Chapter 7 - Sections 7.1-7.6 - Exercises - Problems by Topic - Page 331: 70c

$7.40\times10^{13}\,Hz$

$\Delta E_{atom}= -2.18\times10^{-18}\,J(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$ $= -2.18\times10^{-18}\,J(\frac{1}{4^{2}}-\frac{1}{5^{2}})$ $=-4.905\times10^{-20}\,J$ $E_{photon}=-\Delta E_{atom}=4.905\times10^{-20}\,J$ $E=h\nu\implies \nu=\frac{E}{h}$ $=\frac{4.905\times10^{-20}\,J}{6.626\times10^{-34}\,J\cdot s}$ $=7.40\times10^{13}\,Hz$