Chapter 7 - Quantam Theory of the Atom - Questions and Problems - Page 297: 7.103

$432.3nm$

We know that $\nu=\frac{E}{h}=\frac{4.60\times 10^{-19}}{6.63\times 10^{-34}}=6.938\times 10^{14}s^{-1}$ We also know that $\lambda=\frac{c}{\nu}$ $\lambda=\frac{3.00\times 10^8}{6.93\times 10^{14}}=4.323\times 10^{-7}m=432.3nm$