Topic 7: Atomic, nuclear and particle physics 7.1 – Discrete energy and radioactivity

1. Essential idea: In the microscopic world energy is discrete. Nature of science: Accidental discovery: Radioactivity was discovered by accident when Becquerel developed photographic film that had accidentally been exposed to radiation from radioactive rocks. The marks on the photographic film seen by Becquerel probably would not lead to anything further for most people. What Becquerel did was to correlate the presence of the marks with the presence of the radioactive rocks and investigate the situation further. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity

2. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Understandings: • Discrete energy and discrete energy levels • Transitions between energy levels • Radioactive decay • Fundamental forces and their properties • Alpha particles, beta particles and gamma rays • Half-life • Absorption characteristics of decay particles • Isotopes • Background radiation

3. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Applications and skills: • Describing the emission and absorption spectrum of common gases • Solving problems involving atomic spectra, including calculating the wavelength of photons emitted during atomic transitions • Completing decay equations for alpha and beta decay • Determining the half-life of a nuclide from a decay curve • Investigating half-life experimentally (or by simulation)

4. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Guidance: • Students will be required to solve problems on radioactive decay involving only integral numbers of half-lives • Students will be expected to include the neutrino and antineutrino in beta decay equations Data booklet reference: • E = hf •  = hc/E

5. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity International-mindedness: • The geopolitics of the past 60+ years have been greatly influenced by the existence of nuclear weapons Theory of knowledge: • The role of luck/serendipity in successful scientific discovery is almost inevitably accompanied by a scientifically curious mind that will pursue the outcome of the “lucky” event. To what extent might scientific discoveries that have been described as being the result of luck actually be better described as being the result of reason or intuition?

6. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Utilization: • Knowledge of radioactivity, radioactive substances and the radioactive decay law are crucial in modern nuclear medicine • How to deal with the radioactive output of nuclear decay is important in the debate over nuclear power stations (see Physics sub-topic 8.1) • Carbon dating is used in providing evidence for evolution (see Biology sub-topic 5.1) • Exponential functions (see Mathematical studies SL sub-topic 6.4; Mathematics HL sub-topic 2.4)

7. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Aims: • Aim 8: the use of radioactive materials poses environmental dangers that must be addressed at all stages of research • Aim 9: the use of radioactive materials requires the development of safe experimental practices and methods for handling radioactive materials

8. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases In 1897 British physicist J.J. Thomson discovered the electron, and went on to propose a "plum pudding" model of the atom in which all of the electrons were embedded in a spherical positive charge the size of the atom. The “Plum pudding” model of the atom atomic diameter ( 10-10 m) +7 - 7

9. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases In 1911 British physicist Ernest Rutherford conducted experiments on the structure of the atom by sending alpha particles (which we will study later) through gold leaf. Gold leaf is like tin foil, but it can be made much thinner so that the alpha particles only travel through a thin layer of atoms. FYI An alpha () particle is a doubly-positive charged particle emitted by radioactive materials.

10.    Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases Rutherford proposed that alpha particles would travel more or less straight through the atom without deflection if Thomson’s “Plum pudding” model was correct: scintillationscreen FYI Instead of observing minimal scattering as in the Plum Pudding model, Rutherford observed the scattering as shown on the next slide:

11. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases Here we see that the deflections are much more scattered... Rutherford proposed that the positive charge of the atom was located in the center, and he coined the term nucleus. The Rutherford Model nucleus The atom

12. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases Expected Results Actual Results FYI This experiment is called the Geiger-Marsden scattering experiment.

13. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases Only by assuming a concentration of positive charge at the center of the atom, as opposed to “spread out” as in the Plum Pudding model, could Rutherford’s team explain the results of the experiment. Geiger Marsden

14. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases When a gas in a tube is subjected to a voltage, the gas ionizes, and emits light.

15. 400 450 500 550 600 650 700 750  / 10-9 m ( / nm) Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases We can analyze that light by looking at it through a spectroscope. A spectroscope acts similar to a prism, in that it separates the incident light into its constituent wavelengths. For example, heated barium gas will produce an emission spectrum that looks like this: An emission spectrum is an elemental fingerprint.

16. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Describing the emission and absorption spectrum of common gases Each element also has an absorption spectrum, caused by cool gases between a source of light and the scope. continuous spectrum light source absorption spectrum light source cool gas X emission spectrum hot gas X compare… Same fingerprint!

17. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels In the late 1800s a Swedish physicist by the name of J.J. Balmer observed the spectrum of hydrogen – the simplest of all the elements: His observations gave us clues as to the way the negative charges were distributed about the nucleus.

18. 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels In reality, there are many additional natural groupings for the hydrogen spectrum, two of which are shown here: These groupings led scientists to imagine that the hydrogen’s single electron could occupy many different energy levels, as shown in the next slide:  / nm Balmer Series (Visible) Lyman Series (UV) Paschen Series (IR)

19. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels The first 7 energy levels for hydrogen are shown here: The energy levels are labeled from the lowest to the highest as n = 1 to n = 7 in the picture. n is called the principal quantum number and goes all the way up to infinity ()! In its ground state or unexcited state, hydrogen’s single electron is in the 1st energy level (n = 1): 7 6 5 4 3 2 1

20. 7 6 5 4 3 2 1 Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels As we will see later, light energy is carried by a particle called a photon. If a photon of just the right energy strikes a hydrogen atom, it is absorbed by the atom and stored by virtue of the electron jumping to a new energy level: The electron jumped from the n = 1 state to the n = 3 state. We say the atom is excited.

21. 7 6 5 4 3 2 1 Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels When the atom de-excites the electron jumps back down to a lower energy level. When it does, it emits a photon of just the right energy to account for the atom’s energy loss during the electron’s orbital drop. The electron jumped from the n = 3 state to the n = 2 state. We say the atom is de-excited, but not quite in its ground state.

22. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels The graphic shown here accounts for many of the observed hydrogen emission spectra. The excitation illustrated looked like this: The de- excitation looked like this: Infrared Visible Ultraviolet

23. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels The human eye is only sensitive to the Balmer series of photon energies (or wavelengths): Infrared Visible Ultraviolet

24. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels The previous energy level diagram was NOT to scale. This one is. Note that none of the energy drops of the other series overlap those of the Balmer series, and thus we cannot see any of them. But we can still sense them! n =  0.00 eV Second Excited State n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen Series (IR) First ExcitedState [ HEAT ] n = 2 -3.40 eV Balmer Series (Visible) Ground State n = 1 -13.6 eV Lyman Series (UV) [ SUNBURN ] FYI Transitionenergy is measured in eV because of the tiny amounts involved.

25. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels Because of wave-particle duality, we have discovered that light not only acts like a wave, having a wavelength  and a frequency f, but it acts like a particle (called a photon) having an energy E given by Whereh = 6.6310 -34Js and is called Planck’s constant. energy E of a photon having frequency f E = hf

26. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (a) What series is this de-excitation in? SOLUTION: Find it on the diagram: This jump is contained in the Balmer Series, and produces a visible photon.

27. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (b) Find the atom’s change in energy in eV and in J. SOLUTION: E = Ef – E0 = -3.40 - -1.51 = -1.89 eV. E = (-1.89 eV)(1.6010-19 J / eV) = -3.0210-19 J.

28. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (c) Find the energy (in J) of the emitted photon. SOLUTION: The hydrogen atom lost 3.0210-19 J of energy. From conservation of energy a photon was created having E = 3.0210-19 J.

29. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (d) Find the frequency of the emitted photon. SOLUTION: From E = hf we have 3.0210-19 = (6.6310-34)f, or f = 3.0210-19 / 6.6310-34 f = 4.561014Hz. FYI When finding f, be sure E is in Joules, not eV.

30. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Transitions between energy levels EXAMPLE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (e) Find the wavelength (in nm) of the emitted photon. SOLUTION: From v = f where v = c we have 3.00108 = (4.561014), or  = 6.5810-7 m. Then  = 6.5810-7 m = 65810-9 m = 658 nm.

31. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Discrete energy and discrete energy levels Discrete means discontinuous, or separated. PRACTICE: Which one of the following provides direct evidence for the existence of discrete energy levels in an atom? A. The continuous spectrum of the light emitted by a white hot metal. B. The line emission spectrum of a gas at low pressure. C. The emission of gamma radiation from radioactive atoms. D. The ionization of gas atoms when bombarded by alpha particles. SOLUTION: Just pay attention!

32. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line. (a) What is its frequency? SOLUTION: Use c = f where c = 3.00108 m s-1and  = 434 10-9 m:  3.00108 = (43410-9)f f = 3.00108 / 43410-9 = 6.911014 Hz.

33. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line. (b) What is the energy (in J and eV) of each of its blue- light photons? SOLUTION: Use E = hf:  E = (6.6310-34)(6.911014) E = 4.5810-19 J. E = (4.5810-19 J)(1 eV/ 4.5810-19 J) E = 2.86 eV.

34. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line. (c) What are the energy levels associated with this photon? SOLUTION: Because it is visibleuse the Balmer Series with ∆E = -2.86 eV. Note that E2 – E5 = -3.40 – -0.544 = -2.86 eV. Thus the electron jumped from n = 5 to n = 2.

35. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: The element helium was first identified by the absorption spectrum of the sun. (a) Explain what is meant by the term absorption spectrum. SOLUTION: An absorption spectrum is produced when a cool gas is between a source having a continuous spectrum and an observer with a spectroscope. The cool gases absorb their signature wavelengths from the continuous spectrum. Where the wavelengths have been absorbed by the gas there will be black lines. continuous spectrum absorption spectrum emission spectrum

36. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: One of the wavelengths of the absorption spectrum for helium occurs at 588 nm. (b) Show that the energy of a photon having a wavelength of 588 nm is 3.3810-19 J. SOLUTION: This formula can be used directly: From E = hc /  we see that E = (6.6310-34)(3.00108) / 58810-9) E = 3.3810-19 J. Whereh = 6.6310 -34Js and is called Planck’s constant. E= hc /  energy E of a photon having wavelength 

37. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra PRACTICE: The diagram represents some energy levels of the helium atom. (c) Use the information in the diagram to explain how absorption at 588 nm arises. SOLUTION: We need the difference in energies between two levels to be 3.3810-19 J. Note that 5.80 – 2.42 = 3.38. Since it is an absorption the atom stored the energy by jumping an electron from the -5.8010-19 J level to the -2.4210-19 J level, as illustrated.

38. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Solving problems involving atomic spectra In a later lecture we will discover that the most intense light reaching us from the sun is between 500 nm and 650 nm in wavelength. Evolutionarily our eyes have developed in such a way that they are most sensitive to that range of wavelengths, as shown in the following graphic:

39. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes Recall the mass spectrometer, in which an atom is stripped of its electrons and accelerated through a voltage into a magnetic field. Scientists discovered that hydrogen nuclei had three different masses: Since the charge of the hydrogen nucleus is e, scientists postulated the existence of a neutralparticle called the neutron, which added mass without charge.

40. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes The proton and neutron are called nucleons. For the element hydrogen, it was found that its nucleus existed in three forms: A set of nuclei for a single element having different numbers of neutrons are called isotopes. A particular isotope of an element is called a species or a nuclide. Proton [ Charge = 1e or just +1 ] Nucleons Neutron [ Charge = 0e or just 0 ] Isotopes Hydrogen Deuterium Tritium

41. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes An element’s chemistry is determined by the number of electrons surrounding it. The number electrons an element has is determined by the number of protons in that element’s nucleus. Therefore it follows that isotopes of an element have the same chemical properties. For example there is water, made of hydrogen H and oxygen O, with the molecular structure H2O. But there is also heavy water, made of deuterium D and oxygen O, with the molecular formula D2O. Both have exactly the same chemical properties. But heavy water is slightly denser than water.

42. nucleon relationship A = Z + N Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes A species or nuclide of an element is described by three integers: The nucleon number A is the total number of protons and neutrons in the nucleus. The proton numberZ is the number of protons in the nucleus. It is also known as the atomic number. The neutron numberNis the number of neutrons in the nucleus. It follows that the relationship between all three numbers is just

43. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes In nuclear physics you need to be able to distinguish the different isotopes. NUCLEAR PHYSICS CHEMISTRY Mass Number = A H H N = Neutrons Protons = Z H H H 2 1 3 1 1 1 0 1 2 tritium deuterium hydrogen hydrogen-2 hydrogen-1 hydrogen-3 FYI Since A = Z + N, we need not show N. And Z can be found on any periodic table.

44. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes PRACTICE: Which of the following gives the correct number of electrons, protons and neutrons in the neutral atom 6529Cu? SOLUTION: A = 65, Z = 29, so N = A – Z = 65 – 29 = 36. Since it is neutral, the number of electrons equals the number of protons = Z = 29.

45. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes PRACTICE: Ag-102, Ag-103 and Ag-104 are all isotopes of the element silver. Which one of the following is a true statement about the nuclei of these isotopes? A. All have the same mass. B. All have the same number of nucleons. C. All have the same number of neutrons. D. All have the same number of protons. SOLUTION: Isotopes of an element have different masses and nucleon totals. Isotopes of an element have the same number of protons, and by extension, electrons. This is why their chemical properties are identical.

46. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Isotopes PRACTICE: Track X shows the deflection of a singly-charged carbon-12 ion in the deflection chamber of a mass spectrometer. Which path best shows the deflection of a singly- charged carbon-14 ion? Assume both ions travel at the same speed. SOLUTION: Since carbon-14 is heavier, it will have a bigger radius than carbon-12. Since its mass is NOT twice the mass of carbon-12, it will NOT have twice the radius.

47. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity • Fundamental forces and their properties • Given that a nucleus is roughly 10-15 m in diameterit should be clear that the Coulomb repulsion between protons within the nucleus must be enormous. • Given that most nuclei do NOT spew out their protons, there must be a nucleon force that acts within the confines of the nucleus to overcome the Coulomb force. • We call this nucleon force the strong force. • In a nutshell, the strong force… • counters the Coulomb force to prevent nuclear decay and therefore must be very strong. • is very short-range, since protons located far enough apart do, indeed, repel.

48. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Fundamental forces and their properties PRACTICE: The nucleus of an atom contains protons. The protons are prevented from flying apart by A. The presence of orbiting electrons. B. The presence of gravitational forces. C. The presence of strong attractive nuclear forces. D. The absence of Coulomb repulsive forces at nuclear distances. SOLUTION: It is the presence of the strong force within the nucleus.

49. Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Fundamental forces and their properties PRACTICE: Use Coulomb’s law to find the repulsive force between two protons in a helium nucleus. Assume the nucleus is 1.0010-15 m in diameter and that the protons are as far apart as they can get. SOLUTION: From Coulomb’s law the repulsive force is F = ke2 / r2= 9109(1.610-19)2 / (1.0010-15)2 F = 230 N. FYI From chemistry we know that atoms can be separated from each other and moved easily. This tells us that at the range of about 10-10 m (the atomic diameter), the strong force is zero.

50. + + Topic 7: Atomic, nuclear and particle physics7.1 – Discrete energy and radioactivity Fundamental forces and their properties ELECTRO-WEAK GRAVITY STRONG ELECTROMAGNETIC WEAK nuclear force light, heat and charge freefall radioactivity STRONGEST WEAKEST Range: Extremely Short Range:  Range: Short Range:  Force Carrier: Graviton Force Carrier: Photon Force Carrier: Gluon