Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1112: 5

$\lambda=371.2nm$

We can find the wavelength as follows: $\frac{1}{\lambda}=(R)(\frac{1}{2^2}-\frac{1}{n^2})$ We plug in the known values to obtain: $\frac{1}{\lambda}=(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{15^2})$ Solving for $\lambda$: $\lambda=371.2nm$