in the figure. E4 E3 E2 E1 What is the maximum number of spectral lines obtainable from the transitions of electrons between these energy levels? A 4 C 6 B 5 D 8 9. Figure (a) shows the line spectrum produced by the transitions of electrons between the four energy levels of an atom (Figure (b)). X Increasing wavelength Y E4 E3 E2 E1 The lines X and Y are produced by the transitions A E4 → E1 and E4 → E2 respectively B E4 → E1 and E3 → E2 respectively C E4 → E3 and E3 → E2 respectively D E4 → E3 and E4 → E2 respectively 10. Which of the following is the correct representation of the four lowest energy levels of an atom? A n = 4 n = 3 n = 2 n = 1 Energy B n = 4 n = 3 n = 2 n = 1 Energy C n = 4 n = 3 n = 2 n = 1 Energy D n = 4 n = 3 n = 2 n = 1 Energy 11. A spectrometer is used to observe a beam o f white light that passes through a hydrogen discharge tube and is then at normal incident on a diffraction grating. What is observed? A Line spectrum of hydrogen B Continuous spectrum of white light C Absorption spectrum of hydrogen D A white line for the different orders of diffraction 12. The graph shows the X-ray spectra produced by two X-ray tubes, P and Q. Intensity Wavelength 0 Q P Which statement is correct? A The target metals of the X-ray tubes are different B The potential difference across the X-ray tube P is greater C The anode current in X-ray tube Q is greater D The minimum wavelengths from X-ray tubes are different

Physics Term 3 STPM Chapter 24 Quantum Physics 278 24 13. The figure shows the X-ray spectrum from an X-ray tube with copper target. Intensity Wavelength λmin 0 Kα Kβ P Which of the following does not change for the X-ray spectrum from another X-ray tube using tungsten as target, but operating with the same potential difference between anode and cathode? A The height of the peak P B The wavelength of the Ka-line C The wavelength of the Kb --line D The minimum wavelength, λmin 14. A beam of X-rays of wavelength 3.00 × 10-10 m is directed to a crystal. When the glancing angle is 12.5o , strong first order diffracted beam is detected. What is the interplanar separation of the crystal? A 1.54 × 10–10 m C 6.93 × 10–10 m B 3.07 × 10–10 m D 1.39 × 10–9 m 15. A photon which has the highest energy is the photon of A g-ray C yellow light B X-ray D infrared radiation 16. Photons with a wavelength λ strikes a surface at a rate of n photons per second. What is the expression of the force exerted on the surface if the photons rebound with the same speed? A nh λ C 2nh λ B nλ h D 2nλ h 17. The energy level E of a hydrogen atom is given by En= – ——13.6 n2 eV   n = 1, 2, … An electron in the hydrogen atom is excited to the first excited state. What is the energy of the electron in that state? A – 1.50 eV C – 10.20 eV B – 3.40 eV D – 13.6 eV 18. The production of the characteristic spectrum of X-rays is due to A the transition of an electron to the innermost shell B the transition of an electron to the outermost shell C all the energy of an incident electron being converted to photons D part of the energy of an incident electron being converted to photons 19. The energy levels for an atom is as shown to scale in the figure. The electron transitions give rise to the emission of spectrum of lines of frequencies f 1 , f 2 , f 3 , f 4 , and f 5 Which statement is correct? A f 2 < f 1 B f 3 = f 1 + f 5 C f 4 is the highest D The transition corresponds to the f3 represents the ionization of the atom. 20. An X-ray of wavelength λ experiences maximum reflection of second order when it is incident at an angle of 30° with the surface of a crystal. When an X-ray of wavelength 100 pm is used, maximum reflection of the third order occurs at an angle of 60° with the surface of the crystal. What is the value of λ? A 39 pm C 87 pm B 67 pm D 150 pm 21. In Bohr’s model of the hydrogen atom, an electron of mass m, charge e orbits the nucleus in orbits with discrete radii r. What is the angular momentum of the electron in the orbit closest to the nucleus? A 2π h C h 2π B 2π mh D mh 2π

Physics Term 3 STPM Chapter 24 Quantum Physics 279 24 22. The energy levels of an electron in a hydrogen atom are given by En = – 13.58 n2 eV, n = 1, 2, 3,… What is the wavelength of the photon that would raise an atom from the ground state to the energy level n = 5? A 1.36 × 10–9 m B 9.15 × 10–8 m C 9.54 × 10–8 m D 1.46 × 10–6 m 23. The figure shows part of the line spectrum of hydrogen. 1 2 3 4 Red Violet Which one of the following shows the transitions between levels n = 2 to n = 6 of the hydrogen atom that is responsible for the spectral lines shown. A n = 6 n = 5 n = 4 n = 3 n = 2 B n = 6 n = 5 n = 4 n = 3 n = 2 C n = 6 n = 5 n = 4 n = 3 n = 2 D n = 6 n = 5 n = 4 n = 3 n = 2 24. The figure shows a number of energy levels in the hydrogen atom. E1 E2 E3 E4 E5 – 0.54 eV – 0.85 eV – 1.5 eV – 3.4 eV – 13.6 eV An electron that had been accelerated through a potential difference 13 V collides with a hydrogen atom in the ground state. After collision the possible excited states of the hydrogen atom are A E2 only B E2 and E3 only C E2 , E3 , and E4 only D E2 , E3 , E4 , and E5 25. The intensity of X-rays from an X-ray tube may be increased by A increasing the speed of the colliding electrons. B increasing the current in the electron beam.

Physics Term 3 STPM Chapter 24 Quantum Physics 280 24 C increasing the potential difference across the tube. D increasing the wavelength of the X-rays. 26. The background continuous X-ray spectrum is due to the A deceleration of electrons colliding with the target atom. B collisions of fast electrons with the innermost electron in the target atoms. C transitions of electrons in the target atoms from a higher energy level to a lower energy level. D electron losing all its kinetic energy when colliding with a single atom. 27. The X-ray spectrum of an X-ray tube is represented by the figure below. Wavelength 0 Intensity P Q R If the potential difference across the X-ray tube is increased, which of the following is true? A Only the values of wavelengths of the characteristic lines Q and R increase. B Only the value of the wavelength at P decreases. C The values of the wavelength at P, and that of the characteristic lines Q and R increase. D The values of the wavelength at P, and that of the characteristic lines Q and R decrease. 28. The figure shows the variation of intensity with wavelengths for X-rays from an X-ray tube. λmin 0 Wavelength Intensity If the potential difference across the X-ray tube is increased, what happens to the value of λmin and the intensity of the continuous background spectrum? λmin Intensity of continuous background spectrum A Decreases increases B Increases remains unchanged C Increases increases D Remains decreases unchanged 29. The wavelength of the Kα-line of the X-ray spectrum produced by an X-ray tube increases if A the accelerating potential difference decreases. B the temperature of the filament increases. C the distance of the target from the cathode increases. D the target metal has a lower atom number. 30. (a) State de Broglie’s hypothesis. (b) (i) An electron after been accelerated through a potential difference V has a de Broglie wavelength of 2.5 × 10–11 m. Find the value of V. (ii) A proton and an electron have the same value of de Broglie wavelength. What is the ratio of the kinetic energy of the proton to the kinetic energy of the electron? 31. The accelerating voltage across an X-ray tube is 20 kV and the anode current is 25 mA. (a) Find the speed of the electrons striking the target. (b) What is the minimum wavelength of X-rays produced? (c) If 5% of the power input is transformed into X-ray photons, what is the power dissipated as heat? State two characteristics of the target metal in the X-ray tube. (d) State three characteristics of the spectrum. 32. (a) Distinguish between the excitation energy and the ionisation energy of a hydrogen atom.

Physics Term 3 STPM Chapter 24 Quantum Physics 281 24 (b) The energy at the nth level of a hydrogen atom is given by the expression En = – 13.6 n2 eV, where n is an integer. An electron in a hydrogen atom undergoes a transition from the energy level n = 3 to n = 1 by emitting a photon. (i) Determine the energy of the photon emitted. (ii) Calculate the wavelength of the photon emitted. 33. (a) Explain the difference in the production of the minimum wavelength λmin, and a characteristic of X-ray. (b) An X-ray tube has copper target. The potential difference across the X-ray tube is 28 kV. (i) What is the speed of electrons that strike the copper target? (ii) An electron striking the target gives rise to a photon of X-ray characteristic of wavelength 1.54 × 10-10 m. What is the speed of the electron after striking the target? 34. The cathode of a photocell is radiated with monochromatic light and the current I in the photocell is measured for different anode potential V. The graph of I against V is as shown below. 0 I -V0 V (a) Explain the shape of the graph (i) for positive values of V , and (ii) for negative values of V. (b) The experiment is repeated (i) using the same monochromatic light but the intensity is increased, (ii) using another monochromatic light of shorter wavelength. Copy the above graph and on the same axes, sketch the graphs of I against V for (i) and (ii). Label the graphs clearly. Explain the change in the shape of the graphs. (c) A beam of monochromatic light of wavelength 520 nm from a source of 1.50 W is incident on a metal surface of area of 8.2 mm2 . (i) What is the rate of photons emitted from the source? (ii) One in five of the photons incident of the metal surface is able to eject a photoelectron. What is the rate of photoelectrons from the metal surface? 35. In the Bohr’s model of the hydrogen atom, electron moves round the nucleus in discrete orbits of radius rn given by rn = n2 h2 e0 πme e2 n = 1, 2, 3, … (a) Deduce the expression for (i) electric potential energy of the hydrogen atom, (ii) the energy En (b) Calculate the values of rn , and En for (i) n = 1, and (ii) n = 3. (c) Draw a figure to show the three lowest allowed orbits of the electron. 36. The figure shows the energy levels of molybdenum atom and two transition Lα and Kα. 0.0 keV – 0.5 keV – 2.6 keV – 20.0 keV Lα Kα M L K (a) If molybdenum is used as the target element of an X-ray tube, calculate (i) the wavelength corresponds to the transition Lα. (ii) the minimum potential difference across the X-ray tube required to produce the line associated with the transition Kα.

Physics Term 3 STPM Chapter 24 Quantum Physics 282 24 (b) If electrons in the X-ray tube are accelerated by a potential difference of 100 V, by making the necessary calculation, explain whether X-rays would be produced. 37. A photoelectric effect experiment was conducted using nickel of work function 5.01 eV and the stopping potential of 1.20 V. (a) Determine the maximum kinetic energy of photoelectrons emitted. (b) Calculate the wavelength of the photons used in the experiment. 38. (a) (i) State Bohr's postulates of the hydrogen atom. (ii) State two weaknesses of Bohr's model of the atom. (b) (i) Draw a labelled diagram to show the four lowest energy levels in a hydrogen atom. (ii) An excited hydrogen atom at n = 3 drops to the ground state. Calculate the wavelength of the photon emitted. (iii) Determine the wavelength of the incident photon that would ionize the hydrogen atom. (c) Name two series of spectral lines of the hydrogen atom. Which series is visible? 39. (a) Use Bohr's first postulate of the hydrogen atom, estimate the speed of the electron in the energy level n = 3. (Radius of orbit for n = 3 is 4.78 × 10–10 m) (b) The figure below shows the energy level of a hydrogen atom. E∝= 0 E5 = – 0.54 E4 = – 0.85 eV E3 = – 1.51 eV E2 = – 3.40 eV E1 = – 13.60 eV (i) An electron makes the transition from the energy level n = 3 to n = 1. Determine the wavelength of the emitted photon. (ii) Radiation consisting of photons of energy 10.2 eV and 12.0 eV is directed towards a hydrogen discharged tube. Describe an explain what happens to the photons in the discharge tube on collision with hydrogen atoms in the ground state.. 40. The figure shows three of the electronic energy levels of an atom. What are the frequencies of electromagnetic radiation which can be emitted from transition between these levels? – 9.68 fi 10–19 J – 21.78 fi 10–19 J – 87.12 fi 10–19 J State the type of each of the electromagnetic radiation emitted. 41. Describe the appearance of (a) the spectrum of light from a tungsten filament lamp, (b) the spectrum of light from a sodium vapour lamp, (c) an absorption spectrum. 42. (a) State three properties of X-rays and, (b) give three uses of X-rays based on each of the three properties. 43. The graph below shows the X-ray spectrum from an X-ray tube. λmin 0 Wavelength Intensity For each of the cases below, copy the above figure and on the same axes, sketch the new X-ray spectrum (a) when the current in the filament is increased, (b) when the potential difference across the X-ray tube is increased, (c) when the target metal is one with greater atomic number. Explain the changes in the spectrum for (a), (b), and (c).

Physics Term 3 STPM Chapter 24 Quantum Physics 283 24 44. (a) (i) Write the order of magnitude of the X-ray wavelengths. (ii) State two differences between X-rays and gamma-rays. (b) Explain the production of characteristic spectrum and continuous spectrum of X-rays. (c) The arrangement of atoms in the square lattice of a crystal is shown in the diagram below. Incident X-ray Crystal planes 0.25 nm The interatomic distance is 0.25 nm. An X-ray of wavelength 0.10 nm is incident on a set of crystal planes. (i) Calculate the Bragg’s angle for the first order maximum due to the set of crystal planes. (ii) Sketch two other possible sets of crystal planes. (iii) Determine the interplanar spacing d which gives the minimum Bragg’s angle for the first order maximum. 45. (a) Draw a labelled diagram to show the important structures of an X-ray tube. (b) Explain each of the followings. (i) The X-rays produced by the X-ray tube consists of a range of wavelengths. (ii) The existence of a minimum wavelength. (iii) X-rays of high intensity are produced for discrete wavelengths. 46. (a) (i) What experimental evidence suggests the existence of discrete energy levels in an atom? (ii) Explain why energy of an electron in an atom is quantized. (b) The figure represent five of the lowest energy levels of the hydrogen atom. Energy/J Level – 8.3  10–20 – 1.3  10–19 – 2.3  10–19 – 5.3  10–19 – 2.2  10–18 A B C n = 5 n = 4 n = 3 n = 2 n = 1 (i) Calculate the wavelengths associated with the transitions A, B, and C. (ii) Show that these wavelengths λ fit the formula 1 λ 1 22 1 n2 = R – where R is a constant and n is the quantum number of the upper level involved in the transition. Deduce a value of R. (iii) Calculate the shortest wavelength associated with the Balmer series mentioned above and explain which transition gives rise to it. 47. (a) What is the meaning of each word as used in the term emission line spectrum? (b) Describe how you would carry out an experiment to find the wavelength of light of one of the lines in the hydrogen emission spectrum. (c) The measured wavelengths, λm of selected lines in the hydrogen spectrum are given empirically by 1 λm 1 4 1 m2 = R – where R is a constant and has the value 1.097 × 107 m–1 and m is an integer taking the values 3, 4, 5,… (i) Calculate the value of the wavelength when m = 4. (ii) Calculate the minimum wavelength given by this equation. (iii) Draw a diagram showing the approximate positions of the lines on a horizontal axis of wavelength. Mark the two values you have already calculated and also mark the red and the violet ends of the spectrum.

Physics Term 3 STPM Chapter 24 Quantum Physics 284 24 (iv) Explain why although there is an infinite number of lines in this spectrum, the spectrum is nevertheless seen as a line spectrum. 48. (a) Sketch a simple energy level diagram for a one-electron atom, and use it to explain what is meant by (i) ground state, (ii) ionisation energy, (iii) excitation energies. (b) In a model of the hydrogen atom, an electron of mass m and charge –e goes round a stationary proton of charge +e in an orbit of radius r. (i) Assume that the electrostatic attraction between the electron and proton provides the centripetal force, deduce an expression for the angular velocity ω of the electron in terms of e, r, m, and ε0 the permittivity of a vaccum. (ii) Hence show that the angular momentum L of an electron in orbit is mre2 4πε0 . (iii) Show that the total energy of the electron is given by – L2 2mr2 . (c) According to this model of the atom, the angular momentum of the electron can only take values nh 2π , n = integer and h is Planck constant. (i) Find an expression for a0 the radius of the lowest orbit in terms of e, m, ε0 and h. (ii) If a0 = 5.3 × 10–11 m, what is the energy required to remove completely an electron from this orbit to infi nity. 49. (a) State Bohr’s postulates in relation to the structure of the hydrogen atom. (b) In a hydrogen atom an electron revolves round a proton in an orbit of radius r. (i) Write an expression for the electrostatic force between the electron and the proton. Identify the symbols in your expression. (ii) Use Bohr’s postulate to write an expression for the angular momentum of the electron. (iii) Hence deduce an equation relating the mass of the electron m, its speed v, its radius of orbit r, the quantum number n, and Planck constant h. (iv) Deduce an expression for the radius r of the orbit in terms of the quantities mentioned, and ε0 the permittivity of free space. (v) Calculate the radius of the smallest orbit of the electron. (c) Use the value of the radius of orbit in (b) (v) above to calculate (i) the electrostatic potential energy (ii) the kinetic energy (iii) the total energy of the proton-electron system in the hydrogen atom. 50. (a) Write down a typical wavelength for (i) visible light (ii) X-ray. (b) The basic components of a spectrometer to examine regions of the electromagnetic spectrum consists of a collimator, a dispersion element, and a detector system. (i) What is the function of the collimator? In an X-ray spectrometer, how is the X-ray collimated? (ii) Explain why a piece of diffraction grating is not suitable as the dispersion element in an X-ray spectrometer. (iii) Describe briefl y the detection system used in the X-ray spectrometer. (c) The energy of an electron in the various shells of the nickel atom is: K-shell – 1.36 × 10–15 J L-shell – 0.16 × 10–15 J M-shell – 0.08 × 10–5 J An X-ray tube has a nickel target. (i) What is the minimum potential difference across the tube for it to emit X-ray lines spectrum in the K-series? (ii) What is the wavelength of the Kα-line?

Physics Term 3 STPM Chapter 24 Quantum Physics 285 24 1 1. B 2. B 3. D 4. B 5. A 6. D 7. C 8. C 9. C 10. B 11. B 12. A 13. B 14. B 15. C 16. C 17. A 18. B 19. B 20. C 21. D 22. A 23. A: E = hc λ , λ = hc E = 8.57 × 10–8 m 24. C: Kmax = 1.35 eV = hc λ – hc λmax λmax = 772 nm 25. (a) 4.57 × 10–19 J E = hc λ (b) 1.51 × 10–19 J 26. 2.26 W m–2, E = nhc λ , n = I e Intensity = E Area Assumption: Each photon ejects one electron. 27. 7.1 × 10–19 J Emax = hc λ – W 28. 2.2 eV Apply a reverse potential diff erence until all photoelectrons are stopped. 29. (a) Remains unchanged E = hc λ (b) Remains unchanged. Photon unchanged. (c) Remains unchanged. Work function characteristic of metal surface. (d) Doubled. Rate of emission of photoelectrons doubled. Photon s–1 doubled. 30. (a) I 0 f (c) I 0 p (b) I V 0 31. (a) (i) Planck constant, h = E f . E = energy of photon f = frequency of e.m. radiation (b) Measure stopping potential Vs for various f. Emax = eVs Gradient of graph = h. 0 f Emax f0 (c) (i) Same frequency, but higher intensity of u.v. radiation. (ii) Higher frequency u.v. radiation, but same number of photons per second. 32. (b) (ii) Derive V = h e f – W e (c) (i) 7.2 × 10–19 J E = hc λ (ii) 2.5 eV Emax = hc λ – W (iii) 8.4 × 105 m s–1, 1 2 mv2 = 2.0 eV (iv) Energy 0 Potential energy Distance Total energy Kinetic energy 33. (a) Emax = hf – W Emax : Maximum kinetic energy of photoelectron hf : energy of incident photon W : Work function of metal surface (b) (i) Emax = 0.50 eV Vs = 0.50 V (ii) 2.55 eV Emax = hf – W (iii) 3.50 × 1016 s–1 (iv) 4.95 × 1014 Hz W = hfo (c) (i) No. Frequency of radiations in experiment 1 and 2 are not equal. (ii) Number of photons s–1 in (2) is halved. Higher frequency because stopping potential bigger. 34. (a) Quantum theory: Energy of light is quantized, that is in discrete packets called quanta or photons. ANSWERS

Physics Term 3 STPM Chapter 24 Quantum Physics 286 24 Photon: a discrete packet of energy E of an electromagnetic radiation or light given by E = hf, where h =Planck’s constant, f = frequency of radiation. (b) 1. Threshold frequency - below which no photoemission. Minimum energy required by electrons in the metal to ejected. 2. Instantaneous emission. Electrons are emitted immediately if the photon is greater than the minimum energy required. 3. Maximum kinetic energy of photoelectrons depends on the frequency f of the incident light or em radiation. Energy of photon = hf. 4. Maximum kinetic energy is independent of the intensity of light. Maximum kinetic energy depends on f. 5. Rate of emission increases when intensity of light increases. Intensity increases – number of photons per second increases. 35. (a) Einstein’s equation: Emax = hf – W Emax: maximum K.E. of photoelectrons hf: energy of incident photon W: work function of metal – characteristic of the metal (b) Maximum K.E. Emax is given by Einstein’s equation. – W is the minimum energy required by electrons to escape from the metal surface. – electrons that are further below the metal surface require more energy to escape from the metal surface. (c) (i) λ = hc 1.20 × 10–18 = 5.53 × 10–8 m (ultraviolet radiation) (ii) W = (1.20 × 10–18 - 4.2 × 10–19 ) J = 7.80 × 10-19 J (iii) f 0 = W h = 1.18 × 1015 Hz 2 1. C: de Broglie wavelength = h mv 2. A: λ = h mv , v = h mλ 3. D: Momentum of photon = mc = h λ , independent of intensity. 4. A: λ = h mv , v = h p 5. B: Diffraction – wave property 6. A: Apply F = e(v × B) 7. C: E = 1 2 mv2 , v = 2E m , λ = h mv = h 2mE 8. D: λ = h mv , v = h mλ , speed v is greatest when m is smallest. 9. D: E = eV, λ = h 2mE , λe λp = mp me 10. B: E = 3 2 kT, λ = h 2mE = h 3mkT 11. B: Shorter λ, better resolving power. 12. (a) 3.96 × 10–19 J (b) 1.32 × 10–27 kg m s–1 13. λ = 4.9 × 10–10 m  size of atom 14. Momenteum: 4.32 × 10–24 kg m s–1 Potential difference: 67 V 15. (a) Refer to page 241 (b) (Derive: λ = h 2me E ) (c) (i) 3.88 × 10–11 m (ii) 2 1 2 mv2 = eV (iii) 3° 32′ 2d sin θ = nλ, θ1 = 1 2 (5° ) 16. (a) Refer to page 242 (b) V = h2 2meλ2 1 2 mv2 = eV, mv = h λ (c) 1 × 10–10 m to 1 × 10–9 m (7 × 10–25 to 7 × 10–24) kg m s–1 p = h λ 3 1. B: E1 – E2 = hc λ , λ = hc E1 – E2 2. D: E = hc λ = (6.63 × 10–34)(3.00 × 108 ) (488 × 10–9)(1.60 × 10–19) eV = 2.55 eV = (-0.85) – (-3.40) eV 3. B 4. A 5. A: Kinetic energy after collision = 7.0 – [(-3.7) – (-10.4)] eV = 0.3 eV 6. C: Energy required = E2 – E1 = (– 13.6 22 ) – (–13.6) eV = 10.2 eV 7. B: 1 2 mv2 = E – hc λ , v = 2 m (E – hc λ ) 4 1. B: Sodium emits line spectrum 2. C: Absorption spectrum 3. D: Red photon < green photon 4. A: Lyman series

Physics Term 3 STPM Chapter 24 Quantum Physics 287 24 5. B: Dark lines- absorption spectrum of sodium, background continue spectrum of white light. 6. A: Separation between lines is proportional to energy difference. 7. B: 1 2 mv2 = –En = –( – me4 8e2 0h2 n2 ) λ = h mv = 2e0 h2 n me4 8. (a) A B C D (b) 634 nm EA – EC = hc λ (c) B to C (d) ultraviolet 9. 2.66 eV E = hc λ 10. (a) 3.05 × 10–8 m (b) ultraviolet 11. (a) E3 E 8 = 0 eV = –1.0 eV E2 = –3.00 eV E1 = –9.50 eV 6.50 eV 8.50 eV 2.00 eV (b) E3 E 8 = 0 eV = –1.0 eV E2 = –3.00 eV E1 = –9.50 eV 191 nm 146 nm 131 nm 12. (a) (i) (iii) (ii) (b) 4.09 10–19 J E = hc λ 13. (a) (i) Electron in the hydrogen atom is excited. When the excited electron drops from the higher energy level to a lower energy level, the difference in energy is radiated as a photon. (ii) Discrete wavelength – energy of electron in the various energy levels is quantized. Difference in energy between two energy levels has discrete value. (iii) Intensity of a line depends on the rate of emission of photons associated with the line. Intensity is high if the rate of emission is greater. (b) (i) After switch S is closed – starter switch is heated and opens – a large emf is induced in the coil due to electromagnetic induction. (ii) UV radiation emitted when electron drop back into the mercury atom. (iii) – UV radiation is absorbed by the atoms of the powder. – atoms are excited – EM radiations, including light are emitted when the excited atoms drop from the excited state to lower energy states. 14. (a) (i) Energy level: Sum of kinetic energy and potential energy of electron when in one of the allowed orbits. Negative sign indicates the force between the hydrogen nucleus and the electron is attractive. (ii) Existence of line spectrum of hydrogen. (b) (i) Refer to page 251 (ii) 13.6 42 eV = 0.971 eV (c) (i) K = 2.10 × 10–18 J – [13.6 12 – 13.6 23 ]eV = 1.66 × 10–18 J (ii) λ 1 λ 2 λ 3 n = 3, 2nd excited state n = 2, 1st excited state n = 1, ground state hc λ1 = [13.6 22 – 13.6 32 ] eV, λ1 = 658 nm hc λ2 = [13.6 12 – 13.6 22 ] eV, λ1 = 122 nm hc λ2 = [13.6 12 – 13.6 32 ] eV, λ1 = 103 nm 5 1. C: λmin = hc eV 2. B: λmin = hc eV , q = e 3. D: λmin = hc eV, λ decreases when V increases, 4. B: Power = IV = CdT dt ,

Physics Term 3 STPM Chapter 24 Quantum Physics 288 24 dT dt = (8 ×10–3)(20 × 103 ) 40 = 4.0 K s-1 5. A: Penetration power increases when f increases or λ decreases. 6. (a) Maximum energy of X-ray photon = energy of accelerated electron = 159 keV (b) hf = 159 keV f = (159 × 103 )(1.60 × 10–19) 6.63 × 10–34 = 3.84 × 1019 Hz 6 1. C 2. B 3. C 4. B 5. C 6. D 7. D 8. D 7 1. C 2. A 3. C 4. 3 n  2d λ 5. (a) 0.30 nm (b) 4.14 kV eV = hc λ 6. (c) (i) 0.255 nm (2d)2 = a2 + a2 (ii) 4 (iii) 0.74 4 × 4 3 πr3 /a3 (iv) 6.1 × 1023 mol–1 8 1. Nanoscience is the sciences of developing materials at the atomic and molecular level. 2. • Nanotube transistors • Create a transistor known as a NOMFET (Nanoparticle Organic Memory Field-Effect Transistor). STPM Practice 24 1. B: Einstein’s equation: Maximum kinetic energy, Kmax = hf – W 2. B: Kmax = h (3f 0 ) – hf0 = eVs , Vs = 2hf0 e 3. C: E = hc λ 4. A 5. B: E = hc λ = (6.63 × 10–34)(3.00 × 108 ) (540 × 10–9 (1.60 × 10–19) eV = 2.30 eV > WP and WQ 6. D: Kinetic energy E = 1 2 mv2 , v = 2E m , λ = h mv = h 2mE λ is smallest when m is largest. 7. C: Diffraction is a wave property. 8. C E4 E3 E2 E1 9. A: X: shortest wavelength: largest energy difference. Y: 4th in energy difference. 10. A: Differences in energy get smaller when n increases. 11. C 12. A: Wavelengths of characteristic line X-rays are different. 13. D: λmin = hc λ . Peak P is higher for tungsten as it is more efficient. 14. C: 2d sin 12.5o = (1)( 3.00 × 10–10) d = 6.93 × 10–10 m 15. A: E = hf, g-ray has the highest frequency. 16. C: Momentum of a photon = h λ Force = rate of change of momentum = n[ h λ – (– h λ )] = 2nh λ 17. B: First excited state, n = 2. E = – eV = 3.40 eV 18. A 19. B: hf3 = hf1 + hf5 20. C: Bragg’s equation: 2d sin 30° = (2) λ and 2d sin 60° = (3)(100 pm) 2 λ (3)(100) = sin 30° sin 60° , λ = 87 pm 21. C 22. C: Bohr’s postulate L = nh 2π , n = 1 23. A: Separation between two lines in spectrum is proportional to difference in the length of the transition lines. 24. C: Energy required for each transition < 13 eV. 25. B 26. A 27. B 28. A 29. D 30. (a) Refer to page 242 (b) (i) Use 1 2 mv2 = eV, l = h mv Then V = h2 2mel2 = 2.41 kV (ii) K = eV = h2 2ml2 Then Kp Ke = me mp = 5.46 × 10–4 31. (a) v = 2eV m = 8.4 × 107 m s–1 (b) lmin = hc eV = 6.2 × 10–11 m (c) (95%)(IV) = 475 W High melting point, high specific heat capacity. (d) Has a minimum wavelength, lmin Continuous background spectrum. Discrete lines with high intensity. 32. (a) Excitation energy of hydrogen atom: energy required to raise the hydrogen atom from the ground state (E1 ) to the first excited state E2 or E3 , E4 , …

Physics Term 3 STPM Chapter 24 Quantum Physics 289 24 Ionisation energy: energy required to remove completely the electron from the hydrogen atom in the ground state, E1 . (b) (i) Energy of emitted photon = E3 - E1 = (– 13.6 32 ) – (–13.6) eV = 12.06 eV (ii) hc λ = 12.06 eV λ = (6.63 × 10–34)(3.00 × 108 ) (12.06)(1.60 × 10–19) m = 103 nm 33. (a) λmin: The energy of an accelerated electron from the cathode is completely transformed into a photon of X-ray of minimum wavelength, eV = hc λmin . Characteristic X-ray: A fast electron from the cathode collides with the electron in the K-shell of the target atom. The electron is ejected from the K-shell. When the vacancy is the K-shell is filled by an electron from the L-shell, the energy (EL – EK) is radiated as a photon of characteristic X-ray. (b) (i) 1 2 mu2 = eV Speed, u = 2eV m = 2(1.60 × 10–19)(28 × 103 ) 9.11 × 10–31 = 9.92 × 107 m s–1 (ii) 1 2 mv2 = eV – hc λ v = 2 m (eV – hc λ ) = = 8.37 × 107 m s–1 34. (a) (i) As +V increases, I increases because more photoelectrons reach the anode. When V is further increased, I is constant because all emitted photoelectron reach the anode. Current is saturated. (ii) When V is negative, anode is at a negative potential. As V becomes more negative number of photoelectrons that reach the anode decreases. At –V0 even photoelectrons with the maximum kinetic energy are stopped from reaching the anode. Maximum kinetic energy of photoelectrons = eV0 2 9.11 × 10–31 [(1.60 × 10–19)(28 × 103 ) – (6.63 × 10–34)(3.00 × 108 ) 1.54 × 10–10 ] (b) 0 (ii) (i) I -V0 V (i) Higher intensity, rate of photons incident on the cathode increases. Rate of photoelectrons emitted increases. Current I increases. (ii) Shorter wavelength, photon energy is greater. Photoelectrons emitted with greater maximum kinetic energy. A greater negative potential is required to stop photoelectrons. (c) (i) 1.50 W = nhc λ n = (1.50)(520 × 10–9) (6.63 × 10–34)(3.00 × 108 ) s-1 = 3.92 × 1018 s-1 (ii) Rate of emission of photoelectrons = 1 5(3.92 × 1018) s–1 = 7.84 × 1017 s–1 35. (a) (i) U = (+e)(–e) 4πe0 rn = – e2 4πe0 ( πme e2 n2 h2 e0 ) = – me e4 4n2 h2 e0 2 (ii) En = 1 2 U = – me e4 8n2 h2 e0 2 (b) (i) n = 1, r1 = h2 e0 πme e2 = (6.63 × 10–34)2 (8.85 × 10–12) π(9.11 × 10–31)(1.60 × 10–19)2m = 5.31 × 10–11 m E1 = – me e4 8h2 e2 0 = – (9.11 × 10–31)(1.60 × 10–19)4 8(6.63 × 10–34)2 (8.85 × 10–12)2 J = –2.17 × 10-18 J (ii) n = 3, r3 = (32 ) r1 = 4.78 × 10–10 m E3 = 1 32 E1 = –2.41 × 10–19 J (c) +e n = 2 n = 1 n = 3 -e

Physics Term 3 STPM Chapter 24 Quantum Physics 290 24 36. (a) (i) 5.92 × 10–10 m hc λ = EM – EL (ii) 20.0 keV eV = E∞ – EK (b) No. When V = 100 V λ = hc eV = 1.2 × 10–8 m > λ of X-ray 37. (a) Kmax = eVs = 1.20 eV or (1.20)(1.60 × 10–19) J = 1.92 × 10–19 J (b) hc λ = Kmax + W = (1.20 + 5.01) eV λ = (6.63 × 10–34)(3.00 × 108 ) (6.21)(1.60 × 10–19) m = 2.00 × 10–7 m 38. (a) (i) Refer to page 248 (ii) Bohr's model could not explain – why electron is able to stay in circular motion – the spectral lines of poly-electron atoms – the splitting of spectral lines in magnetic or electric field. (Any two) (b) (i) Refer Figure 24.15 (ii) hc λ = E3 – E1 , En = – 13.6 n2 eV λ = 1.03 × 10–7m (iii) hc λ = E∞ – E1 λ = 9.14 × 10–8m (c) Series of line spectra – Lyman series – Paschen series and – Balmer series which is visible 39. (a) Angular momentum = (mv) r = (3)h 2π v = (3)(6.63 × 10–34) 2π(9.11 × 10–31)(4.78 × 10–10) m s–1 = 7.30 × 105 m s–1 (b) (i) hc λ = E3 – E1 λ = 1.03 × 10–7 m (ii) E2 – E1 = –3.400 – (–13.6) eV = 10.2 eV E3 – E1 = 12.1 eV – 10.2 eV-photons is absorbed and atoms are excited to n = 2. – When excited atoms return to ground state, 10.2 eV-photons are emitted. – 2.0 eV photon is not absorbed. 40. 1.83 × 1015 Hz 9.86 × 1015 Hz 1.17 × 1016 Hz All u.v. radiation. 41. (a) Continuous spectrum of various colours from red to violet. (b) Line spectrum – 2 yellow lines (c) Dark lines with coloured background 42. (a) Properties (b) Uses • Darkened photographic plates • Ionising • High penetrating power • X-ray photography • Destroy cancerous cells • To detect flaw in welding. 43. (a) 0 λ New I Number of electrons per second increases (b) 0 λ New I λmin = hc e (c) 0 λ New I λ for line spectrum decreases. 44. (a) (i) 10–10 m (ii) Differences: 1. Source: X-rays from X-ray tubes G a m m a - r a y f r o m d e c a y o f radioisotopes. 2. Frequency or Wavelength: X-rays have lower frequency, or longer wavelength compared to gamma-ray. (b) A characteristic line is produced due to transitions of electrons in the innermost shells of the atoms of the target metal of the X-ray tube. Continue spectrum is produced because electrons from the cathode have different

Physics Term 3 STPM Chapter 24 Quantum Physics 291 24 decelerations when striking the target. Energy of the electrons are converted into X-rays of different wavelengths. (c) (i) 0.25 nm X Y XY = 2d = 0.252 + 0.252 nm = 0.3536 nm Bragg’s equation: 2d sin θ1 = (1) λ sin θ1 = 0.10 0.3536 , θ1 = 16.4° (ii) Crystal planes 2 Crystal planes 1 0.25 nm (iii) From Bragg’s equation for n = 1, 2d sin θ = (1) λ, θ is minimum when d is maximum. Maximum value of d for the different crystal planes is 0.25 nm 45. (a) Refer to page 263 (b) Minimum wavelength λmin is produced when the energy of a accelerated electron in completely transformed into an X-ray photon. Wavelength > λmin are produced because different fraction of the enegry of the accelerated electrons are transformed into X-ray photons. 46. (a) (i) Emission line spectrum with discrete wavelengths (ii) Angular momentum of electron quantized. (b) (i) 4.43 × 10–7 m 4.95 × 10–7 m 6.60 × 10–7 m (ii) Show that the value of R is the same for each n. R = 1.1 × 10–7 m–1 (iii) λ = 3.7 × 10–7 m Substitute n = ∞ into equation. From n = ∞ to n = 2. 47. (a) Refer to page 255 (b) Use optical spectrometer with diffraction grating. Use λ = d sin θ (c) (i) 4.86 × 10–7 m (ii) 3.65 × 10–7 m (m = ∞). (iii) 3.65 4.86 Violet Red fi 10–7 m (iv) For large m, spectral lines very closed, cannot be resolved by the eye (or spectrometer). 48. (b) (i) ω = √ e 4πε0 mr3 (ii) Use L = Iω = (mr2 ) ω (iii) Total energy = K + U K = 1 2 Iω2 U = –e2 4πε0 r (c) (i) a0 = ε0 h2 πme2 (ii) 2.2 × 10–18 J Use L = mre2 4πε0 = h 2π 49. (a) L = mvr = nh 2π E2 – E1 = hf (b) (i) F = e2 4πε0 r2 (ii) L = nh 2π n = 1, 2, 3, … (iii) mvr = nh 2π (iv) r = ε0 n2 h2 πme2 (v) 5.31 × 10–11 m (c) (i) –4.36 × 10–18 J U = –e2 4πε0 r1 (ii) +2.18 × 10–18 J K = – 1 2 U (iii) –2.18 × 10–18 J E = K+ U = 1 2 U 50. (a) (i) 4 × 10–7 m to 7 × 10–7 m (ii) 10–11 m to 10–9 m (b) (i) To produce a narrow beam. Using lead plates with slit. (ii) Lines separation very much larger than wavelength of X-rays. (iii) GM tube or ionisation chamber. (c) (i) 8.5 kV eV = 1.36 × 10–15 J (ii) 1.66 × 10–10 m

Physics Term 3 STPM Chapter 25 Nuclear Physics 292 25 Physics Term 3 STPM CHAPTER NUCLEAR PHYSICS 25 Concept Map 292 Bilingual Keywords Activity: Keaktifan Alpha-particle: Zarah-alfa Binding energy: Tenaga pengikat Bombardment: Bedilan Carbon dating: Pentarikhan karbon Critical size: Saiz genting Decay: Reputan Element: Unsur Exponential: Eksponen Half-life: Setengah hayat Isotope: Isotop Mass defect: Cacat jisim Neutron: Neutron Nuclear fi ssion: Pembelahan nukleus Nuclear fusion: Pelakuran nukleus Nuclear reactor: Reaktor nuklear Nucleon: Nukleon Proton: Proton Radioactivity: Keradioaktifan Random: Rawak Spontaneous: Spontan Stability: Kestabilan Thermal neutron: Neutron terma Thermonuclear: Termonuklear Tracer: Pengesan Unifi ed atomic mass: Jisim atom disatukan Nucleus Nuclear Reactions Fission Fusion Radioactivity Structure of the nucleus Mass-energy Conservation Exponential law dN dt = –λN N = N0 e–λt Binding energy ΔE = Δmc 2 Half-life t 1/2 = 1n2 λ Nuclear Physics

Physics Term 3 STPM Chapter 25 Nuclear Physics 293 25 25.1 Nucleus Learning Outcomes Students should be able to: • describe the discovery of protons and neutrons (experimental details are not required) • explain mass defect and binding energy • use the formula for mass-energy equivalence ΔE = Δmc2 • relate and use the units u and eV • sketch and interpret a graph of binding energy per nucleon against nucleon number 1. The nucleus is positively charged because of the protons. 2. The proton was discovered by Rutherford in 1919 when he bombarded nitrogen gas with alphaparticles. He deduced that the protons must have come from within the nitrogen atoms. 3. From the defl ections of the proton in magnetic fi eld and electric fi eld, Rutherford found that it is positively charged. 4. In 1932, Chadwick showed that besides the protons, the nucleus contains particles which are about the same mass as the protons but uncharged. These neutral particles are known as neutrons. 5. Chadwick discovered the neutron when he used α-particles from polonium to bombard beryllium. 6. The resulting radiation was found to have high penetrating power. Some scientists thought that the radiation was X-rays or γ-rays. 7. Further experiments carried out by Chadwick showed that the radiations are neutral particles which have the same mass as protons, that is neutrons. 8. To determine the mass of neutron, Chadwick used neutrons to bombard nitrogen gas, and paraffi n wax. Protons were ejected from paraffi n wax. The speeds of nitrogen and proton after collision with neutrons were estimated from the lengths of their tracks in a cloud chamber. 9. From the results of his experiments, Chadwick deduced that the mass of the neutron is slightly greater than the mass of the proton. 2009/P2/Q48 INTRODUCTION 1. At the centre of each atom is the nucleus which is positively charged and has almost the whole mass of the atom. 2. In this chapter, we will study the characteristics of the nucleus, how the nucleons are bounded together, and the usage of the mass spectrometer to measure the mass of different isotopes. 3. In 1896, Becquerel discovered radioactivity when he found that photographic plates placed closed to some salts of uranium were blackened. 4. Becquerel discovered that the radiations from uranium salts were able to penetrate aluminium and glass. These radiations had identical properties to that of X-rays which were discovered by Rontgen in 1895. 5. Marie and Pierre Curie were able to separate radium and polonium from pitch blend.

Physics Term 3 STPM Chapter 25 Nuclear Physics 294 25 10. From the discovery of the neutron by Chadwick, the constituents of the nucleus are protons and neutrons, except for the hydrogen nucleus which has only one proton. 11. Protons and neutrons in a nucleus are known as nucleons. The number of nucleons in a nucleus is known as the nucleon number or mass numbers, A. 12. The proton number or atomic number Z is the number of proton in the nucleus. 13. A nuclide is a nuclear species with a given number of protons and neutrons, and is represented by the symbol A Z X where Z = proton number A = nucleon number Mass Defect and Binding Energy 2008/P1/Q47, 2011/P1/Q47, 2015/P3/Q14, 2016/P3/Q14 2017/P3/Q14 1. Definition: The atomic mass unit (u) is —– 1 12 the mass of an atom of carbon-12. This means that the mass of a carbon-12 atom is 12 atomic mass unit or 12u. 2. In one mole of C-12, there are 6.02 × 1023 atoms and its mass is 0.012 kg. Hence, the mass of one C-12 atom, 12 u = 0.012 6.02 × 1023 kg The atom mass unit, u = 0.012 12 × 6.02 × 1023 kg = 1.66 × 10–27 kg 3. Definition: The relative atomic mass (mr ) of an atom is the ratio of the mass of the atom to the atomic mass unit. mr = mass of an atom atomic mass unit 4. From the definition of relative atom mass, the relative atomic mass of C-12 is 12. 5. The mass of the nucleus is less than the total mass of the nucleons. 6. The difference between the total mass of the nucleons and the mass of the nucleus is known as the mass defect of the nucleus. Mass defect, Δm = (Total mass of nucleons) – Mass of nucleus 7. There is a mass defect because the nuclear forces that bind the nucleons in the nucleus results in the nucleons being at a lower energy state compared to when the nucleons are free. 8. The energy equivalent of the mass defect is equal to the binding energy of the nucleus. Using Einstein mass-energy relation, E = mc2 Binding energy of nucleus = (mass defect) × c 2 E = (Δm) c 2 where Δm = mass defect c = speed of light in vacuum