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Chapter 28 Atoms Alexandra Reed Rebecca Ray Chelsea Mitchell Lena Kral Niels Bohr - history - Neils Bohr full name: Niels Henrick David Bohr born on October 7, 1885. Bohr attended the local University of Copenhagen partly because his father was a professor of physiology there and mostly because his childhood was filled with the presence of scientists. Niels Bohr Master’s degree in 1909 doctorate in 1911 After graduation he moved to Cambridge to study with J. J. Thomson. They didn’t agree. After a meeting with Earnest Rutherford, he decided to move with him to Manchester in 1912. Niels Bohr Bohr felt it greatly extended the understanding of nuclear fission. He thought, the development of the fission bomb was a great achievement. However with every great scientific discovery there were limits that need to be set upon it. He was deeply concerned about the control of these highly powerful weapons and urge for the need of international cooperation. Chapter 28.1 The energy of an orbiting electron in an atom is the sum of the kinetic energy of the electron and the potential energy resulting from the attractive force between the electron and the nucleus. The energy of an electron in an orbit near the nucleus is less than that of an electron in an orbit farther away because work must be done to move an electron to orbits farther away from the nucleus. The electrons in excited states have larger orbits and correspondingly higher energies. Bohr postulated that the change in the energy of an atomic electron when a photon is absorbed is equal to the energy of the photon. That is: ΔE = hf = Eexcited – Eground Predictions of the Bohr Model Bohr’s calculations start with Newton’s law, F=ma, applied to an electron of mass m and charge –q, in a circular orbit of radius r about a massive particle, a proton, of charge q. Fc = mac Kq2/r2 = mv2/r K= constant (9x109 Nm2/C2) q= charge r= radius m= mass v= centripetal velocity Bohr Model Formulas The angular momentum can only have certain values, given by the equation: mvr = nh/2 π h = Planck’s constant (6.63x10-34 m2kg/s) n = integer *Because the quantity can only be integer values, it is said to be quantized. -Combining Newton’s Law with the quantization of angular momentum gives the predicted radius: rn = (h2n2)/(4 π 2Kmq2) *When you substitute in numerical values for the constants, you get: rn= (5.3x10-11m)n2 -The total energy of the electron in its orbit is given by the equation: En = (-2 π 2K2mq4)/(h2n2) *When you substitute in numerical values for the constants, you get: En = -13.6 eV/n2 = -2.17x10-18J/n2 Example Problems For the hydrogen atom, determine a. the energy of the innermost energy level (n = 1), b. the energy of the second level, and c. the energy difference between the first and second energy levels. a. E1 = -13.6eV / (1)2 = -13.6 eV b. E2 = -13.6eV / (2)2 = -3.4 eV c. ΔE = Ef – Ei = E2 – E1 = -3.4 eV – (-13.6 eV) = 10.2 eV More Example Problems! An electron in an excited hydrogen atom drops from the second energy level to the first energy level. A. Determine the energy of the photon emitted. B. Calculate the frequency of the photon emitted. C. Calculate the wavelength of the photon emitted A. hf = Ei – Ef = E2 – E1 =-3.4 eV – (-13.6 eV) = 10.2 eV B. f = E/h =((10.2 eV)(1.6x10-19 J/eV)) / (6.63x10-34 J/Hz) = 2.46x1015 Hz C. λ= c/f = (3x108 m/s )/(2.46x1015 Hz) = 1.22x10-7m Even More Example Problems ;) Calculate the radius of the orbital associated with the energy levels E3 of the hydrogen atom. E3 = -13.6eV/(3)2 = -1.51eV r3 =(h2n2)/(4π2Kmq2) = (5.3x10-11m)n2 = (5.3x10-11m)(3)2 =4.77x10-10 Wake Up!! Background Info Section 28.2 In 1926, the German physicist Erwin Schroedinger used de Brogli’s wave model to create a quantum theory of atom based on waves. The theory does not provide a simple planetary picture of an atom as in the Bohr model. In particular, the radius of the electron orbit is not like the radius of the orbit of a planet about the sun. In wave particle nature matter means that it is impossible to know both the position and momentum of an electron at the same time. Thus the modern Quantum Model of the atom predicts only the probability that an electron is at a specific location. Background Info Cont. The most probable distance of the electron from the nucleus in hydrogen is found to be the same as the radius of the Bohr orbit. The probability that the electron is at any radius can be calculated, and a threedimensional plot can be constructed that shows regions of equal probability. The region in which there is a high probability of finding the electron is called the Electron Cloud. Background Info Cont. Again!!! Even though the quantum model of the atoms is difficult to visualize, quantum mechanics, which uses the Bohr model, has been extremely successful in predicting many details of the structure of the atom. These details are very difficult to calculate exactly for all but the simplest atom. Guided by quantum mechanics, chemists have been able to create new and useful molecules not otherwise available. Lasers Light Amplification by Stimulated Emission of Radiation Lasers produce light that is directional, powerful, monochromatic, and coherent. Uses for Lasers Medical Eye Surgery Plastic Surgery Cut Steel Destroying the World Formula for Section 28.2 The Bohr quantization condition: nλ = 2 π r n = whole number multiple of λ λ = wavelength of the electron r = radius Example Problem If the wavelength of an electron is 450 nm, and there are 2 electrons, what is the radius of the electron? nλ = 2 π r (2)(450 nm) = (2)(π)(r) 2π 2π 143.24 nm = r Every Day needs an Explosion!! http://www.youtube.co m/watch?v=bw85r24W W3s