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Unformatted text preview: Physics 208, Assignment 13, Solutions, 4/27/09, RT Problem 1. Density of nuclear matter, = 2 10 17 kg / m 3 , which is huge . (a) Mass of V =5 cubiccm is M = nucl . V = (2 10 17 ) (5 10 6 ) = 10 12 kg. weight = Mg = 9 . 8 10 12 N = 2 . 2 10 12 lbf 10 9 ton . (b) Radius of sun is R sun = 6 . 96 10 8 m. Density of sun is sun = 1 . 41 10 3 . Compression of the sun from its present radius to radius R at density nucl . requires R = R sun parenleftbigg sun nucl . parenrightbigg 1 / 3 = 6 . 96 10 8 parenleftbigg 1 . 41 10 3 2 10 17 parenrightbigg 1 / 3 = 1 . 34 10 4 m . This is about 8.3miles, or the distance from here to Dryden. Problem 2. An alpha particle has atomic number Z = 2 and kinetic energy K = 7 . 10 6 eV. A gold nucleus has atomic number Z Au = 79 and mass number A Au = 197. Using Eq. (423) from the text, the radius of the gold nucleus is r Au = 1 . 2 10 15 (197) 1 / 3 = 0 . 7 10 14 m . The potential energy U ( r ) of an particle that had kinetic energy K when far away, when at rest at distance r from the center of a gold nucleus, satisfies K /e = U ( r ) /e = Z Z gold e 4 r . Solving for r , r = Z Z gold e 4 K [ eV ] = 2 79 (1 . 6 10 19 ) 4 (0 . 89 10 11 ) (7 . 10 6 ) = 3 . 2 10 14 m . Since r > r Au the particle never touches the gold nucleus. 1 Problem 3. We can represent particles by twoelement column vectors with mass number as the upper component and charge number as the lower component; parenleftbigg mass number charge number parenrightbigg . A neutron is parenleftbigg 1 parenrightbigg , an particle is parenleftbigg 4 2 parenrightbigg...
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This note was uploaded on 09/01/2009 for the course PHYS 208 taught by Professor Amadeuri during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 AMADEURI
 Mass

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