Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1114: 68

a) $13.6eV$ b) $\Delta E=10.2eV$

(a) We know that the total energy possessed by an electron of hydrogen atom in the ground state is $-13.6eV$, therefore, the highest-energy photon this system can absorb without dissociating the electron from the proton is just below $13.6eV$. (b) We know that $\Delta E=E_2-E_1$ We plug in the known values to obtain: $\Delta E=\frac{-13.6}{2^2}-\frac{-13.6}{1^2}$ $\Delta E=10.2eV$