1136_PartUniversity Physics Solution

1136_PartUniversity Physics Solution - Nuclear Physics...

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Nuclear Physics 43-3 43.11. (a) IDENTIFY: Find the energy equivalent of the mass defect. SET UP: A 11 5 B atom has 5 protons, 11 5 6 −= neutrons, and 5 electrons. The mass defect therefore is 11 pne 5 565 ( B ) . Mmmm M Δ= + + − EXECUTE: 5(1.0072765 u) 6(1.0086649 u) 5(0.0005485799 u) 11.009305 u 0.08181 u. M Δ= + + = The energy equivalent is B (0.08181 u)(931.5 MeV/u) 76.21 MeV. E == (b) IDENTIFY and SET UP: Eq.(43.11): 2/3 1/3 2 B1 2 3 4 (1 ) / (2 ) / EC AC A C Z Z A CAZA =− The fifth term is zero since Z is odd but N is even. 11 and 5. AZ EXECUTE: 2/3 1/3 2 B (15.75 MeV)(11) (17.80 MeV)(11) (0.7100 MeV)5(4)/11 (23.69 MeV)(11 10) /11. E B 173.25 MeV 88.04 MeV 6.38 MeV 2.15 MeV 76.68 MeV E =+ = The percentage difference between the calculated and measured B E is 76.68 MeV 76.21 MeV 0.6% 76.21 MeV = EVALUATE: Eq.(43.11) has a greater percentage accuracy for 62 Ni. The semi-empirical mass formula is more accurate for heavier nuclei. 43.12. (a) nH C u 34 29 34(1.008665) u 29(1.007825) u 62.929601u 0.592 u, mm m +− = + = which is 551MeV, or 8.75 MeV per nucleon (using 931.5 MeV/u and 63 nucleons). (b) In Eq.(43.11), Z = 29 and N = 34, so the fifth term is zero. The predicted binding energy is 2 3 1 3 2 B (29)(28) (5) (15.75 MeV)(63) (17.80 MeV)(63) (0.7100 MeV) (23.69 MeV) (63) (63) E . B 556 MeV E = . The fifth term is zero since the number of neutrons is even while the number of protons is odd, making the pairing term zero. This result differs from the binding energy found from the mass deficit by 0.86%, a very good agreement comparable to that found in Example 43.4. 43.13. IDENTIFY In each case determine how the decay changes A and Z of the nucleus. The and ββ particles have charge but their nucleon number is 0. A = (a) SET UP: -decay: α Z increases by 2, ANZ decreases by 4 (an particle is a 4 2 He nucleus) EXECUTE: 239 4 235 94 2 92 Pu He U →+ (b) SET UP: β decay: Z increases by 1, remains the same (a particle is an electron, 0 1 e ) EXECUTE: 24 0 24 11 1 12 Na e Mg (c) SET UP + decay: Z decreases by 1, remains the same (a + particle is a positron, 0 +1 e) EXECUTE: 15 0 15 81 7 Oe N + EVALUATE: In each case the total charge and total number of nucleons for the decay products equals the charge and number of nucleons for the parent nucleus; these two quantities are conserved in the decay. 43.14. (a) The energy released is the energy equivalent of 4 npe 8.40 10 u, mmm −−= × or 783 keV. (b) np , > and the decay is not possible. 43.15. IDENTIFY: The energy of the photon must be equal to the difference in energy of the two nuclear energy levels. SET UP: The energy difference is Δ E = hc/ λ . EXECUTE: () ( ) 34 8 15 9 6.626 10 J s 3.00 10 m/s 8.015 10 J = 0.0501 MeV 0.0248 10 J hc E ×⋅ × Δ= = = × × EVALUATE: Since the wavelength of this photon is much shorter than the wavelengths of visible light, its energy is much greater than visible-light photons which are frequently emitted during electron transitions in atoms. This tells us that the energy difference between the nuclear shells is much greater than the energy difference between electron shells in atoms, meaning that nuclear energies are much greater than the energies of orbiting electrons.
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1136_PartUniversity Physics Solution - Nuclear Physics...

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