# UH PHYS 1302 - Ch31 (20 pages)

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## Ch31

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ch 31 study guide

- Pages:
- 20
- School:
- University of Houston
- Course:
- Phys 1302 - Introductory to Physics II

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Chapter 31 Atomic Physics 1 Early Models of the Atom Thomsom Model Plum Pludding J J Thomson discovered the electron in 1897 The electron is much less massive than the atom and negatively charged Since atoms are neutral he proposed that atoms have an internal structure that includes negatively charged electrons and some positively charged matter In Thomson s plum pludding model the electrons are embedded in a uniform distribution of positive charge like raisins spread throughout a pudding The Rutherford Model A Miniature Solar System Ernest Rutherford and colleagues Hans Geiger and Ernest Marsden decided to test Thomson s model They shot a beam of positively charged particles at a thin sheet of gold foil The particles they used were alpha particles which we know now are the nuclei of helium atoms Rutherford expected that the alpha particles would be deflected only a little small angles since the alpha particles are heavy compared to the electron and the positive charged is spread out He also expected all the alphas to be deflected in approximately the same way since the positive charge is spread out everywhere The results did not match the expectations Most of the alpha particles passed through the foil without being deflected at all The few alphas that were deflected were deflected at very large angles In fact some of the alphas deflected almost L Whitehead 1 Phys 1302 perfectly backwards like they bounced off a wall Rutherford said It was almost as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back and hit you Based on these results Rutherford proposed that the atom has a structure similar to the solar system with the negatively charged electrons orbiting a positively charged nucleus that contains most of the atom s mass When the alpha particles bounced back it s because they hit the nucleus directly He calculated that the radius of the nucleus 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 the foil Problems with Rutherford s model 1 An electron in orbit around a positive nucleus would undergo centripetal acceleration We know that any accelerated charge produced electromagnetic radiation So as the electron orbited it would radiate and lose energy As it lost energy it would spiral inward toward the nucleus It would only take around 1 ns for the electron to collapse in to the nucleus Thus a Rutherford atom would not be stable 2 As the electron spiraled inward the frequency of the radiation it emitted would change continuously But experimental observations at the time indicated that the radiated light coming from atoms only has certain discrete frequencies not a continuous distribution L Whitehead 2 Phys 1302 2 The Spectrum of Atomic Hydrogen To study the radiation emitted from a single atom we can use a gas where the atoms are separated from each other and have very few interactions If we have a hot gas under low pressure the gas will emit electromagnetic radiation characteristic of the individual atoms When this radiation is passed through a diffraction grating it will be separated into its various wavelengths colors for visible light So you see a series of bright lines This series of lines can be used to identify what kind of atom was emitting the radiation as the pattern is different for each element This spectrum is called a line spectrum or emission spectrum Below is the emission spectrum of hydrogen If you pass white light through a hydrogen gas some wavelengths will be absorbed by the atoms giving rise to an absoprtion spectrum dark lines against a bright background The dark lines of the absorption spectrum are located at the same location as the bright lines in the emission spectrum In 1885 Johann Jakob Balmer worked out a simple formula that gives the wavelengths of the visible lines in the spectrum 1 1 1 1 R 22 n2 n 3 4 5 L Whitehead 3 Phys 1302 This is called the Balmer series The constant is known as the Rydberg constant R 1 097 107 m 1 The Balmer series gives all the lines in the visible portion of the spectrum but there are lines in non visible parts of the spectrum as well In general the wavelengths of the lines can be found with this formula 1 1 1 R 2 n02 n2 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 called the Lyman series and the wavelengths are all in the ultraviolet range n0 3 is the Paschen series and those lines are all in the infrared The above equation is just an emprical formula Physicists wanted to derive this relationship based on a model of the atom Find the shortest and longest wavelengths in the Paschen n0 3 series For the Paschen series 1 1 1 R 32 n2 n 4 5 6 The longest wavelength can be found with n 4 1 1 1 7 1 1 097 10 m 32 42 1 5 33 105 m 1 1 9 10 6 m 1900nm The shortest wavelength can be found with n 1 1 1 7 1 1 097 10 m 32 1 1 22 106 m 1 8 2 10 7 m 820nm L Whitehead 4 Phys 1302 3 Bohr s Model of the Hydrogen Atom Neils Bohr came up with a model with which he was able to derive the formula for the 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 momentum of the electron is Ln nh 2 n 1 2 3 3 Electrons do not give off electromagnetic radiation when they are in an allowed orbit The orbits are stable 4 EM radiation is given off or absorbed only when an electron changes from one allowed 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 positive nucleus and the negative electron produces the centripetal acceleration of the electron v2 ke2 2 r r 2 ke mv 2 r m 3 L Whitehead 5 Phys 1302 The angular momentum of a particle moving with speed v in a circular path of radius r is L rmv Chapter 11 So the angular momentum condition can be written nh 2 nh rn mvn 2 nh vn 2 rn m n 1 2 3 Ln where rn and vn are the radius and speed of the electron in the nth orbit Combining the two equations ke2 mvn2 rn 2 ke2 nh m 2 rn m rn 2 2 nh ke2 4 2 rn2 m …

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