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  1. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure 7.1.1 Describe a model of the atom that features a small nucleus surrounded by electrons. 7.1.2 Outline the evidence that supports a nuclear model of the atom. 7.1.3 Outline one limitation of the simple model of the nuclear atom. 7.1.4 Outline evidence for the existence of atomic energy levels.

  2. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Describe a model of the atom that features a small nucleus surrounded by electrons. 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 +7

  3. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Describe a model of the atom that features a small nucleus surrounded by electrons. In 1911 British physicist Ernest Rutherford conducted experiments on the structure of the atom by sending alpha particles 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 double-positively charged particle emitted by radioactive materials such as uranium.

  4.   Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Describe a model of the atom that features a small nucleus surrounded by electrons. 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, Rutherford observed the scattering as shown on the next slide:

  5. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline the evidence that supports a nuclear model of the atom. 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

  6. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline the evidence that supports a nuclear model of the atom. Expected Results Actual Results FYI IBO requires you to qualitatively understand the Geiger-Marsden scattering experiment.

  7. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline the evidence that supports a nuclear model of the atom. Only by assum- ing a concentra- tion of positive charge at the center of the atom, as opposed to “spread out” as in the plum pudding model, could Rutherford and his team ex- plain the re- sults of the experiment. Geiger Marsden

  8. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline the evidence that supports a nuclear model of the atom. PRACTICE: In the Geiger-Marsden experiment  particles are scattered by gold nuclei. The experimental results indicate that most  particles are A. Scattered only at small angles. B. Scattered only at large angles. C. Absorbed by the target. D. Scattered back along the original path. SOLUTION: Observing the image… Most  particles scatter at small angles.

  9. A  B  Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline the evidence that supports a nuclear model of the atom. PRACTICE: In 1913 Geiger and Marsden fired alpha particles at gold foil. The diagram shows two such alpha particles () at A and B and two gold nuclei within the foil. Sketch in the likely paths for each alpha particle within the box. SOLUTION: Since  particles and nuclei are both (+) the  particles will be repelled. From A the particle will scatter at a small angle. Remember it is repulsed, not attracted. From B the particle will scatter at a large angle, and perhaps even right back.

  10. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. When a gas in a tube is subjected to a voltage, the gas ionizes, and emits light.

  11. 4000 4500 5000 5500 6000 6500 7000 7500  / 10-10 m Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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.

  12. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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 cool gas X light source emission spectrum hot gas X compare… Same fingerprint!

  13. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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:

  14. 0 200 400 600 800 1000 1200 1400 1600 1800 2000 Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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 next: Balmer Series (Visible) Lyman Series (UV) Paschen Series (IR)  / 10-10 m

  15. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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

  16. 7 6 5 4 3 2 1 Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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.

  17. 7 6 5 4 3 2 1 Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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.

  18. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. The table shown here accounts for all of the observed hydrogen emission spectra. The excita- tion illustra- ted looked like this: The de- excitation looked like this: Infrared Visible Ultraviolet

  19. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. The human eye is only sensitive to the Balmer series of photon energies (or wavelengths): Infrared Visible Ultraviolet

  20. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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 n = 5 -0.544 eV Second Excited State  n = 4 -0.850 eV n = 3 -1.51 eV First Excited State  Paschen Series (IR) n = 2 -3.40 eV [ HEAT ] Balmer Series (Visible) FYI Energy can be measured in eV. Ground State  n = 1 -13.6 eV [ SUNBURN ] Lyman Series (UV)

  21. energy of a photon E = hf Whereh = 6.6310-34Js and is called Planck’s constant. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. Later when we study of Option B – Quantum Physics, we will find out that light not only acts like a wave, having a wavelength and a frequency, but it acts like a particle (called a photon) having an energy given by n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen Series (IR) n = 2 -3.40 eV [ HEAT ] Balmer Series (Visible) n = 1 -13.6 eV [ SUNBURN ] Lyman Series (UV)

  22. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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 energy diagram: This jump is contained in the Balmer Series, and produces a visible photon.

  23. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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.

  24. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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.

  25. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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.

  26. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic 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)  = 6.5810-7 m. Then  = 6.5810-7 m = 65810-9 m = 658 nm.

  27. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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: Dude, just pay attention!

  28. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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.

  29. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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.

  30. n =  0.00 eV n = 5 -0.544 eV n = 4 -0.850 eV n = 3 -1.51 eV Paschen n = 2 -3.40 eV Balmer n = 1 -13.6 eV Lyman Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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.

  31. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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 and remove them from the continuous spectrum. Where the wavelengths have been absorbed by the gas there will be black lines. continuous spectrum absorption spectrum emission spectrum

  32. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. 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: E = hf but c = f so that f = c/. Thus E = hc/ so that E = (6.6310-34)(3.00108)/58810-9) E = 3.3810-19 J.

  33. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. PRACTICE: The diagram represents a few 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 elec- tron from the -5.8010-19 J level to the -2.4210-19 J level, as illustrated.

  34. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline evidence for the existence of atomic energy levels. In a later option 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:

  35. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline one limitation of the simple model of the nuclear atom. By classical electromagnetic theory, charged particles in accelerated motion radiate energy in the form of electromagnetic radiation. Electrons in atoms are in uniform circular motion, thus having a centripetal acceleration. But if an electron in an atom radiates energy, it will ultimately lose its energy and fall into its nucleus. Since this is not observed, the simple model of the atom is flawed. Is not observed 1 Is observed 1

  36. Topic 7: Atomic and nuclear physics7.1 The atom Atomic structure Outline one limitation of the simple model of the nuclear atom. Another limitation is that the “orbitals” are really not as simple as circular. On the following slide are images of what the orbitals really look like surrounding a hydrogen atom:

  37. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure 7.1.5 Explain the terms nuclide, isotope and nucleon. 7.1.6 Define nucleon number A, proton number Z and neutron number N. 7.1.7 Describe the interactions in a nucleus.

  38. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Explain the terms nuclide, isotope and nucleon. Recall the mass spectrometer in which an atom is stripped of its elec- trons and accelerated through a voltage into a magnetic field. Scientists determined through the use of such a device that hydrogen nuclei came in three different masses: Since the charge of the hydrogen nu-cleus is e, scientists postulated the existence of a neutral particle called the neutron.

  39. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Explain the terms nuclide, isotope and nucleon. 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

  40. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Explain the terms nuclide, isotope and nucleon. 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.

  41. nucleon relationship A = Z + N Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Define nucleon number A, proton number Z and neutron number N. 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 the atomic number. The neutron numberN is the number of neutrons in the nucleus. It follows that the relationship between all three numbers is just

  42. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Define nucleon number A, proton number Z and neutron number N. In nuclear physics you need to be able to distinguish the different isotopes. NUCLEAR PHYSICS CHEMISTRY H Mass Number = A 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.

  43. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Define nucleon number A, proton number Z and neutron number N. PRACTICE: Which of the following gives the correct number of electrons, protons and neutrons in the neutral atom 6529Cu? 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.

  44. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Define nucleon number A, proton number Z and neutron number N. 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.

  45. B-field Source A B X C D Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Define nucleon number A, proton number Z and neutron number N. 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.

  46. Topic 7: Atomic and nuclear physics7.1 The atom • Nuclear structure • Describe the interactions in a nucleus. • Given that a nucleus is roughly 10-15 m in diameter is 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 force that acts within the confines of the nucleus to overcome the Coulomb force. • We call this 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.

  47. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Describe the interactions in a nucleus. 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.

  48. Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Describe the interactions in a nucleus. 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.

  49. + + Topic 7: Atomic and nuclear physics7.1 The atom Nuclear structure Describe the interactions in a nucleus. ELECTROMAGNETIC STRONG WEAK GRAVITY ELECTRO-WEAK nuclear force light, heat and charge radioactivity freefall STRONGEST WEAKEST Range: Extremely Short Range:  Range: Short Range:  Force Carrier: Graviton Force Carrier: Photon Force Carrier: Gluon