12. We reproduce the calculation of Exercise 6: Combining Eqs. 42-2, 42-3 and 42-4, the number density of conduction electrons in gold is (19.3g/cm3)(6.02 × 1023/mol) n = =5.90 × 1022 cm-3 =59.0 nm-3 . (197 g/mol) Now, using the result of Exercise 3 in Chapter 39, Eq. 42-9 leads to 0.121(hc)2 0.121(1240 eV·nm)2 EF = n2/3 = (59.0nm-3)2/3 =5.52 eV . (mec2) 511 × 103 eV