Lec29 Quantum Atom

Information about Lec29 Quantum Atom

Published on October 15, 2007

Author: Nellwyn

Source: authorstream.com

Content

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:  Positively Charged Nucleus Corpuscles (Electrons) Electrons circle the nucleus due to the Coulomb force Rutherford’s Picture of the Atom ~10-14 m ~10-11 m Rutherford Scattering: http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford.html This model was inspired by the results of scattering alpha-particles off of heavy nuclei (like gold, silver, etc). See previous lecture. James Chadwick and the Neutron :  James Chadwick and the Neutron Awarded the Nobel Prize in 1935 1891-1974 Circa 1925-1935 Applying energy and momentum conservation he found that the mass of this new object was ~1.15 times that of the proton mass. 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. ***Electron-Volts (eV)***:  ***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 gains when it is accelerated across a voltage of 1 [V]. Your TV tube accelerates electrons using 20,000 [V] = 20 [kV]. More on [eV]:  More on [eV] How much energy does an electron gain when it is accelerated across a 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] ! Even more on [eV]:  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)” Example 1:  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 Example 2:  Example 2 An electron has a mass of 9.1x10-31 [kg]. E = mc2 = 9.1x10-31*(3x108)2 = 8.2x10-14 [J] Now convert to [eV] What is an electron’s rest mass? m = E / c2 = 0.51 [MeV/c2] According to Einstein, m = E/c2, that is: [mass] = [Energy] / c2 What is it’s rest mass energy in [J] and in [eV]. Example 3:  Example 3 A proton has a mass of 1.67x10-27 [kg]. E = mc2 = 1.67x10-27 *(3x108)2 = 1.5x10-10 [J] Now convert to [eV] What is a proton’s rest mass? m = E / c2 = 940 [MeV/c2] According to Einstein, m = E/c2, that is: [mass] = [Energy] / c2 What is it’s rest mass energy in [J] and in [eV]. Proton vs Electron Mass:  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… Neils Bohr and the Quantum Atom:  Neils Bohr and the Quantum Atom 1885-1962 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 energy difference is given off as a radiation. Awarded the Nobel Prize in 1922 Bohr’s Picture of the Atom:  Electrons circle the nucleus due to the Electric force Bohr’s Picture of the Atom Note: There are many more energy levels beyond n=5, they are omitted for simplicity Atomic Radiation:  Atomic Radiation The difference in energy, DE, is given by: DE = E5 – E1 = hn = Ephoton h = 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 to the wavelength by: c = n l (n = c / l) So, higher frequency  higher energy  lower wavelength This is why UV radiation browns your skin but visible light does not ! 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 One example could be: Hydrogen atom energy “levels”:  Hydrogen atom energy “levels” Quantum physics provides the tools to compute the values of E1, E2, E3, etc…The results are: En = -13.6 / n2 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. These results DO DEPEND ON THE TYPE OF ATOM OR MOLECULE Example 4:  Example 4 In the preceding example, what is the frequency, wavelength of the emitted 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 ! Some Other Quantum Transitions:  Some Other Quantum Transitions This completed the picture, or did it…:  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 atomic electrons fall from a state of high energy  low energy. Yields photons in the visible/ X-ray region. A nucleus can also be excited, and when it “de-excites” it also gives off radiation  Typically gamma rays !

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