Ph 312 Problems: Nuclear Physics (Serway et al. p.560) Answers (R. Egerton)

 

1. (a) r = r0 A^(1/3) = (1.2E-15) (4)^(1/3) = 1.9E-15 m = 1.9 fm

  1. r = (1.2E-15) (238)^(1/3) = (1.2E-15)(6.20) = 7.4E-15 m = 7.4 fm

 

2. Mass = (volume)(density of nuclear matter) = (10E-6)(2.3E17) = 2.3E12 kg

 

12. Eb = (mass of constituent protons + mass of neutrons – mass of Fe nucleus)(c^2)

= [(26)(1.0073u) + (56-26)(1.0087u) – 55.934939u](c^2)

= [56.4508u – 55.934939u] (c^2)

= [0.5159 u](c^2) = (8.57E-28 kg)(9E16) = 7.7E-11 J = 480.5 MeV

Eb/A = 8.58 eV per nucleon, which agrees with Fig. 13.10

 

20. (a) for Cu, binding energy Eb in eV is:

(15.7)(64) –(17.8)[(64)^(2/3)] – (0.71)(29)(28)/[64^(1/3)] – (23.6)[(64-29-29)^2] / 64

= 1004.8 – 284.8 – 144.13 – 13.275 = 562.6 MeV

For Zn, binding energy Eb in eV is:

(15.7)(64) –(17.8)[(64)^(2/3)] – (0.71)(30)(29)/[64^(1/3)] – (23.6)[(64-30-30)^2] / 64

= 1004.8 – 284.8 – 154.4 – 5.9 = 559.7 MeV

(b) For Zn, using Eq(13.4) and values on p. 538 and in Table 13.6,

Eb(MeV) = [(30)(1.007825u) + (64-30)(1.008665u) – 63.929145u](931.494 MeV/u)

= (0.6002)(931.494) MeV = 559.1 MeV