Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1112: 7

$121.5nm,102.6nm,97.23nm$

We can find the three longest wavelength lines of Lyman series as follows: $\lambda=1/R(1-\frac{1}{n^2})$ Now we plug in the known values to obtain: $\lambda_2=(1)/1.097\times 10^7(1-\frac{1}{2^2})=121.5nm$ $\lambda_3=(1)/1.097\times 10^7(1-\frac{1}{3^2})=102.6nm$ $\lambda_4=(1)/1.097\times 10^7(1-\frac{1}{4^2})=97.23nm$