p44_011

# p44_011 - momentum vector K = 1 2 mv 2 = p 2 2 m which...

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11. (a) If X represents the unknown fragment, then the reaction can be written 235 92 U+ 1 0 n 83 32 Ge + A Z X where A is the mass number and Z is the atomic number of the fragment. Conservation of charge yields 92+0 = 32+ Z ,so Z = 60. Conservation of mass number yields 235+1 = 83+ A ,so A = 153. Looking in Appendix F or G for nuclides with Z = 60, we ±nd that the unknown fragment is 153 60 Nd. (b) We neglect the small kinetic energy and momentum carried by the neutron that triggers the ±ssion event. Then, Q = K Ge + K Nd ,where K Ge is the kinetic energy of the germanium nucleus and K Nd is the kinetic energy of the neodymium nucleus. Conservation of momentum yields ~p Ge + ~p Nd = 0. Now, we can write the classical formula for kinetic energy in terms of the magnitude of the
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Unformatted text preview: momentum vector: K = 1 2 mv 2 = p 2 2 m which implies that K Nd = ( m Ge /m Nd ) K Ge . Thus, the energy equation becomes Q = K Ge + M Ge M Nd K Ge = M Nd + M Ge M Nd K Ge and K Ge = M Nd M Nd + M Ge Q = 153 u 153 u + 83 u (170 MeV) = 110 MeV . Similarly, K Nd = M Ge M Nd + M Ge Q = 83 u 153 u + 83 u (170 MeV) = 60 MeV . (c) The initial speed of the germanium nucleus is v Ge = r 2 K Ge M Ge = s 2(110 × 10 6 eV)(1 . 60 × 10 − 19 J / eV) (83 u)(1 . 661 × 10 − 27 kg / u) = 1 . 60 × 10 7 m / s . The initial speed of the neodymium nucleus is v Nd = r 2 K Nd M Nd = s 2(60 × 10 6 eV)(1 . 60 × 10 − 19 J / eV) (153 u)(1 . 661 × 10 − 27 kg / u) = 8 . 69 × 10 6 m / s ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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