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13.2.1-8 Nuclear Physics These notes were typed in association with Physics by Michael Dickinson. For further reading and explanation see:Physics, Tsokos (purple): Ch 6.4 , Physics, Giancoli (mountain): Ch 27 13.2.1 Explain how the radii of nuclei may be estimated from charged particle scattering experiments. Remember back to the gold foil experiment! Some of the alpha particles will be deflected because of electrostatic repulsion. We can use this to estimate the size of a nucleus. If an alpha particle is fired exactly at the center of a nucleus, its velocity will reduce as its kinetic energy is converted into electrostatic potential energy. Eventually the alpha particle’s velocity reaches zero. Then it is projected back from where it came from. This is all due to electrostatic repulsive forces. It’s very similar to the motion of a ball thrown vertically into the air. Kinetic energy is converted to gravitational potential energy Kinetic Energy before = Electrostatic Potential Energy at closest approach E = W = Fs =Fr (r distance between Felectric = (kq1q2)/ r2 E = ((kq1q2)/ r2 ) x r E = (kq1q2)/ r Example B E = (kq1q2)/ r What is the distance of closest approach for an alpha particle (charge = +2e) aimed at a gold foil (charge = +79e target). The alpah particle has an initial kinetic energy of 5MeV (or 8 x 10-13J). Answer: 4.55 x 10-14m It is assumed that the alpha particle never actually touches the nucleus. The more initial energy given to the particle the closer it will get before it is repelled. That’s how we know the size is around 10-15m Practice 9 Calculate the distance of closest approach for alpha particles aimed at a piece of gold foil with and initial kinetic of 8.2MeV. Answer: 2.7 x 10 -14m 13.2.2 Describe how the masses of nuclei may be determined using a Bainbridge mass spectrometer. ***Special Note - Students should be able to draw a schematic diagram of the Bainbridge mass spectrometer, but the experimental details are not required Students should appreciate that nuclear mass values provide evidence for the existence of isotopes. This stuff can get kind of confusing. First I want to you watch the video and then we’ll talk about it. http://www.youtube.com/watch?v=eZnd_gyTwuE&list=PL80C5AF536A5A90DF First: Source of ion Second: Two slits This focuses them into a “beam” or straight line. Third: Velocity selector An electric field and magnetic field are set perpendicular to one another. Allows only ions with specific velocities to pass. Those that are to fast or slow will be pulled to either plate. Electric force = magnetic force qE = qvB1 v = qE/qB1 v = E/B1 Fourth: Second magnetic field Produces a centripetal force on ions. Centripetal force = magnetic force mv2/r = qvB2 m = qvB2 r / v2 m = qB2r / v Fifth: Ions hit detection plate Since they all have the same speed and charge the only difference is their mass. Heavier isotopes will produce larger radius Lighter isotopes will produce smaller radius Substitute the velocities out and you get: v = E/B1 m = qB2B1 / E 13.2.3 Describe one piece of evidence for the existence of nuclear energy levels. . 13.2.3 Describe one piece of evidence for the existence of nuclear energy levels. When atoms undergo radioactive decay the energy of the decay particle are actually discreet. Meaning they have specific levels. Example Radium-226 to Radium-222, an alpha particle is released. We can observe that the energy carried by the alpha particle can be either 4.59MeV OR 4.78MeV