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Solutions13

# Solutions13 - Physics 208 Assignment 13 Solutions RT...

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Physics 208, Assignment 13, Solutions, 4/27/09, RT Problem 1. Density of nuclear matter, ρ = 2 × 10 17 kg / m 3 , which is huge . (a) Mass of V =5 cubic-cm is M = ρ 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 ρ sun = 1 . 41 × 10 3 . Compression of the sun from its present radius to radius R at density ρ nucl . requires R = R sun parenleftbigg ρ sun ρ 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 8.3 miles, or the distance from here to Dryden. Problem 2. An alpha particle has atomic number Z α = 2 and kinetic energy K α = 7 . 0 × 10 6 eV. A gold nucleus has atomic number Z Au = 79 and mass number A Au = 197. Using Eq. (42-3) 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 α -particle that had kinetic energy K α when far away, when at rest at distance r from the center of a gold nucleus, satisfies K α /e = U ( r ) /e = Z α Z gold e 4 πǫ 0 r . Solving for r , r = Z α Z gold e 4 πǫ 0 K α [ eV ] = 2 × 79 × (1 . 6 × 10 - 19 ) 4 π (0 . 89 × 10 - 11 ) × (7 . 0 × 10 6 ) = 3 . 2 × 10 - 14 m . Since r > r Au the α particle never touches the gold nucleus. 1

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Problem 3. We can represent particles by two-element 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 0 parenrightbigg , an α particle is parenleftbigg 4 2 parenrightbigg , and an electron is parenleftbigg 0 - 1 parenrightbigg . The reaction can be represented by a vector
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