Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1112: 8

$1875nm, 1282nm,1094nm$

We can find the three longest wavelength lines of Paschen series as follows: $\lambda=1/R(\frac{1}{3^2}-\frac{1}{n^2})$ Now we plug in the known values to obtain: $\lambda_4=(1)/1.097\times 10^7(\frac{1}{3^2}-\frac{1}{4^2})=1875nm$ $\lambda_5=(1)/1.097\times 10^7(\frac{1}{3^2}-\frac{1}{5^2})=1282nm$ $\lambda_6=(1)/1.097\times 10^7(\frac{1}{3^2}-\frac{1}{6^2})=1094nm$