Chapter 7 - Quantam Theory of the Atom - Questions and Problems - Page 298: 7.117

Please see the work below.

We know that $\nu=\frac{E}{h}$ $\nu=\frac{3.9687\times 10^{-19}}{6.626\times 10^{-34}}=5.9897\times 10^{14}s^{-1}$ We also know that $\lambda=\frac{c}{\nu}$ We plug in the known values to obtain: $\lambda=\frac{2.998\times 10^8}{5.9897\times 10^{14}}=5.0052\times 10^{-7}m=500.52nm \space$ visible region