Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 80

a) $Z\approx 7$ b) $E_3=-74.04eV$

(a) We know that $Z=\frac{n^2h^2}{4\pi^2mKr_ne^2}$ We plug in the known values to obtain: $Z=\frac{(6)^2(6.63\times 10^{-34})^2}{4(3.14)^2(9.11\times 10^{-31})(9\times 10^9)(2.72\times 10^{-10})(1.6\times 10^{-19})^2}$ $Z=7.028$ $\implies Z\approx 7$ (b) We know that $E_n=-13.6\frac{Z^2}{n^2}$ We plug in the known values to obtain: $E_3=-13.6\frac{7^2}{3^2}$ $E_3=-74.04eV$