## University Physics with Modern Physics (14th Edition)

The energy levels are given by $E_n=(n+\frac{1}{2})\hbar\omega$. The ground state energy is when n=0. The energy of the photon equals the energy difference between the two states. Find the energy. $$\Delta E=\hbar \omega=\frac{hc}{\lambda}$$ $$\Delta E =\frac{1.24\times10^{-6}eV\cdot m}{6.35\times10^{-6}m }=0.195eV$$ The ground state energy is $\frac{1}{2}\hbar\omega$, which is half the spacing between energy levels. $$\frac{1}{2}\hbar\omega=0.0976 eV$$