Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 63

$0.195nm$

We know that $\Delta E=E_K-E_L$ $\implies \Delta E=8500-2125$ $\Delta E=6375eV$ $\Delta E=(6375)(1.6\times 10^{-9}J)=10200\times 10^{-19}J$ Now we can determine the required wavelength as follows: $\lambda=\frac{(6.63\times 10^{-34})(3\times 10^8)}{10200\times 10^{-19}}$ $\lambda=0.195\times 10^{-9}m$ $\lambda=0.195nm$