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The Modern Quantum Atom

The Modern Quantum Atom. The nucleus and the discovery of the neutron What are electron-volts ? The Quantum atom. Announcements HW#8 posted (sorry) Prof. Artuso giving Monday’s class. I may allow a calc. For Exam 3. Rutherford’s Picture of the Atom. Electrons circle the nucleus

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The Modern Quantum Atom

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  1. The Modern Quantum Atom The nucleus and the discovery of the neutron What are electron-volts ? The Quantum atom Announcements HW#8 posted (sorry)Prof. Artuso giving Monday’s class.I may allow a calc. For Exam 3

  2. Rutherford’s Picture of the Atom Electrons circle the nucleus due to the Coulomb force Corpuscles (Electrons) ~10-14 m ~10-11 m Positively Charged Nucleus This model was inspired by the results of scattering alpha-particlesoff of heavy nuclei (like gold, silver, etc). See previous lecture. Rutherford Scattering: http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford.html

  3. Performed a series of scattering experiments • with a-particles (recall a particles are He nucleus), • 42 He + 9 Be 12 C +10 n James Chadwick and the Neutron Circa 1925-1935 Picked up where Rutherford left off with more scattering experiments… (higher energy though!) • Chadwick postulated that the emergent radiation • was from a new, neutral particle, the neutron. • Applying energy and momentum conservation he found that the mass of this new object was ~1.15 times that of the proton mass. 1891-1974 Awarded the Nobel Prize in 1935

  4. ElectricPotential -20 [kV] GPE 1 kg 10[J] 0 [V] + 1 m 0 [J] - -20 [kV] 0 [kV] ***Electron-Volts (eV)*** • When talking about subatomic particles, and individual photons,energies are very small (~10-12 or smaller). • It’s cumbersome to always deal with these powers of 10. • We introduce a new unit of energy, called the electron-volt (eV). • An [eV] is equivalent to the amount of energy a single electron gainswhen it is accelerated across a voltage of 1 [V]. • Your TV tube accelerates electrons using 20,000 [V] = 20 [kV].

  5. =1 More on [eV] How much energy does an electron gain when it is accelerated acrossa voltage of 20,000 [V] ? E = 20,000 [eV] [V] is a unit of “Potential”[eV] is a unit of Energy (can be converted to [J]) How can you convert [eV] to [J] ? Not too hard… the conversion is: 1 [eV] = 1.6x10-19 [J] So, let’s do an example ! Convert 20 [keV] to [J]. Since the “k” == kilo = 1000 = 103, 20 [keV] = 20,000 [eV] = 2x104 [eV] It’s a lot easier to say “20 [keV]” than 3.2x10-15 [J] !

  6. Even more on [eV] So, [eV] IS A UNIT OF ENERGY; It’s not a “type” of energy (such as light, mass, heat, etc).When talking about energies of single photons, or of subatomic particles,we often use this unit of energy, or some variant of it.So, 1 [eV] = 1.6x10-19 [J] (can be used to go back & forth between these two energy units) 1 [keV] = 1000 [eV] = 103 [eV] “k = kilo (103)”” 1 [MeV] = 1,000,000 [eV] = 106 [eV] “M = mega (106)” 1 [GeV] = 1,000,000,000 [eV] = 109 [eV] “G = giga (109)”

  7. Example 1 A Cobalt-60 nucleus is unstable, and undergoes a decay where a 1173 [keV] photon is emitted. From what region of the electromagnetic spectrum does this come? The energy is 1173 [keV], which is 1173 [keV] = 1173x103 [eV] = 1.173x106 [eV]. * First convert this energy to [J], E = 1.173x106 [eV] * (1.6x10-19 [J] / 1 [eV]) = 1.88x10-13 [J] * Now, to get the wavelength, we use: E = hc/l, that is l = hc/E. So, l = 6.63x10-34[J s]*3x108[m/s]/1.88x10-13 [J] = 1.1 x 10-12 [m] * Now, convert [m] to [nm], 1.1 x 10-12 [m] * (109 [nm] / 1 [m]) = 1.1x10-3 [nm] It’s a GAMMA Ray

  8. Example 2 An electron has a mass of 9.1x10-31 [kg]. What is it’s rest mass energy in [J] and in [eV]. E = mc2 = 9.1x10-31*(3x108)2 = 8.2x10-14 [J] Now convert to [eV] What is an electron’s rest mass? According to Einstein, m = E/c2, that is: [mass] = [Energy] / c2 m = E / c2 = 0.51 [MeV/c2]

  9. Example 3 A proton has a mass of 1.67x10-27 [kg]. What is it’s rest mass energy in [J] and in [eV]. E = mc2 = 1.67x10-27 *(3x108)2 = 1.5x10-10 [J] Now convert to [eV] What is a proton’s rest mass? According to Einstein, m = E/c2, that is: [mass] = [Energy] / c2 m = E / c2 = 940 [MeV/c2]

  10. Proton vs Electron Mass How much more massive is a proton than an electron ? Ratio = proton mass / electron mass = 940 (MeV/c2) / 0.51 (MeV/c2) = 1843 times more massive You’d get exactly the same answer if you used: electron mass = 9.1x10-31 [kg] Proton mass = 1.67x10-27 [kg]Using [MeV/c2] as units of energy is easier…

  11. Neils Bohr and the Quantum Atom Circa 1910-1925 • Pointed out serious problems with • Rutherford’s atom • Electrons should radiate as they orbit the nucleus, and in doing so, lose energy, until they spiral into the nucleus. • Atoms only emit quantized amounts of energy (i.e., as observed in Hydrogen spectra) • He postulated • Electric force keeps electrons in orbit • Only certain orbits are stable, and they do not radiate energy Radiation is emitted when an e- jumps from an outer orbit to an inner orbit and the energydifference is given off as a radiation. 1885-1962 Awarded the Nobel Prize in 1922

  12. Before After Radiatedphoton n = Electronin lowest“allowed”energy level (n=1) 5 5 4 4 3 3 2 2 1 1 Electronin excitedstate (n=5) Electron falls to the lowest energy level Allowed Orbits Bohr’s Picture of the Atom Electrons circle the nucleus due to the Electric force Note: There are many more energy levels beyond n=5, they are omitted for simplicity

  13. Energy Electronin excitedstate(higher PE) Energy Electronin loweststate(lower PE) E5 E5 n = 5 n = 5 E4 E4 n = 4 n = 4 E3 E3 n = 3 n = 3 E2 E2 n = 2 n = 2 E1 E1 n = 1 n = 1 Before After Atomic Radiation It is now “known” that when an electron is in an “excited state”,it spontaneously decays to a lower-energy stable state. E5 > E4 > E3 > E2 > E1 • The difference in energy, DE, is given by:DE = E5 – E1 = hn = Ephotonh = Planck’s constant = 6.6x10-34 [J s] • = frequency of light [hz] The energy of the light is DIRECTLY PROPORTIONAL to the frequency, n.Recall that the frequency, n, is related tothe wavelength by:c = n l (n = c / l)So, higher frequency  higher energy  lower wavelengthThis is why UV radiation browns your skinbut visible light does not ! One example could be:

  14. 5 4 3 2 1 Hydrogen atom energy “levels” Quantum physics provides the tools to compute the values ofE1, E2, E3, etc…The results are: En = -13.6 / n2 These results DO DEPEND ON THE TYPE OF ATOM OR MOLECULE So, the difference in energy between the 3rd and 1st quantum state is: Ediff = E3 – E1 = -1.51 – (-13.6) = 12.09 (eV) When this 3 1 atomic transition occurs, this energy is released in the form of electromagnetic energy.

  15. Example 4 In the preceding example, what is the frequency, wavelength of theemitted photon, and in what part of the EM spectrum is it in? E = 12.1 [eV]. First convert this to [J]. Since E = hn n = E/h, so: n = E/h = 1.94x10-18 [J] / 6.6x10-34 [J s] = 2.9x1015 [1/s] = 2.9x1015 [hz] l = c/n = (3x108 [m/s]) / (2.9x1015 [1/s]) = 1.02x10-7 [m] = 102 [nm] This corresponds to low energy X-rays !

  16. Some Other Quantum Transitions

  17. This completed the picture, or did it… • Electrons were discovered ~1900 by J. J. Thomson • Protons being confined in a nucleus was put forth ~1905 • Neutrons discovered 1932 by James Chadwick • Quantum theory of radiation had become“widely accepted”, although even Einstein had his doubts • Radiation is produced when atomicelectrons fall from a state of highenergy  low energy. Yields photonsin the visible/ X-ray region. • A nucleus can also be excited, andwhen it “de-excites” it also givesoff radiation  Typically gamma rays !

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