Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 61

$0.167nm$

We can find the required wavelength as follows: $\frac{1}{\lambda}=R_Hz^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ We plug in the known values to obtain: $\frac{1}{\lambda}=(1.097\times 10^7)(28)^2(\frac{1}{1^2}-\frac{1}{2^2})$ $\frac{1}{\lambda}=645.036\times 10^7$ $\lambda=0.167nm$