Chapter 31 - Atomic Physics1 Early Models of the AtomThomsom Model: Plum PluddingJ.J. Thomson discovered the electron in 1897. The electron is much less massivethan the atom, and negatively charged. Since atoms are neutral, he proposed thatatoms have an internal structure that includes negatively-charged electrons and somepositively charged matter. In Thomson’s “plum pludding” model, the electrons areembedded in a uniform distribution of positive charge, like raisins spread throughouta pudding.The Rutherford Model: A Miniature Solar SystemErnest Rutherford and colleagues (Hans Geiger and Ernest Marsden) decided to testThomson’s model. They shot a beam of positively charged particles at a thin sheetof gold foil. The particles they used were alpha particles, which we know now arethe nuclei of helium atoms. Rutherford expected that the alpha particles would bedeflected only a little (small angles), since the alpha particles are heavy (comparedto the electron) and the positive charged is spread out. He also expected all thealphas to be deflected in approximately the same way, since the positive charge isspread out everywhere.The results did not match the expectations. Most of the alpha particles passedthrough the foil without being deflected at all. The few alphas that were deflectedwere deflected at very large angles. In fact, some of the alphas deflected almostL. Whitehead 1 Phys 1302perfectly backwards - like they bounced off a wall. Rutherford said, “It was almostas incredible as if you fired a 15-inch shell at a piece of tissue paper and it came backand hit you.”Based on these results, Rutherford proposed that the atom has a structure similar tothe solar system, with the negatively-charged electrons orbiting a positively-chargednucleus that contains most of the atom’s mass. When the alpha particles bouncedback, it’s because they hit the nucleus directly. He calculated that the radius of thenucleus must be smaller than the diameter of the atom by a factor of around 10,000.If the nucleus were the size of the Sun, an electron would orbit at the radius of Pluto!The atom is mostly empty space, which is why most alphas passed right through thefoil.Problems with Rutherford’s model:1. An electron in orbit around a positive nucleus would undergo centripetal ac-celeration. We know that any accelerated charge produced electromagneticradiation. So as the electron orbited, it would radiate, and lose energy. Asit lost energy, it would spiral inward toward the nucleus. It would only takearound 1 ns for the electron to collapse in to the nucleus. Thus a Rutherfordatom would not be stable.2. As the electron spiraled inward, the frequency of the radiation it emitted wouldchange continuously. But experimental observations at the time indicated thatthe radiated light coming from atoms only has certain discrete frequencies, nota continuous distribution.L. Whitehead 2 Phys 13022 The Spectrum of Atomic HydrogenTo study the radiation emitted from a single atom, we can use a gas, where the atomsare separated from each other and have very few interactions. If we have a hot gasunder low pressure, the gas will emit electromagnetic radiation characteristic of theindividual atoms. When this radiation is passed through a diffraction grating, it willbe separated into its various wavelengths (colors for visible light). So you see a seriesof bright lines. This series of lines can be used to identify what kind of atom wasemitting the radiation, as the pattern is different for each element. This spectrumis called a “line spectrum” or “emission spectrum’.” Below is the emission spectrumof hydrogen.If you pass white light through a hydrogen gas, some wavelengths will be absorbedby the atoms, giving rise to an “absoprtion spectrum” - dark lines against a brightbackground. The dark lines of the absorption spectrum are located at the samelocation as the bright lines in the emission spectrum.In 1885, Johann Jakob Balmer worked out a simple formula that gives the wave-lengths of the visible lines in the spectrum:1λ= R122−1n2(1)n = 3, 4, 5, ...L. Whitehead 3 Phys 1302This is called the “Balmer series.”The constant is known as the Rydberg constant: R = 1.097 × 107m−1.The Balmer series gives all the lines in the visible portion of the spectrum, but thereare lines in non-visible parts of the spectrum as well. In general, the wavelengths ofthe lines can be found with this formula:1λ= R1n02−1n2(2)n0= 1, 2, 3, ...n = n0+ 1, n0+ 2, n0+ 3, ...So the Balmer series corresponds to n0= 2. There are other names: n0= 1 is calledthe Lyman series, and the wavelengths are all in the ultraviolet range. n0= 3 is thePaschen series, and those lines are all in the infrared.The above equation is just an emprical formula. Physicists wanted to derive thisrelationship based on a model of the atom.Find the shortest and longest wavelengths in the Paschen (n0= 3) series.For the Paschen series:1λ= R132−1n2n = 4, 5, 6, ...∞The longest wavelength can be found with n = 4:1λ= 1.097 × 107m−1132−1421λ= 5.33 × 105m−1λ = 1.9 × 10−6m = 1900nmThe shortest wavelength can be found with n = ∞:1λ= 1.097 × 107m−1132−1∞1λ= 1.22 × 106m−1λ = 8.2 × 10−7m = 820nmL. Whitehead 4 Phys 13023 Bohr’s Model of the Hydrogen AtomNeils Bohr came up with a model with which he was able to derive the formula forthe spectral lines of hydrogen. His model was based on four assumptions:1. The electron in a hydrogen atom moves in a circular orbit about the nucleus.2. Only certain circular orbits are allowed. In these orbits, the angular momentumof the electron is Ln= nh/2π, n = 1, 2, 3, ....3. Electrons do not give off electromagnetic radiation when they are in an allowedorbit. (The orbits are stable.)4. EM radiation is given off or absorbed only when an electron changes from oneallowed orbit to another. The frequency of the photon that is emitted satisfies|∆E| = hf, where ∆E is the energy difference between the orbits.If the electron is moving in a circular orbit, the Coulomb force between the posi-tive nucleus and the negative electron produces the centripetal acceleration of theelectron:mv2r=ke2r2mv2=ke2r(3)L. Whitehead 5 Phys 1302The angular momentum of a particle moving with speed v in a circular path of radiusr is L = rmv (Chapter 11). So the angular momentum condition can be written:Ln=nh2πrnmvn=nh2πvn=nh2πrnmn = 1, 2, 3, ...where rnand vnare the radius and