Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 64

$78$

We know that $\Delta E=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{0.0205\times 10^{-9}m}=9.702\times 10^{-15}J$ $\implies \Delta E=(9.702\times 10^{-15}J)(\frac{1eV}{1.6\times 10^{-19}J})=60640eV$ We also know that $\frac{hc}{\lambda}=E_L-E_K$ We plug in the known values to obtain: $60640eV=[(-3.4eV)(Z-1)^2]-[(-13.6eV)(Z-1)^2]$ This simplifies to: $Z=78$ Thus, the atomic number of the atom is $78$.