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1146_PartUniversity Physics Solution

# 1146_PartUniversity Physics Solution - Nuclear Physics...

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Nuclear Physics 43-13 Solving for t gives () He H ln 1 / NN t λ + = . Using the given numbers and 1/2 ln 2 T = , we have ln 2 ln 2 0.0563/ y 12.3 y T == = and ln 1 4.3 0.0563/ y t + = = 30 years. EVALUATE: One limitation on this method would be that after many years the ratio of H to He would be too small to measure accurately. 43.73. (a) IDENTIFY and SET UP: Use Eq.(43.1) to calculate the radius R of a 2 1 H nucleus. Calculate the Coulomb potential energy (Eq.23.9) of the two nuclei when they just touch. EXECUTE: The radius of 2 1 H is 15 1/3 15 (1.2 10 m)(2) 1.51 10 m. R −− The barrier energy is the Coulomb potential energy of two 2 1 H nuclei with their centers separated by twice this distance: 21 9 2 92 2 1 4 15 0 1( 1 . 6 0 2 1 0 C ) (8.988 10 N m /C ) 7.64 10 J 0.48 MeV 42 ( 1 . 5 1 1 0 m ) e U r π × × = × = × P (b) IDENTIFY and SET UP: Find the energy equivalent of the mass decrease. EXECUTE: 22 3 1 11 2 0 HH H en +→ + If we use neutral atom masses there are two electrons on each side of the reaction equation, so their masses cancel. The neutral atom masses are given in Table 43.2. H H has mass 2(2.014102 u) 4.028204 u += 31 20 He n has mass 3.016029 u 1.008665 u 4.024694 u ++ = The mass decrease is 3 4.028204 u 4.024694 u 3.510 10 u. −= × This corresponds to a liberated energy of 3 61 9 1 3 (3.510 10 u)(931.5 MeV/u) 3.270 MeV, or (3.270 10 eV)(1.602 10 J/eV) 5.239 10 J. ×=× × = × (c) IDENTIFY and SET UP: We know the energy released when two 2 1 H nuclei fuse. Find the number of reactions obtained with one mole of 2 1 H. EXECUTE: Each reaction takes two 2 1 H nuclei. Each mole of 23 2 D has 6.022 10 × molecules, so 23 6.022 10 × pairs of atoms. The energy liberated when one mole of deuterium undergoes fusion is 23 13 (6.022 10 )(5.239 10 J) ×× = 11 3.155 10 J/mol. × EVALUATE: The energy liberated per mole is more than a million times larger than from chemical combustion of one mole of hydrogen gas. 43.74. In terms of the number N of cesium atoms that decay in one week and the mass 1.0 kg, m = the equivalent dose is 13 ee 3.5Sv ((RBE) E (RBE) E ) ((1)(0.66 MeV) (1.5)(0.51MeV)) (2.283 10 J), γγ N mm m =+ = + = × so 13 13 (1.0 kg)(3.5Sv) 1.535 10 (2.283 10 J) N × × . The number 0 N of atoms present is related to λ 0 by . t NNN e = 10 1 7 2 ln 2 0.693 7.30 10 sec (30.07 yr)(3.156 10 sec yr) Ty × × . Then 10 1 4 λ 13 (7.30 10 s )(7 days)(8.64 10 s day) 13 0 (1.535 10 ) 1.536 10 . t e e × = × 43.75. (a) cm m vv mM = + . m v v mM mM ⎛⎞ ′=− = ⎜⎟ ⎝⎠ . M vm v ′ = + . 2 2 2 2 1 1 1 2 2 2( ) ) ) mM Mm M mM m Km v M v v v v mM mM mM ′′ = + = + +++ + + . 2 cm 1 2 MM v K K K = =≡ . (b) For an endoergic reaction cm 0 KQ Q =− < at threshold. Putting this into part (a) gives th th Mm M QK K Q M −+ = +

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43-14 Chapter 43 43.76. M KK Mm α = + , where K is the energy that the -particle would have if the nucleus were infinitely massive. Then, () 2 Os Os 186 2.76 MeV 181.94821u 182 MM M K M M c αα =− = . 43.77. ()( )() 235 140 94 92 54 38 n UX e S r mM M M m Δ= () ( ) 2 235.043923 u 139.921636 u 93.915360 u 1.008665 u 0.1983 u 0.1983 u 931.5 MeV u 185 MeV. m Em c = = = 43.78. (a) A least-squares fit of the log of the activity vs. time for the times later than 4.0 h gives a fit with correlation 6 121 0 −−× and decay constant of 1 0.361h , corresponding to a half-life of 1.92 h. Extrapolating this back to time 0 gives a contribution to the rate of about 2500/s for this longer-lived species. A least-squares fit of the log of the activity
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1146_PartUniversity Physics Solution - Nuclear Physics...

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