Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 81

$0.0585nm$

We know that $|\Delta E|=|E_K-E_M|$ We plug in the known values to obtain: $|\Delta E|=|-13.6\frac{(42-1)^2}{1^2}-\frac{-13.6(42-9)^2}{3^2}|$ This simplifies to: $|\Delta E|=33945.6\times 10^{-19}J$ Now $\lambda=\frac{(6.63\times 10^{-34})(3\times 10^8)}{33945.6\times 10^{-19}}$ $\lambda=0.0585\times 10^{-9}m$ $\lambda=0.0585nm$