1141_PartUniversity Physics Solution

# 1141_PartUniversity Physics Solution - 43-8 Chapter 43 The...

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43-8 Chapter 43 The Coulomb potential energy of the two reactant nuclei at this separation is 12 00 11 ( 4 ) 44 qq e e U rr ππ == PP 19 2 92 2 15 19 4(1.602 10 C) (8.988 10 N m /C ) 1.4 MeV (4.0 10 m)(1.602 10 J/eV) U −− × = ×× This is an estimate of the threshold energy for this reaction. EVALUATE: The reaction releases energy but the total initial kinetic energy of the reactants must be 1.4 MeV in order for the reacting nuclei to get close enough to each other for the reaction to occur. The nuclear force is strong but is very short-range. 43.44. IDENTIFY and SET UP: 0.7% of naturally occurring uranium is the isotope 235 U . The mass of one 235 U nucleus is about 235 m p . EXECUTE: (a) The number of fissions needed is 19 29 61 9 1.0 10 J 3.13 10 (200 10 eV)(1.60 10 J/eV) × . The mass of 235 U required is 29 5 p (3.13 10 )(235 ) 1.23 10 kg m ×= × . (b) 5 7 2 1.23 10 kg 1.76 10 kg 0.7 10 × × EVALUATE: The calculation assumes 100% conversion of fission energy to electrical energy. 43.45. IDENTIFY and SET UP: The energy released is the energy equivalent of the mass decrease. 1 u is equivalent to 931.5 MeV. The mass of one 235 U nucleus is 235 m p . EXECUTE: (a) 235 1 144 89 1 92 0 56 36 0 Un B a K r3 n +→ + + We can use atomic masses since the same number of electrons are included on each side of the reaction equation and the electron masses cancel. The mass decrease is ( ) () ( ) ( ) () 235 1 144 89 1 92 0 56 36 0 B a K r 3 n Mm m m m m ⎡⎤ Δ= + + + ⎣⎦ 235.043930 u 1.0086649 u 143.922953 u 88.917630 u 3(1.0086649 u) M + 0.1860 u M . The energy released is (0.1860 u)(931.5 MeV/u) 173.3 MeV = . (b) The number of 235 U nuclei in 1.00 g is 3 21 p 1.00 10 kg 2.55 10 235 m × . The energy released per gram is 21 23 (173.3 MeV/nucleus)(2.55 10 nuclei/g) 4.42 10 MeV/g × . 43.46. (a) 28 24 14 12 Si Mg X. 24 28 so 4. 12 14 so 2. A Z AA ZZ X γ + ++ = = + = = is an α particle. (b) 2 (23.985042 u 4.002603 u 27.976927 u) (931.5 MeV u) 9.984 MeV γ Em c =−Δ = + = 43.47. The energy liberated will be 347 224 ( He) ( He) ( Be) (3.016029 u 4.002603 u 7.016929 u)(931.5 MeV u) 1.586 MeV. MMM +−= + = 43.48. (a) 3205 a n d 4711 0 . ZA =+−= (b) The nuclide is a boron nucleus, and 3 He Li n B 3.00 10 u, mmm m +−−= − × and so 2.79 MeV of energy is absorbed. 43.49. Nuclei: 4(2 ) 4 2 22 XYH e AZ A Z +−− + + →+ . Add the mass of Z electrons to each side and we find: (X ) ( Y ) (H e ) , mM M M where now we have the mass of the neutral atoms. So as long as the mass of the original neutral atom is greater than the sum of the neutral products masses, the decay can happen. 43.50. Denote the reaction as 1 XY e . + The mass defect is related to the change in the neutral atomic masses by XeY e eX Y [] [ ( 1 ) ] ( ) , mZ m m Z mm mm +− = where X m and Y m are the masses as tabulated in, for instance, Table (43.2). 43.51. (1 ) 1 A Z β + + . Adding ( Z –1) electrons to both sides yields 1 . So in terms of masses: ()( ) ( ) () ( ) 1e e 1 e X Y X Y 2 . A A A A Z Z Z Z m M M m M M m + Δ = = = So the decay will occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses. 43.52. IDENTIFY and SET UP: .

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1141_PartUniversity Physics Solution - 43-8 Chapter 43 The...

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