Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1115: 82

$0.3276nm$

We know that $\Delta E=|E_L-E_M|$ We plug in the known values to obtain: $\Delta E=|-13.6\frac{(42-2)}{2^2}-\frac{-13.6(42-9)^2}{3^2}|$ This simplifies to: $\Delta E=6071.04\times 10^{-19}J$ Now $\lambda=\frac{(6.63\times 10^{-34})(3\times 10^8)}{6071.04\times 10^{-19}}$ $\lambda=0.3276\times 10^{-9}m=0.3276nm$